# Galois Correspondence for Partial Groupoid Actions

Wesley G. Lautenschlaeger and Thaísa Tamusiunas  Instituto de Matemática,
Universidade Federal do Rio Grande do Sul, Av. Bento Gonçalves, 9500,
91509-900. Porto Alegre-RS, Brazil<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

We prove a Galois correspondence theorem for groupoids acting orthogonally and
partially on commutative rings. We also consider partial actions that are not
orthogonal, presenting two correspondences in this case: one for strongly
Galois partial groupoid actions and one for global groupoid actions (without
restriction). Some examples are illustrated.

2020 AMS Subject Classification: Primary. 13B05. Secondary. 20L05, 18B40.

Keywords: groupoid, Galois theory, partial actions, Galois correspondence,
orthogonal action

##  1\. Introduction

Chase, Harrison and Rosenberg developed, in 1965, a Galois Theory for finite
groups acting on commutative rings, in where they exhibit a generalization of
the Fundamental Theorem of the Galois theory [5]. Right after, Villamayor and
Zelinsky [18] generalized one of the constructions of [5] to extensions of
commutative rings with finitely many idempotents. Although they did not
explicitly defined groupoid actions, this was the intermediate object that
they worked to prove their correspondence theorem. This was the first time
that groupoids appeared in a Galois context.

A Galois correspondence theorem for groupoids acting on commutative rings was
showed in [15]. Given $B\subseteq A$ a commutative Galois extension with
finite Galois groupoid $\mathcal{G}$, it was proved a one-to-one
correspondence between the wide subgroupoids of $\mathcal{G}$ and the
separable, $\beta$-strong $B$-subalgebras of $A$, where $\beta$ is a (global)
action of $\mathcal{G}$ on $A$. Other Galois theories concerning global action
of groupoids were also developed in [6], [9], [11], [14], [16] and [17].

The study of partial group actions is recent in mathematical terms and its
first objective was to classify $C^{*}$-algebras [8]. In a groupoid context,
many questions concerning partial actions were investigated. For instance, in
[1] it was introduced the concept of partial action of ordered groupoids on
rings; in [12], it was showed that there is a biunivocal correspondence
between partial groupoid actions and global inverse semigroupoid actions; in
[2], a globalization theorem was proved and a characterization theorem for
Galois extensions was showed. We highlight the special case in [3], where the
authors related a class of partial groupoid actions to partial group actions,
calling them as _group-type_ partial groupoid actions. The relation groupoid-
group presented in [3] was used in [4] to construct a Galois correspondence
for partial actions of groupoids, but only restricted to group-type partial
actions, and the mechanisms used there cannot be applied to any partial
action. Therefore, a different approach needs to be thought of.

So, the first objective of this work is to extend the Galois correspondence
presented in [4] to a partial groupoid action without asking the group-type
property. For this, we will relate a partial Galois extension to the Galois
extension regarding its globalization. The partial-to-global relation will
give us a more general partial Galois correspondence, as we will present in
Section 3.

The second objective is to improve the assumption of Galois correspondence
theorems for groupoid actions in general. In fact, all the Galois
correspondences that are found in the literature asks that the action must be
orthogonal, that is, $A=\bigoplus_{e\in\mathcal{G}_{0}}A_{e}$, where
$\alpha=(A_{g},\alpha_{g})_{g\in\mathcal{G}}$ is a partial action of a
groupoid $\mathcal{G}$ on a ring $A$ and $\mathcal{G}_{0}$ is the set of
identities of $\mathcal{G}$. This hypothesis covers a broad class of actions,
but is still restrictive because it does not cover them all. Every action can
be orthogonalized, but the problem is that if we have a Galois extension
$A|_{A^{\varepsilon}}$ regarding an orthogonalization $\varepsilon$, it does
not implies that the extension $A|_{A^{\alpha}}$ is Galois regarding the
original action $\alpha$. In Section 4 we present some examples, as well as we
prove two Galois correspondences without the orthogonalization hypothesis.

We give now a brief outline of the paper. In the second section we present
basic definitions and results about partial groupoid actions and Galois
theory. In the third section we prove our first main result, which states that
there is a one-to-one correspondence between the wide subgroupoids
$\mathcal{H}$ of the groupoid $\mathcal{G}$ and determined separable,
$\alpha$-strong $A^{\alpha}$-subalgebras $C$ of the algebra $A$, where
$\alpha$ is a partial action of $\mathcal{G}$ on $A$ (Theorem 3.6). In Section
4 we define orthogonalization of a given partial action and we study which
properties of Galois theory remain valid under orthogonalization. We also
define strongly $\alpha$-Galois extensions and we present in Theorem 4.11 a
Galois correspondence in this case, which is our second main result. Some
examples are illustrated. We end the paper with our third main result, which
is a Galois correspondence for global actions without restrictions on the
action (Theorem 4.16).

##  2\. Preliminaries

We recall that a _groupoid_ $\mathcal{G}$ is a small category in which every
morphism is an isomorphism. We denote by $\mathcal{G}_{0}$ the set of objects
of $\mathcal{G}$. Observe that
$\text{id}:\mathcal{G}_{0}\rightarrow\mathcal{G}$, given by
$\text{id}(x)=\text{id}_{x}$, is an injective map and whence we identify
$\mathcal{G}_{0}\subseteq\mathcal{G}$. Given $g\in\mathcal{G}$, the _domain_
and the _range_ of $g$ will be denoted by $d(g)$ and $r(g)$, respectively.
Hence, $d(g)=g^{-1}g$ and $r(g)=gg^{-1}$. For all $g,h\in\mathcal{G}$, we
write $\exists gh$ whenever the product $gh$ is defined. We denote
$\mathcal{G}_{2}=\\{(g,h)\in\mathcal{G}\times\mathcal{G}:\exists gh\\}$.

Given $e\in\mathcal{G}_{0}$, consider
$\mathcal{G}(e)=\\{g\in\mathcal{G}:r(g)=d(g)=e\\}$. It is easy to verify that
$\mathcal{G}(e)$ is a group for all $e\in\mathcal{G}_{0}$, called the isotropy
group associated to $e$. A groupoid $\mathcal{G}$ is said to be _connected_ if
given any $e_{1},e_{2}\in\mathcal{G}_{0}$ there exists $g\in\mathcal{G}$ with
$d(g)=e_{1}$ and $r(g)=e_{2}$.

Given $e_{1},e_{2}\in\mathcal{G}_{0}$ we set
$\mathcal{G}(e_{1},e_{2}):=\\{g\in\mathcal{G}:d(g)=e_{1}\,\,\text{and}\,\,r(g)=e_{2}\\}$.
It is well-known that any groupoid is a disjoint union of connected
subgroupoids. Indeed, we define the following equivalence relation in
$\mathcal{G}_{0}$: for all $e_{1},e_{2}\in\mathcal{G}_{0}$,

$\displaystyle e_{1}\sim e_{2}$ $\displaystyle\text{ if and only if
}\mathcal{G}(e_{1},e_{2})\neq\emptyset.$

Every equivalence class $\bar{e}\in\mathcal{G}_{0}/\\!\\!\sim$ determines a
connected subgroupoid $\mathcal{G}_{\bar{e}}$ of $\mathcal{G}$, whose set of
identities is $\bar{e}$. The subgroupoid $\mathcal{G}_{\bar{e}}$ is called the
connected component of $\mathcal{G}$ associated to $\bar{e}$. It is clear that
$\mathcal{G}=\dot{\cup}_{\bar{e}\in\mathcal{G}_{0}/\\!\sim}\mathcal{G}_{\bar{e}}$.

Given a nonempty set $S$, we recall that the _coarse groupoid_ associated to
$S$ is the groupoid $S^{2}=S\times S$, where $\exists(s,t)(u,v)$ if and only
if $s=v$ and in this case $(s,t)(u,v)=(u,t)$. If $\mathcal{G}$ is a connected
groupoid, we have $\mathcal{G}\simeq\mathcal{G}_{0}^{2}\times\mathcal{G}(e)$
(c.f. [13, Proposition 3.3.6]), and this isomorphism does not depend on the
choice of $e\in\mathcal{G}_{0}$. Notice that every two finite coarse groupoids
with the same amount of identities are isomorphic. Hence we will just denote
by $\mathcal{A}_{n}$ the coarse groupoid with $n$ identities. Thus, if
$\mathcal{G}$ is a finite connected groupoid with cardinality
$|\mathcal{G}_{0}|=n$ and $e\in\mathcal{G}_{0}$, we have
$\mathcal{G}\simeq\mathcal{A}_{n}\times\mathcal{G}(e)$.

###  2.1. Partial Groupoid Actions

We begin with the definition of partial groupoid action.

######  Definition 2.1.

Let $\mathcal{G}$ be a groupoid and $A$ be a ring. We say that
$\alpha=(A_{g},\alpha_{g})_{g\in\mathcal{G}}$ is a _partial action_ of
$\mathcal{G}$ on $A$ if, for all $g\in\mathcal{G}$ $A_{g}\triangleleft
A_{r(g)}\triangleleft A$, $\alpha_{g}:A_{g^{-1}}\to A_{g}$ is an isomorphism
of rings, and

  1. (P1)

$\alpha_{e}=\text{Id}_{A_{e}}$, for all $e\in\mathcal{G}_{0}$, and
$A=\sum_{e\in\mathcal{G}_{0}}A_{e}$;

  2. (P2)

$\alpha_{h}^{-1}(A_{g^{-1}}\cap A_{h})\subset A_{(gh)^{-1}}$, for all
$(g,h)\in\mathcal{G}_{2}$;

  3. (P3)

$\alpha_{g}\circ\alpha_{h}(x)=\alpha_{gh}(x)$, for all
$(g,h)\in\mathcal{G}_{2},x\in\alpha_{h}^{-1}(A_{g^{-1}}\cap A_{h})$.

The first thing to note is that condition (P1) is not the same as reference
[2], because we ask that $A=\sum_{e\in\mathcal{G}_{0}}A_{e}$, and in [2] this
condition was not requested. When $\mathcal{G}$ is a group with identity
element $1_{\mathcal{G}}$, this condition translates into
$A=A_{1_{\mathcal{G}}}$. If we don’t ask for this, it may happen that a
partial groupoid action is not a partial group action, even when the groupoid
is a group.

Naturally, a partial action is said to be _global_ if $A_{g}=A_{r(g)}$, for
all $g\in\mathcal{G}$.

######  Proposition 2.2.

[2, Lemma 1.1] Let $\alpha=(A_{g},\alpha_{g})_{g\in\mathcal{G}}$ be a partial
action of the groupoid $\mathcal{G}$ on the ring $A$. Then it hold:

  1. (i)

$\alpha_{g}^{-1}=\alpha_{g^{-1}}$, for all $g\in\mathcal{G}$.

  2. (ii)

$\alpha_{g}(A_{g^{-1}}\cap A_{h})=A_{g}\cap A_{gh}$, for all
$(g,h)\in\mathcal{G}_{2}$.

We say that a partial action $\alpha$ is _preunital_ if $A_{e}$ is a unital
ring with identity $1_{e}$ central and idempotent, so that
$A_{e}=A1_{e}=1_{e}A$, for all $e\in\mathcal{G}_{0}$. We say that a partial
action $\alpha$ is _unital_ if $A_{g}$ is a unital ring with identity $1_{g}$
central and idempotent, so that $A_{g}=A1_{g}=1_{g}A$, for all
$g\in\mathcal{G}$.

We say that a partial action $\alpha$ is _orthogonal_ if

  1. (PO)

$A=\bigoplus_{e\in\mathcal{G}_{0}}A_{e}$

holds.

######  Definition 2.3.

Let $\alpha=(A_{g},\alpha_{g})_{g\in\mathcal{G}}$ be a partial action of a
groupoid $\mathcal{G}$ on a ring $A$. A global action
$\beta=(B_{g},\beta_{g})_{g\in\mathcal{G}}$ of $\mathcal{G}$ on a ring $B$ is
said to be a _globalization_ of $\alpha$ if, for all $e\in\mathcal{G}_{0}$,
there is $\varphi_{e}:A_{e}\to B_{e}$ monomorphism of rings such that

  1. (G1)

$\varphi_{e}(A_{e})\triangleleft B_{e}$;

  2. (G2)

$\varphi_{r(g)}(A_{g})=\varphi_{r(g)}(A_{r(g)})\cap\beta_{g}(\varphi_{d(g)}(A_{d(g)}))$,
for all $g\in\mathcal{G}$;

  3. (G3)

$\beta_{g}\circ\varphi_{d(g)}(a)=\varphi_{r(g)}\circ\alpha_{g}(a)$, for all
$a\in A_{g^{-1}}$, $g\in\mathcal{G}$;

  4. (G4)

$B_{g}=\sum_{r(h)=r(g)}\beta_{h}\left(\varphi_{d(h)}(A_{d(h)})\right)$, for
all $g\in\mathcal{G}$.

We have a characterization for partial actions that have globalizations.

######  Theorem 2.4.

[2, Theorem 2.1] Let $\alpha=(A_{g},\alpha_{g})_{g\in\mathcal{G}}$ be a
preunital partial action of a groupoid $\mathcal{G}$ on a ring $A$. Then
$\alpha$ has a globalization $\beta$ if and only if $\alpha$ is unital.
Moreover, such $\beta$ is unique up to isomorphism.

###  2.2. Galois Theory

In this subsection we present some basic concepts regarding Galois theory.

######  Definition 2.5.

Let $\alpha=(A_{g},\alpha_{g})_{g\in\mathcal{G}}$ be a partial action of a
groupoid $\mathcal{G}$ on a ring $A$. We define

$\displaystyle A^{\alpha}=\\{a\in A:\alpha_{g}(a1_{g^{-1}})=a1_{g},\text{ for
all }g\in\mathcal{G}\\},$

the _subring of invariants of_ $A$ _under_ $\alpha$.

We say that $A$ is a _partial $\alpha$-Galois extension of_ $A^{\alpha}$ if
there is $\\{x_{i},y_{i}\\}_{i=1}^{n}\subseteq A$ such that

$\displaystyle\sum_{i=1}^{n}x_{i}\alpha_{g}(y_{i}1_{g^{-1}})=\sum_{e\in\mathcal{G}_{0}}\delta_{e,g}1_{e},$

for all $g\in\mathcal{G}$. In this case, we say that the set
$\\{x_{i},y_{i}\\}_{i=1}^{n}$ is a set of _partial $\alpha$-Galois coordinates
of_ $A$ _over_ $A^{\alpha}$.

######  Definition 2.6.

Let $\mathcal{G}$ be a groupoid, $\alpha=(A_{g},\alpha_{g})_{g\in\mathcal{G}}$
be a partial action of $\mathcal{G}$ on a ring $A$ and $\mathcal{H}$ be a
subgroupoid of $\mathcal{G}$. We define the _restriction of $\alpha$ to_
$\mathcal{H}$ as

$\displaystyle\alpha|_{\mathcal{H}}=(A_{h},\alpha_{h})_{h\in\mathcal{H}}.$

######  Definition 2.7.

Let $\mathcal{G}$ be a groupoid, $\alpha=(A_{g},\alpha_{g})_{g\in\mathcal{G}}$
a partial action of $\mathcal{G}$ on a ring $A$ and $C$ a
$A^{\alpha}$-subalgebra of $A$. We define

$\displaystyle\mathcal{G}_{C}=\\{g\in\mathcal{G}:\alpha_{g}(a1_{g^{-1}})=a1_{g},\text{
for all }a\in C\\}.$

Clearly, $\mathcal{G}_{0}\subseteq\mathcal{G}_{C}$, for all $C$. However,
$\mathcal{G}_{C}$ is not always a subgroupoid of $\mathcal{G}$. An example of
that can be found at [7, Example 6.3].

######  Definition 2.8.

Let $C$ be an $A^{\alpha}$-subalgebra of a ring $A$, where $\alpha$ is a
partial action of a groupoid $\mathcal{G}$ on $A$. We say that $C$ is:

  1. (i)

$A^{\alpha}$-_separable_ if $C$ is a projective
($C\otimes_{A^{\alpha}}C^{op}$)-module [10, Ch. III $\S$1];

  2. (ii)

$\alpha$-_strong_ if for all $g,h\in\mathcal{G}$ with $r(g)=r(h)$ and
$g^{-1}h\notin\mathcal{G}_{C}$, and for all nonzero idempotent $e\in A_{g}\cup
A_{h}$ there is an element $a\in C$ such that
$\alpha_{g}(a1_{g^{-1}})e\neq\alpha_{h}(a1_{h^{-1}})e$ [4, Proposition 5.4].

######  Remark 2.9.

Concerning $C$ being $A^{\alpha}$-separable, we have the following equivalent
definition (c.f. [10, Théorème 1.4]): $C$ is $A^{\alpha}$-separable if there
is an idempotent $e=\sum_{i=1}^{n}x_{i}\otimes y_{i}\in
C\otimes_{A^{\alpha}}C$ such that
$\displaystyle\sum_{i=1}^{n}x_{i}y_{i}=1_{C}$ and $e(1_{C}\otimes x-x\otimes
1_{C})=0$, for all $x\in S$. We say that $e$ is an _idempotent of
separability_ of the extension $C|_{A^{\alpha}}$.

##  3\. Correspondence Theorem

This section is intended to prove a Galois correspondence for partial groupoid
actions.

Let $e\in\mathcal{G}_{0}$. Consider
$\mathcal{G}(-,e)=\\{g\in\mathcal{G}:r(g)=e\\}$. Suppose that
$\mathcal{G}(-,e)=\\{e=g_{1,e},g_{2,e}\ldots,g_{n_{e},e}\\}$, where $n_{e}$ is
the cardinality $|\mathcal{G}(-,e)|$ of $\mathcal{G}(-,e)$. Consider the
following assertion:

(1) $\displaystyle\text{For each
}g\in\mathcal{G},g_{r(g),i}^{-1}gg_{d(g),i}\in\mathcal{G}_{0}\text{ implies
that }g\in\mathcal{G}_{0},\text{ for all }1\leq i\leq n_{r(g)}=n_{d(g)}.$

In [2, Theorem 5.1], it was proved that if (1) holds, then $A$ being a partial
$\alpha$-Galois extension of $A^{\alpha}$ implies that $B$ is a $\beta$-Galois
extension of $B^{\beta}$. We shall prove that this hypothesis is superfluous.
So we start this section by improving [2, Theorem 5.1].

######  Lemma 3.1.

Let $\alpha$ be a unital, orthogonal partial action of a finite groupoid
$\mathcal{G}$ on a ring. Consider $\beta$ globalization of $\alpha$, global
action of $\mathcal{G}$ on a ring $B$. The following statements are
equivalent:

  1. (i)

$A$ is a partial $\alpha$-Galois extension of $A^{\alpha}$.

  2. (ii)

$B$ is a $\beta$-Galois extension of $B^{\beta}$.

###### Proof.

The implication (ii) $\Rightarrow$ (i) was proved in [2, Theorem 5.1].

(i) $\Rightarrow$ (ii): Assume that $\mathcal{G}$ is connected. For the
general case, it is enough to obtain a convenient enumeration for each
connected component of $\mathcal{G}$.

Since $\mathcal{G}$ is finite and connected,
$\mathcal{G}\simeq\mathcal{A}_{n}\times G$, where $\mathcal{A}_{n}$ is the
coarse groupoid with $n$ identities and $G$ is a finite group with $m$
elements [13, Proposition 3.3.6]. Observe that $n_{e}=|\mathcal{G}(-,e)|=nm$,
for all $e\in\mathcal{G}_{0}$. Now enumerate $G$ as the following:
$G=\\{1_{G}=g_{0},\ldots,g_{m-1}\\}$. Let $\mathbb{S}_{n}$ be the group of
permutation of $n$ elements. Consider $\sigma\in\mathbb{S}_{n}$ given by
$\sigma(i)=i-1$, for all $1<i\leq n$, and $\sigma(1)=n$. Hence, given $1\leq
j\leq mn$, there are unique $0\leq q\leq m-1$ and $0\leq r\leq n-1$ such that
$j=qn+r+1$. We define, then, $g_{e_{i},j}=(e_{\sigma^{r}(i)},e_{i},g_{q})$.

In this enumeration,
$g_{e_{i},1}=(e_{\sigma^{0}(i)},e_{i},g_{0})=(e_{i},e_{i},1_{G})=e_{i}$. Given
$\ell=(e_{i},e_{j},g_{k})\in\mathcal{G}$ and given $1\leq u=qn+r\leq mn$, we
have:

$\displaystyle g_{r(\ell),u}^{-1}\ell g_{d(\ell),u}$
$\displaystyle=g_{e_{j},u}^{-1}\ell
g_{e_{i},u}=(e_{\sigma^{r}(j)},e_{j},g_{q})^{-1}(e_{i},e_{j},g_{k})(e_{\sigma^{r}(i)},e_{i},g_{q})$
$\displaystyle=(e_{j},e_{\sigma^{r}(j)},g_{q}^{-1})(e_{i},e_{j},g_{k})(e_{\sigma^{r}(i)},e_{i},g_{q})=(e_{\sigma^{r}(i)},e_{\sigma^{r}(j)},g_{q}^{-1}g_{k}g_{q}).$

Therefore

$\displaystyle g_{r(\ell),u}^{-1}\ell g_{d(\ell),u}\in\mathcal{G}_{0}$
$\displaystyle\Leftrightarrow\sigma^{r}(i)=\sigma^{r}(j)\text{ and
}g_{q}^{-1}g_{k}g_{q}=1_{G}$ $\displaystyle\Leftrightarrow i=j\text{ and
}g_{k}=g_{q}g_{q}^{-1}=1_{G}\Leftrightarrow\ell\in\mathcal{G}_{0},$

which ends the proof. ∎

From now on, consider $\alpha$ an orthogonal partial action of a finite
groupoid $\mathcal{G}$ on a commutative ring $A$. We fix the notation
$\beta=(B_{g},\beta_{g})_{g\in\mathcal{G}}$ to denote the globalization of
$\alpha$, which is a global action on a ring $B$. According to [2, Theorem
2.1], it follows that $B=\prod_{e\in\mathcal{G}_{0}}B_{e}$, so that $\beta$ is
an orthogonal action. To shorten the notation, we will assume $A_{e}\subseteq
B_{e}$, for all $e\in\mathcal{G}_{0}$, instead of $\varphi_{e}(A_{e})\subseteq
B_{e}$. We will also denote by $1_{g}$ the identity of $A_{g}$ and by
$1_{g}^{\prime}$ the identity of $B_{g}$, for all $g\in\mathcal{G}$.

Now we shall prove the first part of the Galois correspondence.

######  Theorem 3.2.

Let $A$ be a partial $\alpha$-Galois extension of $A^{\alpha}$ and
$\mathcal{H}$ a wide subgroupoid of $\mathcal{G}$. Then
$\alpha|_{\mathcal{H}}$ is a partial action of $\mathcal{H}$ on $A$ and $A$ is
a partial $\alpha|_{\mathcal{H}}$-Galois extension of
$C=A^{\alpha|_{\mathcal{H}}}$. Furthermore, $C$ is $A^{\alpha}$-separable,
$\alpha$-strong and $\mathcal{G}_{C}=\mathcal{H}$.

###### Proof.

It is easy to see that $\alpha|_{\mathcal{H}}$ is an orthogonal partial action
of $\mathcal{H}$ on $A$ and that $A$ is a partial
$\alpha|_{\mathcal{H}}$-Galois extension of $C$ with the same partial Galois
coordinates of $A$ over $A^{\alpha}$.

Consider the globalization $\beta$ of $\alpha$. Define
$B_{g}^{\prime}=\sum_{\begin{subarray}{c}h\in\mathcal{H}\\\
r(h)=r(g)\end{subarray}}\beta_{h}(A_{d(h)})$ and
$\beta_{g}^{\prime}=\beta_{g}|_{B_{g^{-1}}^{\prime}}$. Then
$\beta^{\prime}=(B_{g}^{\prime},\beta_{g}^{\prime})_{g\in\mathcal{H}}$ is a
global action of $\mathcal{H}$ on
$B^{\prime}=\prod_{e\in\mathcal{H}_{0}}B_{e}^{\prime}$. In fact,

$\displaystyle\beta_{h}^{\prime}(B_{h^{-1}}^{\prime})$
$\displaystyle=\beta_{h}^{\prime}\left(\sum_{\begin{subarray}{c}k\in\mathcal{H}\\\
r(k)=d(h)\end{subarray}}\beta_{k}(A_{d(k)})\right)=\sum_{\begin{subarray}{c}k\in\mathcal{H}\\\
r(k)=d(h)\end{subarray}}\beta_{h}(\beta_{k}(A_{d(k)}))$
$\displaystyle=\sum_{\begin{subarray}{c}k\in\mathcal{H}\\\
r(k)=d(h)\end{subarray}}\beta_{hk}(A_{d(k)})=\sum_{\begin{subarray}{c}\ell\in\mathcal{H}\\\
r(\ell)=r(h)\end{subarray}}\beta_{\ell}(A_{d(\ell)})=B_{h}^{\prime}.$

Since $\beta_{h}$ is an isomorphism for all $h\in\mathcal{H}$, it follows that
$\beta_{h}^{\prime}$ is also an isomorphism. Clearly
$\beta_{e}^{\prime}=\beta_{e}|_{B_{e}^{\prime}}=\text{Id}_{B_{e}}|_{B_{e}^{\prime}}=\text{Id}_{B_{e}^{\prime}}$,
for all $e\in\mathcal{H}_{0}$, and given $(h,k)\in\mathcal{H}_{2}$,
$\beta_{h}^{\prime}\circ\beta_{k}^{\prime}=\beta_{h}|_{B_{d(h)}^{\prime}}\circ\beta_{k}|_{B_{d(k)}^{\prime}}=\beta_{h}|_{B_{r(k)}^{\prime}}\circ\beta_{k}|_{B_{d(k)}^{\prime}}=\beta_{h}\circ\beta_{k}|_{B_{d(k)}^{\prime}}=\beta_{hk}|_{B_{d(hk)}^{\prime}}=\beta_{hk}^{\prime}$.

Moreover, $\beta^{\prime}$ is the globalization of $\alpha|_{\mathcal{H}}$.
Indeed, $A_{e}\triangleleft B_{e}^{\prime}$ by definition, for all
$e\in\mathcal{H}_{0}$. Furthermore, $A_{h}=A_{r(h)}\cap\beta_{h}(A_{d(h)})$,
for all $h\in\mathcal{H}$. In particular, since $A_{d(h)}\subseteq
B_{d(h)}^{\prime}$, it follows that
$\beta_{h}(A_{d(h)})=\beta_{h}^{\prime}(A_{d(h)})$, from where
$A_{h}=A_{r(h)}\cap\beta_{h}^{\prime}(A_{d(h)})$, for all $h\in\mathcal{H}$.
By the same reason it follows that $\beta_{h}^{\prime}(a)=\alpha_{h}(a)$, for
all $a\in A_{h^{-1}},h\in\mathcal{H}$. Finally, the condition (G4) is
precisely the definition of $B_{h}$, for all $h\in\mathcal{H}$.

The action $\beta$ of $\mathcal{G}$ on $B$ induces a partial action $\gamma$
of $\mathcal{G}$ on $B^{\prime}$ via restriction, and $\beta$ is the
globalization of $\gamma$. Hence, the proof of [2, Theorem 4.6] implies that
$B^{\beta}1_{B^{\prime}}=(B^{\prime})^{\gamma}$ and
$B^{\beta}1_{A}=A^{\alpha}$. Therefore,
$(B^{\prime})^{\gamma}1_{A}=B^{\beta}1_{B^{\prime}}1_{A}=B^{\beta}1_{A}=A^{\alpha}$.
Furthermore, $(B^{\prime})^{\beta^{\prime}}1_{A}=A^{\alpha|_{\mathcal{H}}}$.

On the other hand, $B=B1_{B^{\prime}}\oplus
B(1_{B}-1_{B^{\prime}})=B^{\prime}\oplus B(1_{B}-1_{B^{\prime}})$. We have
that $\beta_{h}(B(1_{B}-1_{B^{\prime}})1_{h^{-1}}^{\prime})\subseteq
B(1_{B}-1_{B^{\prime}})$, for all $h\in\mathcal{H}$, as

$\displaystyle\beta_{h}(b(1_{B}-1_{B}^{\prime})1_{h^{-1}}^{\prime})$
$\displaystyle=\beta_{h}(b1_{h^{-1}}^{\prime})-\beta_{h}(b1_{B^{\prime}}1_{h^{-1}}^{\prime})=\beta_{h}(b1_{h^{-1}}^{\prime})-\beta_{h}(b1_{h^{-1}}^{\prime})\beta_{h}(1_{B^{\prime}}1_{h^{-1}}^{\prime})$
$\displaystyle=\beta_{h}(b1_{h^{-1}}^{\prime})-\beta_{h}(b1_{h^{-1}}^{\prime})\beta_{h}^{\prime}(1_{h^{-1}}^{\prime\prime})=\beta_{h}(b1_{h^{-1}}^{\prime})-\beta_{h}(b1_{h^{-1}}^{\prime})1_{h}^{\prime\prime}$
$\displaystyle=\beta_{h}(b1_{h^{-1}}^{\prime})1_{B}-\beta_{h}(b1_{h^{-1}}^{\prime})1_{B^{\prime}}=\beta_{h}(b1_{h^{-1}}^{\prime})(1_{B}-1_{B^{\prime}}).$

Thus,
$B^{\beta|_{\mathcal{H}}}=(B^{\prime})^{\beta|_{\mathcal{H}}}\oplus(B(1_{B}-1_{B^{\prime}}))^{\beta|_{\mathcal{H}}}$
and
$B^{\beta|_{\mathcal{H}}}1_{B^{\prime}}=(B^{\prime})^{\beta|_{\mathcal{H}}}1_{B^{\prime}}=(B^{\prime})^{\beta|_{\mathcal{H}}}=(B^{\prime})^{\beta^{\prime}}$
and hence
$C=A^{\alpha|_{\mathcal{H}}}=(B^{\prime})^{\beta^{\prime}}1_{A}=B^{\beta|_{\mathcal{H}}}1_{B^{\prime}}1_{A}=B^{\beta|_{\mathcal{H}}}1_{A}$.

By Lemma 3.1, we have that $B$ is a $\beta$-Galois extension of $B^{\beta}$.
So, by [15, Theorem 4.1], the $B^{\beta}$-subalgebra
$B^{\beta|_{\mathcal{H}}}$ is separable, $\beta$-strong and
$\mathcal{H}=\mathcal{G}_{B^{\beta|_{\mathcal{H}}}}^{\prime}:=\\{g\in\mathcal{G}:\beta_{g}(b1_{g^{-1}}^{\prime})=b1_{g}^{\prime},\text{
for all }b\in B^{\beta|_{\mathcal{H}}}\\}$. If $e^{\prime}\in
B^{\beta|_{\mathcal{H}}}\otimes_{B^{\beta}}B^{\beta|_{\mathcal{H}}}$ is the
idempotent of separability of $B^{\beta|_{\mathcal{H}}}$ over $B^{\beta}$,
then $e=e^{\prime}(1_{A}\otimes 1_{A})\in
B^{\beta|_{\mathcal{H}}}1_{A}\otimes_{B^{\beta}1_{A}}B^{\beta|_{\mathcal{H}}}1_{A}=A^{\alpha|_{\mathcal{H}}}\otimes_{A^{\alpha}}A^{\alpha|_{\mathcal{H}}}=C\otimes_{A^{\alpha}}C$
is the idempotent of separability of $C$ over $A^{\alpha}$.

Now we prove that $C$ is $\alpha$-strong. Let $g,h\in\mathcal{G}$ be such that
$\exists g^{-1}h$, $g^{-1}h\notin\mathcal{G}_{C}$ and $e\in A_{g}\cup A_{h}$ a
nonzero idempotent. Since $B^{\beta|_{\mathcal{H}}}$ is $\beta$-strong and
$g^{-1}h\notin\mathcal{G}_{C}\supseteq\mathcal{H}=\mathcal{G}_{B^{\beta|_{\mathcal{H}}}}^{\prime}$,
if $e1_{g}1_{h}\neq 0$, there is $b\in B^{\beta|_{\mathcal{H}}}$ such that
$\beta_{g}(b1_{g^{-1}}^{\prime})e1_{g}1_{h}\neq\beta_{h}(b1_{h^{-1}}^{\prime})e1_{g}1_{h}$.
Hence $b1_{A}\in B^{\beta|_{\mathcal{H}}}1_{A}=C$ and
$\alpha_{g}(b1_{A}1_{g^{-1}})e1_{g}1_{h}=\alpha_{g}(b1_{g^{-1}})e1_{g}1_{h}=\beta_{g}(b1_{g^{-1}}^{\prime})e1_{g}1_{h}\neq\beta_{h}(b1_{h^{-1}}^{\prime})e1_{g}1_{h}=\alpha_{h}(b1_{h^{-1}})e1_{g}1_{h}=\alpha_{h}(b1_{A}1_{h}^{-1})e1_{g}1_{h}$,
from where we obtain
$\alpha_{g}(b1_{A}1_{g^{-1}})e\neq\alpha_{h}(b1_{A}1_{h^{-1}})e$. If
$e1_{g}1_{h}=0$ and $e\in A_{g}$, then
$\alpha_{g}(1_{A}1_{g^{-1}})e=1_{g}e=e\neq
0=e1_{g}1_{h}=e1_{h}=\alpha_{h}(1_{A}1_{h^{-1}})e$. The case such that $e\in
A_{h}$ is similar.

Finally, it only remains to us to show that
$\mathcal{G}_{C}\subseteq\mathcal{H}$, since we already have that
$\mathcal{H}\subseteq\mathcal{G}_{C}$. Assume, by contradiction, that there is
$g\in\mathcal{G}_{C}\setminus\mathcal{H}$. Since $B^{\beta|_{\mathcal{H}}}$ is
$\beta$-strong, there is $b\in B^{\beta|_{\mathcal{H}}}$ such that
$\beta_{g}(b1_{g^{-1}}^{\prime})1_{g}\neq\beta_{r(g)}(b1_{r(g)}^{\prime})1_{g}=b1_{g}$.
Hence,
$\alpha_{g}(b1_{A}1_{g^{-1}}^{\prime})=\beta_{g}(b1_{g^{-1}}^{\prime})1_{g}\neq
b1_{g}=b1_{A}1_{g}$, and it is a contradiction because $g\in\mathcal{G}_{C}$
and $b1_{A}\in B^{\beta|_{\mathcal{H}}}1_{A}=C$. Thus $\mathcal{H}$ must
coincide with $\mathcal{G}_{C}$, ending the proof. ∎

This result proves the first part of the Galois Correspondence theorem. We
also observe that Theorem 3.2 shows that if $A|_{A^{\alpha}}$ is a partial
$\alpha$-Galois extension, then $A$ is $A^{\alpha}$-separable and
$\alpha$-strong. We will prove in the next result that the converse also
holds.

######  Lemma 3.3.

Let $\alpha=(A_{g},\alpha_{g})_{g\in\mathcal{G}}$ be a unital, orthogonal
partial action of a groupoid $\mathcal{G}$ on a ring $A$. Then
$A|_{A^{\alpha}}$ is a partial $\alpha$-Galois extension if and only if $A$ is
$A^{\alpha}$-separable and $\alpha$-strong.

###### Proof.

By the discussion above, it only remains to us to prove that if $A$ is
$A^{\alpha}$-separable and $\alpha$-strong, then $A$ is partial
$\alpha$-Galois over $A^{\alpha}$.

For that, let $e=\sum_{i=1}^{n}x_{i}\otimes y_{i}\in A\otimes_{A^{\alpha}}A$
the idempotent of separability of $A$. That is,
$\mu(e)=\sum_{i=1}^{n}x_{i}y_{i}=1_{A}$ and $(a\otimes 1_{A}-1_{A}\otimes
a)e=0$, for all $a\in A$, where $\mu:A\otimes_{A^{\alpha}}A$ is the
multiplication map. For all $g\in\mathcal{G}$, define
$e_{g}=m((\text{Id}_{A}\otimes\alpha_{g}(-1_{g^{-1}}))(e))\in A_{g}$. We have
that $\text{Id}_{A}\otimes\alpha_{g}(-1_{g^{-1}})$ is an endomorphism of
$A\otimes_{A^{\alpha}}A$. Besides that, $\mu$ is a ring homomorphism, since
$A$ is commutative. Hence, for all $g\in\mathcal{G}$ we have

$\displaystyle e_{g}^{2}$
$\displaystyle=\mu((\text{Id}_{A}\otimes\alpha_{g}(-1_{g^{-1}}))(e))\cdot\mu((\text{Id}_{A}\otimes\alpha_{g}(-1_{g^{-1}}))(e))$
$\displaystyle=\mu((\text{Id}_{A}\otimes\alpha_{g}(-1_{g^{-1}}))(e)\cdot(\text{Id}_{A}\otimes\alpha_{g}(-1_{g^{-1}}))(e))$
$\displaystyle=\mu((\text{Id}_{A}\otimes\alpha_{g}(-1_{g^{-1}}))(e^{2}))=\mu((\text{Id}_{A}\otimes\alpha_{g}(-1_{g^{-1}}))(e))=e_{g}.$

That is, $e_{g}$ is an idempotent in $A_{g}$. On the one hand, for all $a\in
A$, we get

$\displaystyle\alpha_{r(g)}(a1_{r(g)})e_{g}$
$\displaystyle=ae_{g}=a\mu((\text{Id}_{A}\otimes\alpha_{g}(-1_{g^{-1}}))(e))=a\mu((\text{Id}_{A}\otimes\alpha_{g}(-1_{g^{-1}}))(e))1_{g}$
$\displaystyle=(a\otimes
1_{g})\cdot\mu((\text{Id}_{A}\otimes\alpha_{g}(-1_{g^{-1}}))(e))=\mu((a\otimes
1_{g})\cdot(\text{Id}_{A}\otimes\alpha_{g}(-1_{g^{-1}}))(e))$
$\displaystyle=\mu((\text{Id}_{A}\otimes\alpha_{g}(-1_{g^{-1}}))(a\otimes
1_{A})\cdot(\text{Id}_{A}\otimes\alpha_{g}(-1_{g^{-1}}))(e))$
$\displaystyle=\mu((\text{Id}_{A}\otimes\alpha_{g}(-1_{g^{-1}}))((a\otimes
1_{A})e))=\mu((\text{Id}_{A}\otimes\alpha_{g}(-1_{g^{-1}}))((1_{A}\otimes
a)e))$
$\displaystyle=\mu((\text{Id}_{A}\otimes\alpha_{g}(-1_{g^{-1}}))(1_{A}\otimes
a)\cdot(\text{Id}_{A}\otimes\alpha_{g}(-1_{g^{-1}}))(e))$
$\displaystyle=\mu((1_{A}\otimes\alpha_{g}(a1_{g^{-1}}))\cdot(\text{Id}_{A}\otimes\alpha_{g}(-1_{g^{-1}}))(e))$
$\displaystyle=\mu((1_{A}\otimes\alpha_{g}(a1_{g^{-1}})))\cdot\mu((\text{Id}_{A}\otimes\alpha_{g}(-1_{g^{-1}}))(e))$
$\displaystyle=\alpha_{g}(a1_{g^{-1}})\mu((\text{Id}_{A}\otimes\alpha_{g}(-1_{g^{-1}}))(e))=\alpha_{g}(a1_{g^{-1}})e_{g}.$

On the other hand, since $A$ is $\alpha$-strong, we have that if
$g\notin\mathcal{G}_{0}$, then we should have $e_{g}=0$. But

$\displaystyle e_{g}$
$\displaystyle=\mu((\text{Id}_{A}\otimes\alpha_{g}(-1_{g^{-1}}))(e))=\sum_{i=1}^{n}x_{i}\alpha_{g}(y_{i}1_{g^{-1}}).$

If $g\in\mathcal{G}_{0}$, then $g=r(g)$ and

$\displaystyle e_{g}$
$\displaystyle=\mu((\text{Id}_{A}\otimes\alpha_{r(g)}(-1_{r(g)}))(e))=\sum_{i=1}^{n}x_{i}\alpha_{r(g)}(y_{i}1_{r(g)})$
$\displaystyle=\sum_{i=1}^{n}x_{i}y_{i}1_{r(g)}=1_{A}1_{r(g)}=1_{r(g)}.$

Therefore $\\{x_{i},y_{i}\\}_{i=1}^{n}$ is a partial $\alpha$-Galois
coordinate system of $A$ over $A^{\alpha}$. ∎

Before we prove the converse of Theorem 3.2, we need a technical lemma.

######  Lemma 3.4.

Let $A$ be a partial $\alpha$-Galois extension of $A^{\alpha}$ and $C$ a
separable, $\alpha$-strong $A^{\alpha}$-subalgebra of $A$. Then there exist
$x_{i},y_{i}\in C$, $1\leq i\leq n$, such that
$\sum_{i=1}^{n}x_{i}y_{i}=1_{A}$ and
$\sum_{i=1}^{n}x_{i}\alpha_{g}(y_{i}1_{g^{-1}})=0$, for all
$g\in\mathcal{G}\setminus\mathcal{G}_{C}$.

###### Proof.

Let $e=\sum_{i=1}^{n}x_{i}\otimes y_{i}\in C\otimes_{A^{\alpha}}C$ be the
idempotent of separability of $C$ over $A^{\alpha}$. By definition we have
that $\sum_{i=1}^{n}x_{i}y_{i}=1_{A}$. Take $g\in\mathcal{G}$ and define
$e_{g}=\sum_{i=1}^{n}x_{i}\alpha_{g}(y_{i}1_{g^{-1}})\in A_{g}$, which is an
idempotent since $e_{g}=\mu(\theta_{g}(e))$, where $\theta_{g}:C\otimes C\to
C\otimes A_{g}$ is the $C$-algebra homomorphism given by
$\theta_{g}\left(\sum_{i=1}^{m}a_{i}\otimes
b_{i}\right)=\sum_{i=1}^{m}a_{i}\alpha_{g}(b_{i}1_{g^{-1}})$ and $\mu:C\otimes
C\to C$ is the product map, which is a homomorphism since $C$ is commutative.

Observe now that $ae_{g}=\alpha_{g}(a1_{g}^{-1})e_{g}$, for all $a\in C$. In
fact,

$\displaystyle ae_{g}$
$\displaystyle=ae_{g}1_{g}=a\mu(\theta_{g}(e))1_{g}=a\mu((\text{Id}_{C}\otimes\alpha_{g}(-1_{g^{-1}}))(e))1_{g}$
$\displaystyle=a\mu((\text{Id}_{C}\otimes\alpha_{g}(-1_{g^{-1}}))(e))1_{g}=(a\otimes
1_{g})\cdot\mu((\text{Id}_{C}\otimes\alpha_{g}(-1_{g^{-1}}))(e))$
$\displaystyle=\mu((a\otimes
1_{g})\cdot(\text{Id}_{C}\otimes\alpha_{g}(-1_{g^{-1}}))(e))$
$\displaystyle=\mu((\text{Id}_{C}\otimes\alpha_{g}(-1_{g^{-1}}))(a\otimes
1_{A})\cdot(\text{Id}_{C}\otimes\alpha_{g}(-1_{g^{-1}}))(e))$
$\displaystyle=\mu((\text{Id}_{C}\otimes\alpha_{g}(-1_{g^{-1}}))((a\otimes
1_{A})\cdot e))$
$\displaystyle=\mu((\text{Id}_{C}\otimes\alpha_{g}(-1_{g^{-1}}))((1_{A}\otimes
a)\cdot e))$
$\displaystyle=\mu((\text{Id}_{C}\otimes\alpha_{g}(-1_{g^{-1}}))(1_{A}\otimes
a)\cdot(\text{Id}_{C}\otimes\alpha_{g}(-1_{g^{-1}}))(e))$
$\displaystyle=\mu((1_{A}\otimes\alpha_{g}(a1_{g^{-1}})\cdot(\text{Id}_{C}\otimes\alpha_{g}(-1_{g^{-1}}))(e))$
$\displaystyle=(1_{A}\otimes\alpha_{g}(a1_{g^{-1}})\cdot\mu((\text{Id}_{C}\otimes\alpha_{g}(-1_{g^{-1}}))(e))$
$\displaystyle=\alpha_{g}(a1_{g^{-1}})\mu((\text{Id}_{C}\otimes\alpha_{g}(-1_{g^{-1}}))(e))=\alpha_{g}(a1_{g^{-1}})e_{g}.$

Since $C$ is $\alpha$-strong, it follows that $e_{g}=0$ unless
$g\in\mathcal{G}_{C}$. ∎

######  Theorem 3.5.

Let $A$ be a partial $\alpha$-Galois extension of $A^{\alpha}$ and $C$ a
separable, $\alpha$-strong $A^{\alpha}$-subalgebra of $A$ such that
$\mathcal{G}_{C}$ is a wide subgroupoid of $\mathcal{G}$. Then
$A^{\alpha|_{\mathcal{G}_{C}}}=C$.

###### Proof.

We always have that $C\subseteq A^{\alpha|_{\mathcal{G}_{C}}}$. Our goal is to
prove the reverse inclusion. Let $\beta$ be the globalization of $\alpha$.
Consider the subalgebra

$\displaystyle
B^{\prime}=\prod_{e\in\mathcal{G}_{0}}\left(\sum_{\begin{subarray}{c}g\in\mathcal{G}_{C}\\\
r(g)=e\end{subarray}}\beta_{g}(A_{d(g)})\right),$

where $\mathcal{G}_{C}$ acts via $\beta^{\prime}$ as in the proof of Theorem
3.2 and $\beta^{\prime}$ is the globalization of the orthogonal partial action
$\alpha|_{\mathcal{G}_{C}}$ of $\mathcal{G}_{C}$ on $A$.

By [2, Theorem 4.6], there is a ring homomorphism $\psi_{\mathcal{G}_{C}}:A\to
B^{\prime}$ such that
$\psi_{\mathcal{G}_{C}}|_{A^{\alpha|_{\mathcal{G}_{C}}}}:A^{\alpha|_{\mathcal{G}_{C}}}\to(B^{\prime})^{\beta^{\prime}}$
is a ring homomorphism and $\psi_{\mathcal{G}_{C}}(a)1_{A}=a$, for all $a\in
A^{\alpha|_{\mathcal{G}_{C}}}$. Let $C^{\prime}=\psi_{\mathcal{G}_{C}}(C)$.

Claim 1: There exist $x_{i}^{\prime},y_{i}^{\prime}\in C^{\prime}$, $1\leq
i\leq n$, such that
$\sum_{i=1}^{n}x_{i}^{\prime}y_{i}^{\prime}=1_{B^{\prime}}$ and such that
$\sum_{i=1}^{n}x_{i}^{\prime}\beta_{g}^{\prime}(y_{i}^{\prime}1_{g^{-1}}^{\prime})=0$,
for all $g\in\mathcal{G}\setminus\mathcal{G}_{C}$.

In fact, by Lemma 3.4, there exist $x_{i},y_{i}\in C$, $i\leq i\leq n$, such
that $\sum_{i=1}^{n}x_{i}y_{i}=1_{A}$ and
$\sum_{i=1}^{n}x_{i}\alpha_{g}(y_{i}1_{g^{-1}})=0$, for all
$g\in\mathcal{G}\setminus\mathcal{G}_{C}$. Since $x_{i},y_{i}\in C\subseteq
A$, given $g\in\mathcal{G}\setminus\mathcal{G}_{C}$, we have that

$\displaystyle\sum_{i=1}^{n}x_{i}\beta_{g}(y_{i}1_{g^{-1}}^{\prime})$
$\displaystyle=\sum_{i=1}^{n}x_{i}\beta_{g}(y_{i}1_{A}1_{g^{-1}}^{\prime})=\sum_{i=1}^{n}x_{i}\beta_{g}(y_{i}1_{g^{-1}})=\sum_{i=1}^{n}x_{i}\alpha_{g}(y_{i}1_{g^{-1}})=0.$

Let $x_{i}^{\prime}=\psi_{\mathcal{G}_{C}}(x_{i})$ and
$y_{i}^{\prime}=\psi_{\mathcal{G}_{C}}(y_{i})$. Then
$\sum_{i=1}^{n}x_{i}^{\prime}y_{i}^{\prime}=\psi_{\mathcal{G}_{C}}\left(\sum_{i=1}^{n}x_{i}y_{i}\right)=\psi_{\mathcal{G}_{C}}(1_{A})=1_{B^{\prime}}$.
Consider now $e\in\mathcal{G}_{0}$ and
$\\{h_{1,e}=e,h_{2,e},\ldots,h_{m_{e},e}\\}=\mathcal{G}_{C}(-,e)=\\{g\in\mathcal{G}_{C}:r(g)=e\\}$,
where $m_{e}>0$. By [2, pg. 15], $1_{e}^{\prime}=v_{1,e}+\cdots+v_{m_{e},e}$,
where $v_{1,e}=1_{e}$ and
$v_{j,e}=(1_{e}^{\prime}-1_{e})(1_{e}^{\prime}-\beta_{h_{2,e}}(1_{d(h_{2,e})}))\cdots(1_{e}^{\prime}-\beta_{h_{j-1,e}}(1_{d(h_{j-1,e})}))\beta_{h_{j,e}}(1_{d(h_{j,e})})$.

Defining $\psi_{e}:A\to B^{\prime}$ by
$\psi_{e}(a)=\sum_{j=1}^{m_{e}}\beta_{h_{j,e}}(a_{d(h_{j,e})}1_{h_{j,e}^{-1}}^{\prime})v_{j,e}$,
we have that $\psi_{\mathcal{G}_{C}}=\sum_{e\in\mathcal{G}_{0}}\psi_{e}$.
Hence, for all $g\in\mathcal{G}\setminus\mathcal{G}_{C}$, we have that

$\displaystyle\sum_{i=1}^{n}x_{i}^{\prime}\beta_{g}(y_{i}^{\prime}1_{g^{-1}}^{\prime})$
$\displaystyle=\sum_{i=1}^{n}\sum_{e\in\mathcal{G}_{0}}\sum_{1\leq j,k\leq
m_{e}}\beta_{h_{j,e}}((x_{i})_{d(h_{j,e})}1_{h_{j,e}^{-1}}^{\prime})v_{j,e}\beta_{g}(\beta_{h_{k,e}}((y_{i})_{d(h_{k,e})}1_{h_{k,e}^{-1}}^{\prime})v_{k,e}1_{g^{-1}}^{\prime})$
$\displaystyle=\sum_{i=1}^{n}\sum_{1\leq j,k\leq
m_{r(g)}}\beta_{h_{j,r(g)}}((x_{i})_{d(h_{j,r(g)})}1_{h_{j,r(g)}^{-1}}^{\prime})v_{j,r(g)}\beta_{g}(\beta_{h_{k,d(g)}}((y_{i})_{d(h_{k,d(g)})}1_{h_{k,d(g)}^{-1}}^{\prime})v_{k,d(g)})$
$\displaystyle=\sum_{1\leq j,k\leq
m_{r(g)}}v_{j,r(g)}\beta_{g}(v_{k,d(g)})\beta_{h_{j,r(g)}}\left(\sum_{i=1}^{n}(x_{i})_{d(h_{j,r(g)})}\beta_{h_{j,r(g)}^{-1}gh_{k,d(g)}}((y_{i})_{d(h_{k,d(g)})}1_{h_{k,d(g)}^{-1}}^{\prime})\right)$
$\displaystyle=\sum_{1\leq j,k\leq
m_{r(g)}}v_{j,r(g)}\beta_{g}(v_{k,d(g)})\beta_{h_{j,r(g)}}\left(\sum_{i=1}^{n}x_{i}\beta_{h_{j,r(g)}^{-1}gh_{k,d(g)}}(y_{i}1_{h_{k,d(g)}^{-1}}^{\prime})\right)=0,$

because if $h_{j,r(g)}^{-1}gh_{k,d(g)}\in\mathcal{G}_{C}$, then there is
$h\in\mathcal{G}_{C}$ such that $h_{j,r(g)}^{-1}gh_{k,d(g)}=h$. But then
$g=h_{j,r(g)}hh_{k,d(g)}^{-1}$, from where it follows that
$g\in\mathcal{G}_{C}$, a contradiction. This completes the proof of Claim 1.

Now, since $\mathcal{G}_{C}\subseteq\mathcal{G}_{C^{\prime}}$ and the elements
$x_{i}^{\prime},y_{i}^{\prime}\in C^{\prime}$, we have, in particular, that
$\mathcal{G}_{C^{\prime}}=\mathcal{G}_{C}$.

Consider again the restriction $\gamma$ of $\beta$ to $B^{\prime}$. It follows
from [2, Theorem 4.6] that there is a ring isomorphism
$B^{\beta}\to(B^{\prime})^{\gamma}$ given by $x\mapsto x1_{B^{\prime}}$.
Moreover, $x\mapsto x1_{A}$ is an isomorphism of $B^{\beta}$ on $A^{\alpha}$.
Hence, there is an isomorphism $(B^{\prime})^{\gamma}\to A^{\alpha}$ given by
$x1_{B^{\prime}}\mapsto x1_{A}$, whose inverse is
$\psi_{\mathcal{G}_{C}}|_{A^{\alpha}}$. Therefore
$\psi_{\mathcal{G}_{C}}(A^{\alpha})=(B^{\prime})^{\gamma}$ and hence
$C^{\prime}=\psi_{\mathcal{G}_{C}}(C)$ is separable over
$(B^{\prime})^{\gamma}$.

Notice that
$B^{\beta|_{\mathcal{G}_{C}}}=B^{\beta|_{\mathcal{G}_{C}}}1_{B^{\prime}}\oplus
B^{\beta|_{\mathcal{G}_{C}}}(1_{B}-1_{B^{\prime}})=(B1_{B^{\prime}})^{\beta|_{\mathcal{G}_{C}}}\oplus
B^{\beta|_{\mathcal{G}_{C}}}(1_{B}-1_{B^{\prime}})=(B^{\prime})^{\beta|_{\mathcal{G}_{C}}}\oplus
B^{\beta|_{\mathcal{G}_{C}}}(1_{B}-1_{B^{\prime}})=(B^{\prime})^{\beta^{\prime}}\oplus
B^{\beta|_{\mathcal{G}_{C}}}(1_{B}-1_{B^{\prime}})$. In particular,
$B^{\beta|_{\mathcal{G}_{C}}}1_{A}=(B^{\prime})^{\beta^{\prime}}1_{A}=A^{\alpha|_{\mathcal{G}_{C}}}$.
Consider the subalgebra $\overline{C}=C^{\prime}\oplus
B^{\beta|_{\mathcal{G}_{C}}}(1_{B}-1_{B^{\prime}})$ of
$B^{\beta|_{\mathcal{G}_{c}}}$.

Claim 2: $\overline{C}$ is $B^{\beta}$-separable and $\beta$-strong.

Indeed, $B^{\beta|_{\mathcal{G}_{C}}}$ is separable over $B^{\beta}$ by [15,
Theorem 4.1(ii)], from where it follows that
$B^{\beta|_{\mathcal{G}_{C}}}(1_{B}-1_{B^{\prime}})$ is separable over
$B^{\beta}(1_{B}-1_{B^{\prime}})$. Since $C^{\prime}$ is
$(B^{\prime})^{\gamma}$-separable and
$(B^{\prime})^{\gamma}=B^{\beta}1_{B^{\prime}}$, it follows that
$\overline{C}$ is separable over $(B^{\prime})^{\gamma}\oplus
B^{\beta}(1_{B}-1_{B^{\prime}})=B^{\beta}$.

To prove that $\overline{C}$ is $\beta$-strong, assume that
$g\in\mathcal{G}\setminus\mathcal{G}_{C}$, that $e\in B_{g}$ is an idempotent
and that $\beta_{g}((x+y)1_{g^{-1}}^{\prime})e=(x+y)e$, for all $x\in
C^{\prime}$ and $y\in B^{\beta|_{\mathcal{G}_{C}}}(1_{B}-1_{B^{\prime}})$.

Write $e=e_{1}+e_{2}$, with $e_{1}=e1_{B^{\prime}}$ and
$e_{2}=e(1_{B}-1_{B}^{\prime})$. Observe that $e_{1},e_{2}\in B_{g}$. Then
multiplying $\beta_{g}((x+y)1_{g^{-1}}^{\prime})e=(x+y)e$ by $1_{B^{\prime}}$,
we obtain $\beta_{g}((x+y)1_{g^{-1}})e_{1}=(x+y)e_{1}=xe_{1}$, for all $x\in
C^{\prime}$ and $y\in B^{\beta|_{\mathcal{G}_{C}}}(1_{B}-1_{B^{\prime}})$. In
particular, when $y=0$ we have that
$\beta_{g}(x1_{g^{-1}}^{\prime})e_{1}=xe_{1}$, for all $x\in C^{\prime}$.

By Claim 1, there are $x_{i}^{\prime},y_{i}^{\prime}\in C^{\prime}$, $1\leq
i\leq n$, such that
$\sum_{i=1}^{n}x_{i}^{\prime}y_{i}^{\prime}=1_{B^{\prime}}$ and such that
$\sum_{i=1}^{n}x_{i}^{\prime}\beta_{g}(y_{i}^{\prime}1_{g^{-1}}^{\prime})=0$,
for all $g\in\mathcal{G}\setminus\mathcal{G}_{C}$. Thus
$0=\sum_{i=1}^{n}x_{i}^{\prime}\beta_{g}(y_{i}^{\prime}1_{g^{-1}}^{\prime})e_{1}=\sum_{i=1}^{n}x_{i}^{\prime}y_{i}^{\prime}e_{1}=1_{B^{\prime}}e_{1}=e_{1}$.
So we obtain $e=e_{2}$ and
$\beta_{g}((x+y)1_{g^{-1}}^{\prime})e_{2}=(x+y)e_{2}=ye_{2}$, for all $x\in
C^{\prime}$ and $y\in B^{\beta|_{\mathcal{G}_{C}}}(1_{B}-1_{B^{\prime}})$.
Considering $x=0$, we obtain $\beta_{g}(y1_{g^{-1}}^{\prime})e_{2}=ye_{2}$,
for all $y\in B^{\beta|_{\mathcal{G}_{C}}}(1_{B}-1_{B^{\prime}})$. Since
$B^{\beta|_{\mathcal{G}_{C}}}$ is $\beta$-strong and separable over
$B^{\beta}$, it follows that there are $u_{j},v_{j}\in
B^{\beta|_{\mathcal{G}_{C}}}$, $1\leq j\leq m$, such that
$\sum_{j=1}^{m}u_{j}v_{j}=1_{B}$ and
$\sum_{j=1}^{m}u_{j}\beta_{g}(v_{j}1_{g^{-1}}^{\prime})=0$, for all
$g\in\mathcal{G}\setminus\mathcal{G}_{C}$. Thus
$0=\sum_{j=1}^{m}u_{j}\beta_{g}(v_{j}(1_{B}-1_{B^{\prime}})1_{g^{-1}}^{\prime})e_{2}=\sum_{j=1}^{m}u_{j}v_{j}\beta_{g}((1_{B}-1_{B^{\prime}})1_{g^{-1}}^{\prime})e_{2}=1_{B}(1_{g}^{\prime}-1_{g}^{\prime\prime})e_{2}=(1_{B}-1_{B^{\prime}})e_{2}=e_{2}$,
from where it follows that $e=0$, that is, Claim 2 holds.

Now, by Claim 2 and [15, Theorem 4.5],
$\overline{C}=B^{\beta|_{\mathcal{G}_{\overline{C}}^{\prime}}}$, where
$\mathcal{G}_{\overline{C}}^{\prime}=\\{g\in\mathcal{G}:\beta_{h}(x1_{h^{-1}}^{\prime})=x1_{h}^{\prime},\text{
for all }x\in\overline{C}\\}$. Besides that, by definition of $\overline{C}$,
$g\in\mathcal{G}$ is in
$\mathcal{G}_{\overline{C}}=\mathcal{G}_{\overline{C}}^{\prime}$ if and only
if $g\in\mathcal{G}_{C^{\prime}}=\mathcal{G}_{C}$, from where
$\mathcal{G}_{C}=\mathcal{G}_{\overline{C}}$. It follows that
$\overline{C}=B^{\beta|_{\mathcal{G}_{C}}}$ and
$A^{\alpha|_{\mathcal{G}_{C}}}=(B^{\prime})^{\beta^{\prime}}1_{A}=B^{\beta|_{\mathcal{G}_{C}}}1_{B^{\prime}}1_{A}=B^{\beta|_{\mathcal{G}_{C}}}1_{A}=\overline{C}1_{A}=C^{\prime}1_{A}=C$.
∎

Now we can state the main result of this section. This theorem generalizes the
Galois Correspondence for orthogonal groupoid actions [15, Theorem 4.6], for
orthogonal group-type partial groupoid actions [4, Theorem 5.7] and for
partial group actions [7, Theorem 5.1].

######  Theorem 3.6 (Galois Correspondence for Orthogonal Partial Groupoid
Actions).

Let $\mathcal{G}$ be a finite groupoid acting partially on a ring $A$ via
unital and orthogonal partial action
$\alpha=(A_{g},\alpha_{g})_{g\in\mathcal{G}}$ such that $A_{g}\neq 0$, for all
$g\in\mathcal{G}$. Assume that $A$ is a partial $\alpha$-Galois extension of
$A^{\alpha}$. Then there is a one-to-one correspondence between the wide
subgroupoids $\mathcal{H}$ of $\mathcal{G}$ and the separable, $\alpha$-strong
$A^{\alpha}$-subalgebras $C$ of $A$ such that $\mathcal{G}_{C}$ is a
subgroupoid of $\mathcal{G}$, given by $\mathcal{H}\mapsto
A^{\alpha|_{\mathcal{H}}}$ with inverse $C\mapsto\mathcal{G}_{C}$.

###### Proof.

It follows by Theorems 3.2 and 3.5. ∎

##  4\. General Actions

Until now, Galois theory for groupoid actions has required the orthogonality
of the action and the reason is to assure the invariance of the trace map.
This hypothesis, even though it covers a large class of actions, is somewhat
restrictive. So the goal of this section is to develop a Galois theory for
nonorthogonal partial actions of groupoids.

###  4.1. Orthogonalization

######  Definition 4.1.

Two partial actions $\alpha=(A_{g},\alpha_{g})_{g\in\mathcal{G}}$ of
$\mathcal{G}$ on a ring $A$ and
$\varepsilon=(E_{g},\varepsilon_{g})_{g\in\mathcal{G}}$ of $\mathcal{G}$ on a
ring $E$ are _equivalent_ (we denote by $\alpha\simeq\varepsilon$) if there is
a collection of ring isomorphisms $\\{\varphi_{e}:A_{e}\to
E_{e}:e\in\mathcal{G}_{0}\\}$ such that

  1. (i)

$\varphi_{r(g)}(A_{g})=E_{g}$, for all $g\in\mathcal{G}$,

  2. (ii)

$\varphi_{r(g)}\circ\alpha_{g}(a)=\varepsilon_{g}\circ\varphi_{d(g)}(a)$, for
all $a\in A_{g^{-1}}$.

Let $\alpha=(A_{g},\alpha_{g})_{g\in\mathcal{G}}$ be a preunital partial
action (not necessarily orthogonal) of a finite groupoid $\mathcal{G}$ on a
ring $A$. We wish to construct a partial action $\varepsilon$ of $\mathcal{G}$
on a ring $E$ which is orthogonal and equivalent to $\alpha$. Indeed, consider
$E=\prod_{e\in\mathcal{G}_{0}}A_{e}$ and let $\varphi_{e}:A_{e}\to E$ be the
natural inclusion given by $\varphi(a)=(a\delta_{e,f})_{f\in\mathcal{G}_{0}}$.
Define $E_{g}=\varphi_{r(g)}(A_{g})$ and
$\varepsilon_{g}=\varphi_{r(g)}\circ\alpha_{g}\circ\varphi_{d(g)}^{-1}|_{E_{g^{-1}}}$,
for all $g\in\mathcal{G}$. Then it is easy to see that
$\varepsilon=(E_{g},\varepsilon_{g})_{g\in\mathcal{G}}$ is an orthogonal
partial action of $\mathcal{G}$ on $E$ and $\varepsilon\simeq\alpha$.

######  Definition 4.2.

If an orthogonal partial action $\varepsilon$ is equivalent to a partial
action $\alpha$, we say that $\varepsilon$ is the _orthogonalization_ of
$\alpha$.

######  Remark 4.3.

Observe that all the orthogonalizations are equivalent, that is,
orthogonalization is unique up to equivalence, and that is why we use the term
“the” orthogonalization instead of “an” orthogonalization.

As well as in the previous section, given a unital partial action $\alpha$, we
fix the notation $\beta=(B_{g},\beta_{g})_{g\in\mathcal{G}}$ to represent its
globalization, which is an orthogonal global action on a ring
$B=\prod_{e\in\mathcal{G}_{0}}B_{e}$ such that $A_{e}$ is an ideal of $B_{e}$,
for all $e\in\mathcal{G}_{0}$. We also fix the notation
$\varepsilon=(E_{g},\varepsilon_{g})_{g\in\mathcal{G}}$ to denote the
orthogonalization of $\alpha$.

######  Proposition 4.4.

Let $\alpha$ be a unital partial action and $\beta$ its globalization. Then
$\beta$ is also the globalization of $\varepsilon$.

###### Proof.

It only takes us to observe that we can realize
$E=\prod_{e\in\mathcal{G}_{0}}A_{e}$ as an ideal of $B$ and that $\varepsilon$
coincides with the standard restriction of $\beta$ to $E$ as defined in [2,
Section 1], that is, $E_{e}=B_{e}\cap E$, for all $e\in\mathcal{G}_{0}$,
$E_{g}=E_{r(g)}\cap\beta_{g}(E_{d(g)})$, for all $g\in\mathcal{G}$, and
$\varepsilon_{g}=\beta_{g}|_{E_{g^{-1}}}$. ∎

Partial Galois extensions also have some properties preserved by
orthogonalizations.

######  Proposition 4.5.

On the above notations, assume that $A|_{A^{\alpha}}$ is a partial
$\alpha$-Galois extension. Then $E|_{E^{\varepsilon}}$ is a partial
$\varepsilon$-Galois extension.

###### Proof.

Let $\\{x_{i},y_{i}\\}_{i=1}^{n}$ be an $\alpha$-Galois partial coordinate
system of $A$ over $A^{\alpha}$, that is,

$\displaystyle\sum_{i=1}^{n}x_{i}\alpha_{g}(y_{i}1_{g^{-1}})=\sum_{e\in\mathcal{G}_{0}}1_{e}\delta_{e,g},$

for all $g\in\mathcal{G}$. Write
$x_{i}^{\prime}=\sum_{e\in\mathcal{G}_{0}}\varphi_{e}(x_{i}1_{e})\in E$ and
$y_{i}^{\prime}=\sum_{e\in\mathcal{G}_{0}}\varphi_{e}(y_{i}1_{e})\in E$, for
all $1\leq i\leq n$. Then

$\displaystyle\sum_{i=1}^{n}x_{i}^{\prime}\varepsilon_{g}(y_{i}^{\prime}\varphi_{d(g)}(1_{g^{-1}}))$
$\displaystyle=\sum_{i=1}^{n}\sum_{e\in\mathcal{G}_{0}}\sum_{f\in\mathcal{G}_{0}}\varphi_{e}(x_{i}1_{e})\varepsilon_{g}(\varphi_{f}(y_{i}1_{f})\varphi_{d(g)}(1_{g^{-1}}))$
$\displaystyle=\sum_{i=1}^{n}\varphi_{r(g)}(x_{i}1_{r(g)})\varepsilon_{g}(\varphi_{d(g)}(y_{i}1_{d(g)})\varphi_{d(g)}(1_{g^{-1}}))$
$\displaystyle=\sum_{i=1}^{n}\varphi_{r(g)}(x_{i}1_{r(g)})\varphi_{r(g)}\circ\alpha_{g}\circ\varphi_{d(g)}^{-1}(\varphi_{d(g)}(y_{i}1_{g^{-1}}))$
$\displaystyle=\sum_{i=1}^{n}\varphi_{r(g)}(x_{i}\alpha_{g}(y_{i}1_{g^{-1}})1_{r(g)})=\varphi_{r(g)}\left(\sum_{i=1}^{n}x_{i}\alpha_{g}(y_{i}1_{g^{-1}})1_{r(g)}\right).$

There are two cases to consider: (1) $g\in\mathcal{G}_{0}$; or (2)
$g\notin\mathcal{G}_{0}$. Suppose (1). Then
$\varphi_{r(g)}\left(\sum_{i=1}^{n}x_{i}\alpha_{g}(y_{i}1_{g^{-1}})1_{r(g)}\right)=\varphi_{r(g)}(1_{r(g)})=1_{C_{r(g)}}$.
Now suppose (2). Thus

$\varphi_{r(g)}\left(\sum_{i=1}^{n}x_{i}\alpha_{g}(y_{i}1_{g^{-1}})1_{r(g)}\right)=\varphi_{r(g)}(0)=0.$

Hence, $\\{x_{i}^{\prime},y_{i}^{\prime}\\}_{i=1}^{n}$ is a
$\varepsilon$-Galois partial coordinate system of $E$ over $E^{\varepsilon}$.
∎

The converse of the above proposition is not true even for the global case, as
we can see in the next example.

######  Example 4.6.

Let $\mathcal{G}=\\{g,g^{-1},r(g),d(g)\\}$, $A=Re_{1}\oplus Re_{2}\oplus
Re_{3}$, where $R\neq 0$ is a commutative ring and the $e_{i}$’s are pairwise
orthogonal, central idempotents such that $e_{1}+e_{2}+e_{3}=1_{A}$.

Define $\beta=(A_{s},\beta_{s})_{s\in\mathcal{G}}$ as

$\displaystyle A_{g^{-1}}=A_{d(g)}=Re_{1}\oplus Re_{2},$ $\displaystyle\quad
A_{g}=A_{r(g)}=Re_{2}\oplus Re_{3},$
$\displaystyle\beta_{g}(ae_{1}+be_{2})=be_{2}+ae_{3},$
$\displaystyle\quad\beta_{e}=\text{Id}_{A_{e}},\text{ for
}e\in\mathcal{G}_{0},\,\,\text{ and }\beta_{g^{-1}}=\beta_{g}^{-1}.$

Hence $A^{\alpha}=R(e_{1}+e_{3})\oplus Re_{2}.$

Claim. $A$ is not a $\beta$-Galois extension of $A^{\beta}$.

Indeed, suppose that $A$ is a $\beta$-Galois extension of $A^{\beta}$, that
is, there are $x_{i}=x_{i1}e_{1}+x_{i_{2}}e_{2}+x_{i_{3}}e_{3}$ and
$y_{i}=y_{i1}e_{1}+y_{i_{2}}e_{2}+y_{i_{3}}e_{3}$ in $A$ and a positive
integer $n$ such that
$\sum_{i=1}^{n}x_{i}\beta_{s}(y_{i}1_{s^{-1}})=\sum_{e\in\mathcal{G}_{0}}\delta_{e,s}1_{e},\text{
for all }s\in\mathcal{G}.$

In particular, for $s=g$, $\sum_{i=1}^{n}x_{i}\beta_{g}(y_{i}1_{g^{-1}})=0.$
Therefore

$\displaystyle\sum_{i=1}^{n}(x_{i1}e_{1}+x_{i2}e_{2}+x_{i3}e_{3})\beta_{g}(y_{i1}e_{1}+y_{i2}e_{2})=0$
$\displaystyle\implies$
$\displaystyle\sum_{i=1}^{n}(x_{i1}e_{1}+x_{i2}e_{2}+x_{i3}e_{3})(y_{i1}e_{3}+y_{i2}e_{2})=0$

$\displaystyle\implies$
$\displaystyle\sum_{i=1}^{n}x_{i2}y_{i2}e_{2}+x_{i3}y_{i1}e_{3}=0$
$\displaystyle\implies$
$\displaystyle\left(\sum_{i=1}^{n}x_{i2}y_{i2}\right)e_{2}+\left(\sum_{i=1}^{n}x_{i3}y_{i1}\right)e_{3}=0.$

Then $\sum_{i=1}^{n}x_{i2}y_{i2}=0=\sum_{i=1}^{n}x_{i3}y_{i1}$. On the other
hand, since $\sum_{i=1}^{n}x_{i}\beta_{r(g)}(y_{i}1_{g^{-1}})=e_{2}+e_{3},$ it
follows that

$\displaystyle\sum_{i=1}^{n}(x_{i1}e_{1}+x_{i2}e_{2}+x_{i3}e_{3})\beta_{r(g)}(y_{i2}e_{2}+y_{i3}e_{3})=e_{2}+e_{3}$
$\displaystyle\implies$
$\displaystyle\sum_{i=1}^{n}(x_{i1}e_{1}+x_{i2}e_{2}+x_{i3}e_{3})(y_{i2}e_{2}+y_{i3}e_{3})=e_{2}+e_{3}$
$\displaystyle\implies$
$\displaystyle\sum_{i=1}^{n}x_{i2}y_{i2}e_{2}+x_{i3}y_{i3}e_{3}=e_{2}+e_{3}$
$\displaystyle\implies$
$\displaystyle\left(\sum_{i=1}^{n}x_{i2}y_{i2}\right)e_{2}+\left(\sum_{i=1}^{n}x_{i3}y_{i3}\right)e_{3}=e_{2}+e_{3}.$

Hence $\sum_{i=1}^{n}x_{i2}y_{i2}=1_{R}=\sum_{i=1}^{n}x_{i3}y_{i1}$. But then
$1_{R}=0$, a contradiction.

Now consider the orthogonalization $\varepsilon$ of $\alpha$ given as follows:
$E=\prod_{e\in\mathcal{G}_{0}}A_{e}=A_{d(g)}\times A_{r(g)}=(Re_{1}\oplus
Re_{2})\times(Re_{2}\oplus Re_{3})$, $E_{g^{-1}}=E_{d(g)}=(Re_{1}\oplus
Re_{2},0)$, $E_{g}=E_{r(g)}=(0,Re_{2}\oplus Re_{3})$,
$\varepsilon_{r(g)}=\text{Id}_{E_{g}}$,
$\varepsilon_{d(g)}=\text{Id}_{E_{g^{-1}}}$,
$\varepsilon_{g}((ae_{1}+be_{2},0))=(0,ae_{2}+be_{3})$ and
$\varepsilon_{g^{-1}}=\varepsilon_{g}^{-1}$. We can easily see that $E$ is
$\varepsilon$-Galois over $E^{\varepsilon}=R(e_{1},e_{2})\oplus
R(e_{2},e_{3})$ with coordinate system
$\\{x_{1}=y_{1}=(e_{1},0),x_{2}=y_{2}=(e_{2},0),x_{3}=y_{3}=(0,e_{2}),x_{4}=y_{4}=(0,e_{3})\\}$.

As a corollary of Proposition 4.5 we have the following.

######  Corollary 4.7.

Assume that $\alpha$ is a unital partial action (not necessarily orthogonal).
If $A$ is $\alpha$-Galois partial over $A^{\alpha}$, then $B$ is
$\beta$-Galois over $B^{\beta}$.

###### Proof.

By Proposition 4.5, $A|_{A^{\alpha}}$ partial $\alpha$-Galois implies
$E|_{E^{\varepsilon}}$ partial $\varepsilon$-Galois. By Lemma 3.1,
$E|_{E^{\varepsilon}}$ partial $\varepsilon$-Galois implies $B|_{B^{\beta}}$
$\beta$-Galois, which concludes the proof. ∎

###  4.2. Strongly $\alpha$-Galois Extension

Keeping the notations of the previous subsection, define

$\displaystyle\varphi:A$ $\displaystyle\to E$ $\displaystyle a$
$\displaystyle\mapsto(\varphi_{e}(a1_{e}))_{e\in\mathcal{G}_{0}}.$

We have that $\varphi$ is a monomorphism. Indeed, $\varphi(a)=0$ if and only
if $\varphi_{e}(a1_{e})=0$, for all $e\in\mathcal{G}_{0}$. But
$\varphi_{e}(a1_{e})=0$ if and only if $a1_{e}=0$, since $\varphi_{e}:A_{e}\to
E_{e}$ is a monomorphism, and from (P1) it follows that $a=0$ if and only if
$a1_{e}=0$, for all $e\in\mathcal{G}_{0}$.

######  Proposition 4.8.

Let $\mathcal{H}$ be a wide subgroupoid of $\mathcal{G}$. The following
assertion holds:
$\varphi(A^{\alpha|_{\mathcal{H}}})=E^{\varepsilon|_{\mathcal{H}}}\cap\varphi(A)$.

###### Proof.

Let $a\in A^{\alpha|_{\mathcal{H}}}$. Given $h\in\mathcal{H}$,

$\displaystyle\varepsilon_{h}(\varphi(a)\varphi_{d(h)}(1_{h^{-1}}))$
$\displaystyle=\varepsilon_{h}(\varphi_{d(h)}(a1_{d(h)})\varphi_{d(h)}(1_{h^{-1}}))=\varepsilon_{h}(\varphi_{d(h)}(a1_{d(h)}1_{h^{-1}}))$
$\displaystyle=\varepsilon_{h}(\varphi_{d(h)}(a1_{h^{-1}}))=\varphi_{r(h)}\circ\alpha_{h}\circ\varphi_{d(h)}^{-1}(\varphi_{d(h)}(a1_{h^{-1}}))$
$\displaystyle=\varphi_{r(h)}\circ\alpha_{h}(a1_{h^{-1}})=\varphi_{r(h)}(a1_{h})$
$\displaystyle=\varphi_{r(h)}(a)\varphi_{r(h)}(1_{h})=\varphi(a)\varphi_{r(h)}(1_{h}).$

Since we chose $h$ arbitrarily, we have that $\varphi(a)\in
E^{\varepsilon|_{\mathcal{H}}}$, proving the first inclusion.

For the reverse one, consider $a\in A$ such that $\varphi(a)\in
E^{\varepsilon|_{\mathcal{H}}}$. Then, given $h\in\mathcal{H}$, we have that
$\varepsilon_{h}(\varphi(a)\varphi_{d(h)}(1_{h^{-1}}))=\varphi(a)\varphi_{r(h)}(1_{h})$.
However

$\displaystyle\varepsilon_{h}(\varphi(a)\varphi_{d(h)}(1_{h^{-1}}))$
$\displaystyle=\varepsilon_{h}(\varphi_{d(h)}(a1_{d(h)})\varphi_{d(h)}(1_{h^{-1}}))=\varepsilon_{h}(\varphi_{d(h)}(a1_{h^{-1}}))=\varphi_{r(h)}\circ\alpha_{h}(a1_{h^{-1}}),$

and
$\varphi(a)\varphi_{r(h)}(1_{h})=\varphi_{r(h)}(a1_{r(h)})\varphi_{r(h)}(1_{h})=\varphi_{r}(h)(a1_{h})$.

Therefore $\varphi_{r(h)}(\alpha_{h}(a1_{h^{-1}}))=\varphi_{r(h)}(a1_{h})$.
Since $\varphi_{r(h)}$ is a monomorphism, it follows that
$\alpha_{h}(a1_{h^{-1}})=a1_{h}$, for all $h\in\mathcal{H}$, that is, $a\in
A^{\alpha|_{\mathcal{H}}}$. ∎

######  Definition 4.9.

Let $\alpha$ be a partial action of the groupoid $\mathcal{G}$ on a ring $A$.
Consider $\varepsilon$ the orthogonalization of $\alpha$ on a ring $E$. We say
that $\alpha$ is a _strongly Galois partial action_ if:

  1. (i)

$A_{g}\neq 0$, for all $g\in\mathcal{G}_{0}$;

  2. (ii)

$A$ is a partial $\alpha$-Galois extension of $A^{\alpha}$;

  3. (iii)

For all wide subgroupoids $\mathcal{H}_{1}\neq\mathcal{H}_{2}$ of
$\mathcal{G}$, it holds that

$E^{\varepsilon|_{\mathcal{H}_{1}}}\cap\varphi(A)\neq
E^{\varepsilon|_{\mathcal{H}_{2}}}\cap\varphi(A).$

In this case, we also say that $A$ is a _strongly $\alpha$-Galois extension
over_ $A^{\alpha}$.

Now we are able to state a Galois correspondence for the nonorthogonal partial
case. Firstly, we set:

$\displaystyle\mathsf{W}$
$\displaystyle=\\{\mathcal{H}\subseteq\mathcal{G}:\mathcal{H}\text{ is a wide
subgroupoid of }\mathcal{G}\\},$ $\displaystyle\mathsf{A}$
$\displaystyle=\\{C\subseteq
A:C=\varphi^{-1}(E^{\varepsilon|_{\mathcal{H}}}\cap\varphi(A)),\text{ for some
}\mathcal{H}\in\mathsf{W}\\}.$

######  Proposition 4.10.

Assume $A|_{A^{\alpha}}$ a partial $\alpha$-Galois extension. If
$C\in\mathsf{A}$, then $A|_{C}$ is a partial $\alpha|_{\mathcal{H}}$-Galois
extension for some wide subgroupoid $\mathcal{H}$ of $\mathcal{G}$.

###### Proof.

If $C\in\mathsf{A}$, then
$C=\varphi^{-1}(E^{\varepsilon|_{\mathcal{H}}}\cap\varphi(A))$, for some
$\mathcal{H}\in\mathsf{W}$. By Proposition 4.8,
$C=\varphi^{-1}(\varphi(A^{\alpha|_{\mathcal{H}}}))=A^{\alpha|_{\mathcal{H}}}$.
Now, consider $\\{x_{i},y_{i}\\}_{i=1}^{n}$ a partial $\alpha$-Galois
coordinate system for $A$ over $A^{\alpha}$, that is,

$\displaystyle\sum_{i=1}^{n}x_{i}\alpha_{g}(y_{i}1_{g^{-1}})=\sum_{e\in\mathcal{G}_{0}}\delta_{e,g}1_{e},$

for all $g\in\mathcal{G}$. Then it is clear that

$\displaystyle\sum_{i=1}^{n}x_{i}\alpha_{h}(y_{i}1_{h^{-1}})=\sum_{e\in\mathcal{G}_{0}}\delta_{e,h}1_{e},$

for all $h\in\mathcal{H}$, that is, $A$ is a partial
$\alpha|_{\mathcal{H}}$-Galois extension of $A^{\alpha|_{\mathcal{H}}}=C$.∎

Now we can prove the main result of this subsection.

######  Theorem 4.11 (Galois Correspondence for Strongly Galois Partial
Groupoid Actions).

Let $\mathcal{G}$ be a finite groupoid acting partially on a commutative ring
$A$ via a unital strongly Galois partial action $\alpha$. Then there is a one-
to-one correspondence between the sets $\mathsf{W}$ and $\mathsf{A}$ given by
$\mathsf{W}\ni\mathcal{H}\mapsto A^{\alpha|_{\mathcal{H}}}$ with inverse
$\mathsf{A}\ni C\mapsto\mathcal{G}_{\varphi(C)}$.

###### Proof.

First apply Theorem 3.6 on the orthogonalization $\varepsilon$ of $\alpha$, so
that we obtain a Galois correspondence between the wide subgroupoids of
$\mathcal{G}$ and the $E^{\varepsilon}$-separable, $\varepsilon$-strong
subalgebras $T$ of $E$ whose $\mathcal{G}_{T}$ is a wide subgroupoid. By
Proposition 4.8 and the strongly Galois hypothesis, there is a one-to-one
correspondence between those subalgebras $T$ and the $A^{\alpha}$-subalgebras
$C$ that appear in the set $\mathsf{A}$, via $T\leftrightarrow
T\cap\varphi(A)=C$. Composing these two correspondences we obtain the one
stated in the theorem, finishing the proof. ∎

In Definition 4.9, clearly if $\mathcal{G}$ is a group or if $\alpha$ is
already orthogonal, the condition (iii) holds. But these are not the only
classes that satisfy (i)-(iii). In the following we can see an example.

######  Example 4.12.

Let $G=\\{1_{G},g,h,gh\\}$ be a subgroup of $\mathbb{S}_{8}$, where

$\displaystyle g=\binom{1\;2\;3\;4\;5\;6\;7\;8}{2\;1\;4\;3\;6\;5\;8\;7},\quad
h=\binom{1\;2\;3\;4\;5\;6\;7\;8}{3\;4\;1\;2\;7\;8\;5\;6},\quad
gh=\binom{1\;2\;3\;4\;5\;6\;7\;8}{4\;3\;2\;1\;8\;7\;6\;5}.$

Consider $\mathcal{G}=\mathcal{A}_{2}\times G$ with
$\mathcal{G}_{0}=\\{f_{1},f_{2}\\}$, that is,
$G=\\{(f_{1},f_{1},1_{G}),(f_{1},f_{1},g),(f_{1},f_{1},h),$
$(f_{1},f_{1},gh),(f_{1},f_{2},1_{G}),(f_{1},f_{2},g),(f_{1},f_{2},h),(f_{1},f_{2},gh),(f_{2},f_{1},1_{G}),(f_{2},f_{1},g),$
$(f_{2},f_{1},h),\qquad$ $(f_{2},f_{1},gh),$ $(f_{2},f_{2},1_{G}),$
$(f_{2},f_{2},g),(f_{2},f_{2},h),(f_{2},f_{2},gh)\\}$. Given
$k=(i,j,\ell)\in\mathcal{G}$ with $i,j\in\\{f_{1},f_{2}\\}$, $\ell\in G$, we
define $\partial_{k}=\ell$.

Consider a commutative ring $R$ and the $R$-algebra
$A=\bigoplus_{i=1}^{12}Re_{i}$, where $\\{e_{i}\\}_{i=1}^{12}$ is a set of
pairwise orthogonal idempotents whose sum is $1_{A}$. Defining
$A_{k_{1}}=\bigoplus_{i=1}^{8}Re_{i}$, $A_{k_{2}}=\bigoplus_{i=5}^{12}Re_{i}$,
for all $k_{1},k_{2}\in\mathcal{G}$ with $r(k_{1})=f_{1}$, $r(k_{2})=f_{2}$,
and

$\displaystyle\beta_{k}\left(\sum_{i=1}^{8}a_{i}e_{i}\right)$
$\displaystyle=\begin{cases}\sum_{i=1}^{8}a_{i}e_{\partial_{k}(i)},\text{ if
}d(k)=r(k)=f_{1},\\\ \sum_{i=1}^{8}a_{i}e_{\partial_{k}(i)+4},\text{ if
}d(k)=f_{1},r(k)=f_{2},\end{cases}$
$\displaystyle\beta_{k}\left(\sum_{i=1}^{8}a_{i}e_{i+4}\right)$
$\displaystyle=\begin{cases}\sum_{i=1}^{8}a_{i}e_{\partial_{k}(i)+4},\text{ if
}d(k)=r(k)=f_{2},\\\ \sum_{i=1}^{8}a_{i}e_{\partial_{k}(i)},\text{ if
}d(k)=f_{2},r(k)=f_{1},\end{cases}$

we have that $\beta=(A_{k},\beta_{k})_{k\in\mathcal{G}}$ is a nonorthogonal
action of $\mathcal{G}$ on $A$.

Moreover, $A$ is strongly $\beta$-Galois over $A^{\beta}=R$ with coordinate
system $\\{x_{i}=y_{i}=e_{i}\\}_{i=1}^{12}$. We will present the Galois
correspondence for this action. For that, we will denote
$\mathcal{G}(f_{1})=G_{1}$, $\mathcal{G}(f_{2})=G_{2}$, and we will use the
following bracket notation to describe the subalgebras: we will represent by
$[i_{1},i_{2},\ldots,i_{n}]$ the $R$-algebra $R(e_{i_{1}}+e_{i_{2}}\cdots
e_{i_{n}})\oplus\bigoplus_{j\notin I}Re_{j}$, where
$I=\\{i_{1},\ldots,i_{n}\\}\subseteq\\{1,\ldots,12\\}$, and we will omit the
direct sum symbol. For example,

$\displaystyle[1,2]$ $\displaystyle\leftrightarrow
R(e_{1}+e_{2})\oplus\bigoplus_{i=3}^{12}Re_{i},$ $\displaystyle[3,4][5,6,7]$
$\displaystyle\leftrightarrow Re_{1}\oplus Re_{2}\oplus R(e_{3}+e_{4})\oplus
R(e_{5}+e_{6}+e_{7})\oplus\bigoplus_{i=8}^{12}Re_{i},$
$\displaystyle[1,2,3,4,5,6,7,8,9,10,11,12]$ $\displaystyle\leftrightarrow
A^{\beta}.$

The Galois correspondence relates the 33 elements of $\mathsf{W}$ with the 33
elements of $\mathsf{A}$:

$\displaystyle\mathcal{G}_{0}$ $\displaystyle\leftrightarrow A$
$\displaystyle\mathcal{G}_{0}\cup\\{(f_{1},f_{1},g)\\}$
$\displaystyle\leftrightarrow[1,2][3,4][5,6][7,8]$
$\displaystyle\mathcal{G}_{0}\cup\\{(f_{1},f_{1},h)\\}$
$\displaystyle\leftrightarrow[1,3][2,4][5,7][6,8]$
$\displaystyle\mathcal{G}_{0}\cup\\{(f_{1},f_{1},gh)\\}$
$\displaystyle\leftrightarrow[1,4][2,3][5,8][6,7]$
$\displaystyle\mathcal{G}_{0}\cup G_{1}$
$\displaystyle\leftrightarrow[1,2,3,4][5,6,7,8]$
$\displaystyle\mathcal{G}_{0}\cup\\{(f_{2},f_{2},g)\\}$
$\displaystyle\leftrightarrow[5,6][7,8][9,10][11,12]$
$\displaystyle\mathcal{G}_{0}\cup\\{(f_{2},f_{2},h)\\}$
$\displaystyle\leftrightarrow[5,7][6,8][9,11][10,12]$
$\displaystyle\mathcal{G}_{0}\cup\\{(f_{2},f_{2},gh)\\}$
$\displaystyle\leftrightarrow[5,8][6,7][9,12][10,11]$
$\displaystyle\mathcal{G}_{0}\cup G_{2}$
$\displaystyle\leftrightarrow[5,6,7,8][9,10,11,12]$ $\displaystyle
H_{1}:=\mathcal{G}_{0}\cup\\{(f_{1},f_{1},g),(f_{2},f_{2},g)\\}$
$\displaystyle\leftrightarrow[1,2][3,4][5,6][7,8][9,10][11,12]$
$\displaystyle\mathcal{G}_{0}\cup\\{(f_{1},f_{1},g),(f_{2},f_{2},h)\\}$
$\displaystyle\leftrightarrow[1,2][3,4][5,6,7,8][9,11][10,12]$
$\displaystyle\mathcal{G}_{0}\cup\\{(f_{1},f_{1},g),(f_{2},f_{2},gh)\\}$
$\displaystyle\leftrightarrow[1,2][3,4][5,6,7,8][9,12][10,11]$
$\displaystyle\mathcal{G}_{0}\cup\\{(f_{1},f_{1},g)\\}\cup G_{2}$
$\displaystyle\leftrightarrow[1,2][3,4][5,6,7,8][9,10,11,12]$
$\displaystyle\mathcal{G}_{0}\cup\\{(f_{1},f_{1},h),(f_{2},f_{2},g)\\}$
$\displaystyle\leftrightarrow[1,3][2,4][5,6,7,8][9,10][11,12]$ $\displaystyle
H_{2}:=\mathcal{G}_{0}\cup\\{(f_{1},f_{1},h),(f_{2},f_{2},h)\\}$
$\displaystyle\leftrightarrow[1,3][2,4][5,7][6,8][9,11][10,12]$
$\displaystyle\mathcal{G}_{0}\cup\\{(f_{1},f_{1},h),(f_{2},f_{2},gh)\\}$
$\displaystyle\leftrightarrow[1,3][2,4][5,6,7,8][9,12][10,11]$
$\displaystyle\mathcal{G}_{0}\cup\\{(f_{1},f_{1},h)\\}\cup G_{2}$
$\displaystyle\leftrightarrow[1,3][2,4][5,6,7,8][9,10,11,12]$
$\displaystyle\mathcal{G}_{0}\cup\\{(f_{1},f_{1},gh),(f_{2},f_{2},g)\\}$
$\displaystyle\leftrightarrow[1,4][2,3][5,6,7,8][9,10][11,12]$
$\displaystyle\mathcal{G}_{0}\cup\\{(f_{1},f_{1},gh),(f_{2},f_{2},h)\\}$
$\displaystyle\leftrightarrow[1,4][2,3][5,6,7,8][9,11][10,12]$ $\displaystyle
H_{3}:=\mathcal{G}_{0}\cup\\{(f_{1},f_{1},gh),(f_{2},f_{2},gh)\\}$
$\displaystyle\leftrightarrow[1,4][2,3][5,8][6,7][9,12][10,11]$
$\displaystyle\mathcal{G}_{0}\cup\\{(f_{1},f_{1},gh)\\}\cup G_{2}$
$\displaystyle\leftrightarrow[1,4][2,3][5,6,7,8][9,10,11,12]$
$\displaystyle\mathcal{G}_{0}\cup G_{1}\cup\\{(f_{2},f_{2},g)\\}$
$\displaystyle\leftrightarrow[1,2,3,4][5,6,7,8][9,10][11,12]$
$\displaystyle\mathcal{G}_{0}\cup G_{1}\cup\\{(f_{2},f_{2},h)\\}$
$\displaystyle\leftrightarrow[1,2,3,4][5,6,7,8][9,11][10,12]$
$\displaystyle\mathcal{G}_{0}\cup G_{1}\cup\\{(f_{2},f_{2},gh)\\}$
$\displaystyle\leftrightarrow[1,2,3,4][5,6,7,8][9,12][10,11]$

$\displaystyle\mathcal{G}_{0}\cup G_{1}\cup G_{2}$
$\displaystyle\leftrightarrow[1,2,3,4][5,6,7,8][9,10,11,12]$ $\displaystyle
K_{0}:=\mathcal{G}_{0}\cup\\{(f_{1},f_{2},e),(f_{2},f_{1},e)\\}$
$\displaystyle\leftrightarrow[1,5,9][2,6,10][3,7,11][4,8,12]$ $\displaystyle
K_{1}:=\mathcal{G}_{0}\cup\\{(f_{1},f_{2},g),(f_{2},f_{1},g)\\}$
$\displaystyle\leftrightarrow[1,6,9][2,5,10][3,8,11][4,7,12]$ $\displaystyle
K_{2}:=\mathcal{G}_{0}\cup\\{(f_{1},f_{2},h),(f_{2},f_{1},h)\\}$
$\displaystyle\leftrightarrow[1,7,9][2,8,10][3,5,11][4,6,12]$ $\displaystyle
K_{3}:=\mathcal{G}_{0}\cup\\{(f_{2},f_{1},gh),(f_{2},f_{1},gh)\\}$
$\displaystyle\leftrightarrow[1,8,9][2,7,10][3,6,11][4,5,12]$ $\displaystyle
K_{0}\cup K_{1}\cup H_{1}$
$\displaystyle\leftrightarrow[1,2,5,6,9,10][3,4,7,8,11,12]$ $\displaystyle
K_{0}\cup K_{2}\cup H_{2}$
$\displaystyle\leftrightarrow[1,3,5,7,9,11][2,4,6,8,10,12]$ $\displaystyle
K_{0}\cup K_{3}\cup H_{3}$
$\displaystyle\leftrightarrow[1,4,5,8,9,12][2,3,6,7,10,11]$
$\displaystyle\mathcal{G}$ $\displaystyle\leftrightarrow A^{\beta}$

######  Remark 4.13.

Until now, we can conclude the following:

  1. (1)

There are properties in Galois theory that are valid for orthogonal actions,
but not for nonorthogonal ones, so the theory cannot be reduced to the
orthogonal case.

  2. (2)

Theorem 4.11 gives us a generalization of the Galois correspondences that
already exist in the literature, but it is still restrictive, once it asks
that the partial action is strongly Galois, and not all partial actions over
Galois extensions belongs to this class, as we can see in the next example.

######  Example 4.14.

Let $\mathcal{G}=\\{g,f_{1}\\}\cup\\{h,f_{2}\\}$ the groupoid given by the
disjoint union of two groups isomorphic to $\mathbb{Z}_{2}$, that is,
$g^{2}=f_{1}=f_{1}^{2}$ and $h^{2}=f_{2}=f_{2}^{2}$. Consider a commutative
ring $R$ and $A=Re_{1}\oplus Re_{2}$, with $e_{1},e_{2}$ orthogonal
idempotents whose sum is $1_{A}$. Define $A_{k}=A$, for all $k\in\mathcal{G}$,

$\displaystyle\beta_{g}(ae_{1}+be_{2})=be_{1}+ae_{2}=\beta_{h}(ae_{1}+be_{2}),$

and $\beta_{f_{1}}=\beta_{f_{2}}=\text{id}_{A}$. We have that $\beta$ is a
(nonorthogonal) global action of $\mathcal{G}$ on $A$. Moreover, $A$ is
$\beta$-Galois over $A^{\beta}=R$ with coordinate system $x_{i}=y_{i}=e_{i}$,
$i=1,2$. However, we have that
$A^{\beta|_{\mathcal{G}(f_{1})}}=A^{\beta|_{\mathcal{G}(f_{2})}}=A^{\beta}$,
from where it follows that $A$ is not strongly $\beta$-Galois over
$A^{\beta}$.

###  4.3. Weakening the hypothesis

To finish the paper, we extend, in the global case, the Galois correspondence
for Galois extensions that are not strongly Galois. The main argument is the
following proposition.

######  Proposition 4.15.

Let $\beta=(A_{g},\beta_{g})$ be a global action of a groupoid $\mathcal{G}$
on a commutative ring $A$. If $\mathcal{H}_{1}$ and $\mathcal{H}_{2}$ are wide
groupoids of $\mathcal{G}$ such that
$A^{\beta|_{\mathcal{H}_{1}}}=A^{\beta|_{\mathcal{H}_{2}}}$, then the wide
subgroupoid of $\mathcal{G}$ generated by $\mathcal{H}_{1}$ and
$\mathcal{H}_{2}$, named $\mathcal{H}$, is such that
$A^{\beta|_{\mathcal{H}}}=A^{\beta|_{\mathcal{H}_{1}}}$.

###### Proof.

It is true that $A^{\beta|_{\mathcal{H}}}\subseteq
A^{\beta|_{\mathcal{H}_{1}}}$, since $\mathcal{H}_{1}\subseteq\mathcal{H}$.
For the reverse inclusion, let $a\in
A^{\beta|_{\mathcal{H}_{1}}}=A^{\beta|_{\mathcal{H}_{2}}}$. Let
$h\in\mathcal{H}$. We have that $h=h_{1}\cdots h_{n}$, for some positive
integer $n$ and $h_{1},\ldots,h_{n}\in\mathcal{H}_{1}\cup\mathcal{H}_{2}$ with
$d(h_{i})=r(h_{i+1})$, for all $1\leq i\leq n-1$. Then

$\displaystyle\beta_{h}(a1_{h^{-1}})$ $\displaystyle=\beta_{h_{1}\cdots
h_{n}}(a1_{d(h_{n})})=\beta_{h_{1}\cdots
h_{n-1}}(\beta_{h_{n}}(a1_{d(h_{n})}))$ $\displaystyle=\beta_{h_{1}\cdots
h_{n-1}}(a1_{r(h_{n})})=\beta_{h_{1}\cdots h_{n-1}}(a1_{d(h_{n-1})})$
$\displaystyle=\beta_{h_{1}\cdots
h_{n-2}}(\beta_{h_{n-1}}(a1_{d(h_{n-1})}))=\cdots$
$\displaystyle=\beta_{h_{1}}(a1_{d(h_{1})})=a1_{r(h_{1})}=a1_{h}.$

∎

Keeping the notations of the previous subsection, we define an equivalence
relation $\sim$ in $\mathsf{W}$ by: $\mathcal{H}_{1}\sim\mathcal{H}_{2}$ if
and only if $A^{\alpha|_{\mathcal{H}_{1}}}=A^{\alpha|_{\mathcal{H}_{2}}}$. If
$\alpha$ is global, then by Proposition 4.15 each $\sim$-class has a unique
maximal element with respect to inclusion. Namely, given
$\mathcal{H}\in\mathsf{W}$, we write $\mathcal{H}^{max}$ to represent the
maximal element of the $\sim$-class of $\mathcal{H}$. Therefore we can define
the set

$\displaystyle\mathsf{W}^{max}=\\{\mathcal{H}^{max}:\mathcal{H}\in\mathsf{W}\\}.$

Now we state the correspondence that finish our paper. We keep the same
notation $\mathsf{W}$ and $\mathsf{W}^{max}$ for a global action $\beta$ on a
ring $A$. When the extension $A|_{A^{\beta}}$ is strongly Galois in the
theorem below, we have exactly Theorem 4.11 for global actions, since in this
case every $\sim$-class consists of only one wide subgroupoid. If $\beta$ is
orthogonal, we have the Galois correspondence for orthogonal groupoid actions
on commutative rings [15, Theorem 4.6], since $\beta$ being orthogonal implies
that the extension $A|_{A^{\beta}}$ is strongly Galois and that $\beta$ is
equal to its own orthogonalization.

######  Theorem 4.16 (Galois Correspondence for Global Groupoid Actions).

Let $\mathcal{G}$ be a finite groupoid acting on a commutative ring $A$ via a
unital global action $\beta$. Consider $\varepsilon$ the orthogonalization of
$\beta$ on a ring $E$. Assume that $A_{g}\neq 0$, for all
$g\in\mathcal{G}_{0}$, and that $A$ is a $\beta$-Galois extension of
$A^{\beta}$. Then there is a one-to-one correspondence between the sets
$\mathsf{W}^{max}$ and $\mathsf{A}$ given by
$\mathsf{W}^{max}\ni\mathcal{H}\mapsto A^{\beta|_{\mathcal{H}}}$ with inverse
$\mathsf{A}\ni C\mapsto(\mathcal{G}_{\varphi(C)})^{max}$.

###### Proof.

Firstly we apply [15, Theorem 4.6] on the orthogonalization $\varepsilon$ of
$\beta$ to obtain a Galois correspondence between all the wide subgroupoids of
$\mathcal{G}$ and all the $E^{\varepsilon}$-separable, $\varepsilon$-strong
subalgebras $C$ of $E$, that is, $\mathcal{H}\leftrightarrow
E^{\varepsilon|_{\mathcal{H}}}$. Restricting this correspondence to
$\mathsf{W}^{max}$, we have a one-to-one correspondence
$\mathcal{H}^{max}\leftrightarrow E^{\varepsilon|_{\mathcal{H}^{max}}}$.

_Claim 1._ There is a one-to-one correspondence
$E^{\varepsilon|_{\mathcal{H}^{max}}}\leftrightarrow
A^{\beta|_{\mathcal{H}^{max}}}=\varphi^{-1}(E^{\varepsilon|_{\mathcal{H}^{max}}}\cap\varphi(A))$.

Indeed, define
$\theta:\\{E^{\varepsilon|_{\mathcal{H}^{max}}}:\mathcal{H}\in\mathsf{W}\\}\to\\{A^{\beta|_{\mathcal{H}^{\max}}}:\mathcal{H}\in\mathsf{W}\\}$
by
$\theta(E^{\varepsilon|_{\mathcal{H}^{max}}})=A^{\beta|_{\mathcal{H}^{\max}}}$.
We have that $\theta$ is injective, since if
$\theta(E^{\varepsilon|_{\mathcal{H}^{max}}})=\theta(E^{\varepsilon|_{\mathcal{K}^{max}}})$,
for $\mathcal{H}^{max},\mathcal{K}^{max}\in\mathsf{W}^{max}$, then
$A^{\beta|_{\mathcal{H}^{max}}}=A^{\beta|_{\mathcal{K}^{max}}}$, from where it
follows by Proposition 4.15 that $\mathcal{H}\sim\mathcal{K}$, that is,
$\mathcal{H}^{max}=\mathcal{K}^{max}$. Hence
$E^{\varepsilon|_{\mathcal{H}^{max}}}=E^{\varepsilon|_{\mathcal{K}^{max}}}$.
Furthermore, Proposition 4.8 implies that $\theta$ is surjective, proving the
claim.

_Claim 2._
$\mathsf{A}=\\{A^{\beta|_{\mathcal{H}^{max}}}:\mathcal{H}^{max}\in\mathsf{W}^{max}\\}$.

Indeed, for each $C\in\mathsf{A}$, there is a wide subgroupoid
$\mathcal{H}\in\mathsf{W}$ such that $C=A^{\beta|_{\mathcal{H}}}$. But
$\mathcal{H}\sim\mathcal{H}^{max}$, that is,
$A^{\beta|_{\mathcal{H}}}=A^{\beta|_{\mathcal{H}^{max}}}$, by Proposition
4.15. So
$\mathsf{A}\subseteq\\{A^{\beta|_{\mathcal{H}^{max}}}:\mathcal{H}^{max}\in\mathsf{W}^{max}\\}$.
The other inclusion is immediate.

So, by Claim 1 and Claim 2, there is a one-to-one correspondence between the
sets $\mathsf{W}^{max}$ and $\mathsf{A}$.

∎

######  Remark 4.17.

The same argument used in Proposition 4.15 cannot be applied in the partial
case to obtain an analogous of Theorem 4.16, since given $a\in A$,
$h,k\in\mathcal{H}_{1}\cup\mathcal{H}_{2}$ such that
$(h,k)\in\mathcal{G}_{2}$, we do not have, in general, that
$a1_{k^{-1}h^{-1}}\in A_{k^{-1}}\cap A_{k^{-1}h^{-1}}$. Therefore we cannot
guarantee that
$\alpha_{hk}(a1_{k^{-1}h^{-1}})=\alpha_{h}\circ\alpha_{k}(a1_{k^{-1}h^{-1}})$.
The existence of a more general Galois correspondence theorem for partial
groupoid actions will require other technique and is still an open question.

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