# Equivalence Between Category of Covering Spaces and Category of Sets with an Action By The Fundamental Groupoid

Let $$\Pi_1(X)$$ be the fundamental groupoid of a locally path-connected topological space $$X$$ and define $$\Pi_1(X)-\mathbf{Sets}$$ to be the category of sets equipped with an action by the fundamental groupoid. Let furthermore $$\mathbf{Cov}(X)$$ be the category of covering spaces over $$X$$. I want to show that we have the following equivalence of categories $$\mathbf{Cov}(X)\simeq \Pi_1(X)-\mathbf{Sets}.$$

So far, I have constructed two functors $$\Phi:\mathbf{Cov}(X)\to\Pi_1(X)-\mathbf{Sets}$$ and (I think and hope this is correct) $$\Psi:\Pi_1(X)-\mathbf{Sets}\to\mathbf{Cov}(X).$$ I now want to show that they give me the equivalence between the categories. However, I find it quite difficult to show that we have natural isomorphisms $$\Phi\Psi\Longrightarrow \mathbf{1}_{\Pi_1(X)-\mathbf{Sets}}$$ and $$\Psi\Phi\Longrightarrow \mathbf{1}_{\mathbf{Cov}_{/X}}$$.

In the first section, I will briefly go through the construction of the functors $$\Phi$$ and $$\Psi$$. After that, I will explain my concerns about the equivalence.

This post will only concern one of the directions about the equivalence. Just to make the post more concentrated and not "too much". I will (maybe) post another question later, depending on if I don't manage to figure out the other direction by myself.

## Construction of Functors

We begin with an explanation of the functor $$\Phi$$, then with $$\Psi$$.

### The Construction of the Functor $$\Phi$$

In the category $$\Pi_1(X)-\mathbf{Sets}$$, an object is a functor $$F:\Pi_1(X)\to\mathbf{Sets}$$. So, we need to construct such a functor using a covering space.

Objects. Let $$p:Y\to X$$ be a covering space, we define a functor $$F:\Pi_1(X)\to\mathbf{Sets}$$ as follows.

Objects of the fundamental groupoid are points of $$X$$, so we define $$F(x)=p^{-1}(x),$$ where $$x\in X$$.

A morphism in $$\Pi_1(X)$$, between two points $$x_0$$ and $$x_1$$, is a path $$\alpha:x_0\to x_1$$ modulo homotopy equivalence. Given such a path, we construct a set-theoretical map $$\alpha_*:p^{-1}(x_0)\to p^{-1}(x_1)$$ (from a point $$y_0\in p^{-1}(x_0)$$ to a point $$y_1\in p^{-1}(x_1)$$) as follows:

We have a covering space $$p:Y\to X$$, which makes it possible to apply the path-lifting property. That is, we lift $$\alpha:x_0\to x_1$$ to a path $$\widetilde{\alpha}:y_0\to y_1$$, such that $$\beta(0)=y_0$$ and $$p_*\widetilde{\alpha}=\alpha$$. Letting $$\widetilde{\alpha}(1)=y_1$$, one can check that $$p_*\widetilde{\alpha}(1)=p(y_1)=\alpha(1)=x_1$$, which gives a well-defined map $$\alpha_*:p^{-1}(x_0)\to p^{-1}(x_1),$$ which is the morphism in $$\mathbf{Sets}$$. So $$F$$ is, indeed, a functor.

Morphisms. A morphism in $$\Pi_1(X)-\mathbf{Sets}$$ between two functors $$F,G:\Pi_1(X)\to\mathbf{Sets}$$ is a natural transformation $$\eta:F\Longrightarrow G$$.

One can construct a natural transformation $$\eta$$ using a covering space homomorphism $$\varphi:(Y_0,p_0)\to (Y_1,p_1)$$, by letting $$\eta_x:F(x)\to G(x)$$ be defined by $$\eta_{x}(y)=\varphi_{|p_0^{-1}(x)}(y)$$. This gives us a commutative diagram $$\require{AMScd}$$ $$\begin{CD} p_0^{-1}(x_0) @>\eta_{x_0}>> p_1^{-1}(x_0)\\ @VVF(\alpha)V @VVG(\alpha)V\\ p_0^{-1}(x_1) @>\eta_{x_1}>> p_1^{-1}(x_1), \end{CD}$$ with $$\alpha:x_0\to x_1$$ a homotopy path.

Hence, we know where to map objects through the functor $$\Phi$$.

### The Construction of the Functor $$\Psi$$

Objects. Factor $$X$$ as follows $$X=\bigcup_i V_i,$$ where $$V_i$$ has the subspace topology and such that $$\Pi_1(V_i)=\pi_1(V_i,x_i)=0$$ (with a basepoint $$x_i$$), which can be done since $$X$$ is locally path-connected.

An object in $$\Pi_1(X)-\mathbf{Sets}$$ is a functor $$G$$, say. The functor $$G$$ is constant at each $$V_i$$. That is, for $$v_i\in V_i$$, we have $$G(v_i)=S_i,$$ where $$S_i$$ is some set, equipped with the discrete topology.

To build a covering space of $$X$$, let us define an equivalence relation $$\sim_a$$ on points in $$\bigsqcup_i V_i$$, we will do it through the collection of inclusions $$h_i:V_i\hookrightarrow X.$$ We say that two points $$x\in V_i$$ and $$y\in V_j$$ are equivalent under $$\sim_a$$ if and only if $$h_i(x)=h_j(y).$$ We then paste it together and express $$X$$ as the quotient space $$X=\bigsqcup_i V_i / \sim_a$$ Furthermore, let $$V_j\times S_j$$ be the topological space, where $$V_j$$ has the subspace topology, $$S_j$$ the discrete topology and their product the product topology. We define another equivalence relation $$\sim_b$$ on the space $$\bigsqcup_i \Big [V_i\times S_i \Big ]$$ as follows $$V_i\times S_i\ni(x,s)\sim_b (y,t)\in V_j\times S_j \iff \\ h_i(x)=h_j(y)\\ s\in G(v_i)\cong G(x)=G(y)\cong G(v_j)\ni t$$ This gives us the space $$\widetilde{X}$$, which covers $$X$$ $$\widetilde{X}=\bigsqcup_i \Big [V_i\times S_i \Big ]\Big /\sim_b.$$

Morphisms. A morphism in $$\Pi_1(X)-\mathbf{Sets}$$, is a natural transformations $$\mu:G_1\Longrightarrow G_2$$ between $$G_1,G_2:\Pi_1(X)-\mathbf{Sets}$$. That is, given a point $$x\in\Pi_1(X)$$, we assign a morphism $$\mu_x:G_1(x)\to G_2(x)$$. And given a path $$\gamma:x\to y$$ in $$X$$, such that we have the following commutativity $$\mu_y\circ G_1(\gamma)=G_2(\gamma)\circ\mu_x$$.

We use that natural transformation to construct a covering space homomorphism.

Let $$f: \bigsqcup_i\Big [ V_i\times G_1(V_i)\Big ] \to \bigsqcup_i\Big [ V_i\times G_2(V_i)\Big ]$$ be defined by $$f(v,s)=(v,\mu_{x}(s))$$, where $$x\in V_j$$ for some $$j$$. This gives us the commutativity of the diagram follows, and hence, we have a covering homomorphism $$\require{AMScd} \require{cancel} \def\diaguparrow#1{\smash{\raise.6em\rlap{\ \ \scriptstyle #1} \lower.6em{\cancelto{}{\Space{2em}{1.7em}{0px}}}}} \begin{CD} && \bigsqcup_i\Big [ V_i\times G_2(V_i)\Big ]\\ & \diaguparrow{f} @VVp_2V \\ \bigsqcup_i\Big [ V_i\times G_1(V_i)\Big ] @>>p_1> \bigsqcup_i\Big [ V_i\Big ] \end{CD}$$ which induces a covering space homomorphism from $$\widetilde{X}$$ onto $$X$$.

## Equivalence of Categories

I now want to show that the categories are equivalent $$\mathbf{Cov}(X)\simeq \Pi_1(X)-\mathbf{Sets}.$$ This means that we have to construct natural isomorphisms of functors $$\Psi\circ \Phi\cong \operatorname{Id}_{\mathbf{Cov}(X)}\qquad \text{and}\qquad \Phi\circ\Psi\cong \operatorname{Id}_{\Pi_1(X)-\mathbf{Sets}}$$

### Constructing a Natural Isomorphism $$\eta:\Psi\circ\Phi\Longrightarrow \operatorname{Id}_{\mathbf{Cov}(X)}$$

What we need is to construct an isomorphism $$\eta_{(Y,p)}:\Psi\circ\Phi(Y,p)\to \mathbf{1}_{\mathbf{Cov}(X)}(Y,p)$$, for every covering space $$(Y,p)$$ of $$X$$, such that the following diagram commutes $$\require{AMScd}$$ $$\begin{CD} \Psi\circ\Phi(Y_0,p_0) @>\eta_{(Y_0,p_0)}>> \operatorname{Id}_{\mathbf{Cov}(X)}(Y_0,p_0)\\ @VV\Psi\circ\Phi(\varphi)V @VV\operatorname{Id}_{\mathbf{Cov}(X)}(\varphi)V\\ \Psi\circ\Phi(Y_1,p_1) @>\eta_{(Y_1,p_1)}>> \operatorname{Id}_{\mathbf{Cov}(X)}(Y_1,p_1) \end{CD}$$

The above diagram reduces to

$$\require{AMScd}$$ $$\begin{CD} \Psi\circ\Phi(Y_0,p_0) @>\eta_{(Y_0,p_0)}>> (Y_0,p_0)\\ @VV\Psi\circ\Phi(\varphi)V @VV\varphi V\\ \Psi\circ\Phi(Y_1,p_1) @>\eta_{(Y_1,p_1)}>> (Y_1,p_1) \end{CD}$$

Question 1. Maybe this is obvious, but how should I define $$\eta_{(Y,p)}$$ to make it both an isomorphism and such that the diagram commutes?

Question 2. It is difficult for me to understand what $$\Psi\circ\Phi(Y,p)$$ and $$\Psi\circ\Phi(\varphi)$$ looks like. Applying $$\Phi$$ to $$(Y,p)$$ gives a functor $$F:\Pi_1(X)-\mathbf{Sets}$$, such that $$F(x)=p^{-1}(x)$$ and which maps paths according to rule we gave earlier.

Given the functor $$F$$, what does $$\Psi(F)$$ look like? Do I need to understand it in order to prove the commutativity of the diagram and/or construct $$\eta_{(Y_0,p_0)}$$?

Likewise, given a covering space homomorphism $$\varphi:(Y_0,p_0)\to (Y_1,p_1)$$, what does $$\Psi\circ\Phi(\varphi)$$ give me? First, I map $$\varphi$$ to a natural equivalence $$\mu:F_0\Longrightarrow F_1$$ such that we have morphisms $$\mu_x:F_0(x)\to F_1(x)$$ and a diagram which commutes $$\require{AMScd}$$ $$\begin{CD} p_0^{-1}(x_0) @>\mu_{x_0}>> p_1^{-1}(x_0)\\ @VVF_0(\alpha)V @VVF_1(\alpha)V\\ p_0^{-1}(x_1) @>\mu_{x_1}>> p_1^{-1}(x_1), \end{CD}$$ where does $$\Psi$$ map this natural transformation?

This is probably easy. But there are a lot of abstract things going on at the same time, which makes me lose myself.

Best wishes,

Joel

## 2 Answers

In simple terms, both a covering map $$p$$ (by looking at preimages) and a $$\Pi_1(X)$$-set are assigning discrete sets to the points of $$X$$, in a path compatible way.

For a given cover $$(Y,p)$$, we first construct the functor that assigns each point of $$x\in X$$ the fibre of $$Y$$ above it $$\Phi(Y,p)$$, then we construct a covering $$(Z,q):=\Psi\circ\Phi\,(Y,p)$$ such that its fibres are $$q^{-1}(x)\cong\Phi(Y,p)\,(x)=p^{-1}(x)$$.

Note that, by construction of $$\Psi$$, for every $$x$$ an index $$i$$ can be chosen such that $$x\in V_i$$ and we should also fix a path $$w_x^i:v_i\leadsto x$$ in $$V_i$$, so that $$s\in G(v_i)\cong G(x)=G(y)\cong G(v_j)\ni t$$ should really mean $$x=y$$ and $$G(w_x^i)\,(s)=G(w_x^j)\,(t)$$, so that when we will lift the path $$w_x^i$$, point $$s$$ over $$v_i$$ goes to the same place where point $$t$$ over $$v_j$$.
(Since $$V_i$$ was chosen to be simply connected, this construction is independent of the actual choices $$w_x^i$$.)

Now, let $$(x,s)\in Z$$ be a point, then $$s$$ actually implicitly refers to an index $$i$$ such that $$x\in V_i$$ and $$s\in \Phi(Y,p)\,(v_i)$$, but $$\Phi(Y,p)\,(v_i)=p^{-1}(v_i)$$, so you receive the correspondent of $$(x,s)$$ by travelling $$s\in p^{-1}(v_i)$$ along $$w_x^i$$.
This correspondence is indeed bijective, for its inverse, travel a point $$y\in p^{-1}(x)$$ along $$(w_x^i)^{-1}$$ for some $$i$$ to obtain an element of $$p^{-1}(v_i)$$.

There are still lot of details to check, e.g. that these maps are well defined, continuous and natural.
I hope it helps. Good luck!

To prove the equivalence of the categories, it suffices to take one of the above functors (e.g. $$\Phi:\text{Cov}(X)→\Pi1(X)−Sets$$) and show that it is:

In other words, you don't need to explicitly construct the inverse functor and the natural isomorphisms. Hope this helps shorten the proof.