TITLE: Functor From Category of Covering Spaces to Category of Sets Equipped With An Action By The Fundamental Groupoid
QUESTION [5 upvotes]: I have some problems with the understanding of the connection between covering spaces and the fundamental groupoid.
Let $X$ be a topological space and let $\Pi_1(X)$ denote the fundamental groupoid of $X$. We can then construct the category $\Pi_1(X)-\mathbf{Sets}$, which has functors $F:\Pi_1(X)\to\mathbf{Sets}$ as its objects and natural transformations $\eta:F\Longrightarrow G$ as its morphisms ($F,G\in\Pi_1(X)-\mathbf{Sets}$). Let furthermore $\mathbf{Cov}(X)$ be the category of covering spaces over $X$. I would like to construct a functor
$$\Phi:\mathbf{Cov}(X)\to\Pi_1(X)-\mathbf{Sets}.$$
I think I have managed to construct the functors correctly, but not the morphisms.

Construction of Functors
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 simply 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
$$\alpha_*:p^{-1}(x_0)\to p^{-1}(x_1).$$
One can then check that this satisfies the functorial properties.
Construction of Morphisms
So, I think I know how to do this step without any reference to functors (it feels like it should be an easy translation to functors, but somehow I cannot figure it out). Let us begin to do this without any functors.
A Construction Without Functors
Let $X$ be a topological space. Without the category-theoretical language, I think this is how the argument goes.
Let $(Y_0,p_0)$ and $(Y_1,p_1)$ be covering spaces of a topological space $X$. A morphism between the two covering spaces is then a map $\varphi:Y_0\to Y_1$ such that $p_1\circ \varphi=p_0$.
We want to use the covering space morphism to construct a $\pi_1(X,x)-\text{Set}$ morphism; That is, maps between sets which preserves the action by the paths. What one can do is to simply combine the covering space morphism and the path-lifting diagram to show this.
Let $\alpha:I\to X$ be a path, by the path lifting property, we can lift $\alpha$ to a path $\widetilde{\alpha}:I\to Y$ such that ${p_0}_*(\widetilde{\alpha})=\alpha$ and $\widetilde{\alpha}(0)=y$, with $y\in p_0^{-1}(x)$. Thus $$\alpha_*(y)=\widetilde{\alpha}(1).$$
We get another path through $\varphi_*(\widetilde{\alpha})$ in $Y$, with initial point $\varphi_*(\widetilde{\alpha})(0)=\varphi(y)$ and terminal point $\varphi_*(\widetilde{\alpha})(1)=\varphi(\alpha_*)(y)$.
The following extended commutative diagram shows that $p_*[\varphi_*(\widetilde{\alpha})]=\alpha$ (with an extra map $p_0:Y_0\to X$, but I didn't knew how to include it, without making a total mess of the diagram).
$\require{AMScd}$
\begin{CD}
Y_0 @>\varphi>> Y_1\\
@AA\widetilde{\alpha}A @VVp_1V\\
I @>\alpha>> X
\end{CD}
Hence $\varphi_*(\widetilde{\alpha})$ is a lifting of $\alpha$. So we get that the terminal point of $\alpha_*(\varphi y)$ is the terminal point of $\varphi_*(\widetilde{\alpha})$, also. So
$$\alpha_*(\varphi(y))=\varphi(\alpha_*(y)).$$
Thus, we have a morphism between path-actions.
A Construction With Functors
Given a covering space homomorphism $\varphi:(Y_0,p_0)\to (Y_1,p_1)$, we want to map it to a morphism in $\Pi_1(X)-\mathbf{Sets}$, by applying $\Phi$. In a category of functors a morphism is a natural transformation.
A natural transformation is constructed as follows. Given two functors $\Gamma,\Delta:\mathcal{A}\to\mathcal{B}$ between two categories $\mathcal{A},\mathcal{B}$, we have morphism $\mu_X:\Gamma(X)\to\Delta(X)$, and a commutative diagram
$\require{AMScd}$
\begin{CD}
\Gamma(X) @>\mu_X>> \Delta(X)\\
@VV\Gamma(f)V @VV\Delta(f)V\\
\Gamma(Y) @>\mu_Y>> \Delta(Y),
\end{CD}
where $f:X\to Y$ is a morphism in $\mathcal{A}$.
In our case, we have two functor $F,G:\Pi_1(X)\to\mathbf{Sets}$, morphisms $\eta_x:F(x)\to G(x)$ and a commutative diagram
$\require{AMScd}$
\begin{CD}
F(x_0) @>\eta_{x_0}>> G(x_0)\\
@VVF(\alpha)V @VVG(\alpha)V\\
F(x_1) @>\eta_{x_1}>> G(x_1),
\end{CD}
with $\alpha:x_0\to x_1$ a path.

Question 1. What does the natural transformation diagram look like? Is it correct to rewrite it as
$\require{AMScd}$
\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}
or do we get something like
$\require{AMScd}$
\begin{CD}
p_0^{-1}(x_0) @>\eta_{x_0}>> p_0^{-1}(x_0)\\
@VVF(\alpha)V @VVG(\alpha)V\\
p_1^{-1}(x_1) @>\eta_{x_1}>> p_1^{-1}(x_1)?
\end{CD}
Question 2. How do we define $\eta_x:F(x)\to G(x)$? If $F(x)=p_0^{-1}(x)$ and $G(x)=p_1^{-1}(x)$, then we want to construct a map $\eta_x:p_0^{-1}(x)\to p_1^{-1}(x)$. I would like to construct this map through $\varphi$, I guess. But I'm not really sure how to do it.
Question 3. If $F(x_0)=G(x_0)=p_0^{-1}(x_0)$. Don't we have the following equality then $F(\alpha)=G(\alpha)$? Which I think maybe talks against on of my suggestions in Question 1.
I think all I should do, is to try to mimic what we did in the previous subsection. Or well, if not, more importantly just how to construct the natural transformation. But I don't really know what I am doing at all, there are just so much abstract things happening at the same time right now - which makes me confused.
I would be really happy if someone could help me out with the expression of the natural transformation.
Best wishes,
Joel

REPLY [2 votes]: For question 1, your first diagram is correct.
For question 2, the map $\eta_x: p_0^{-1}(x) \to p_1^{-1}(x)$ is just the restriction of the covering space homomorphism $\phi: Y_0 \to Y_1$ to $p_0^{-1}(x)$.  The image lands in $p_1^{-1}(x)$ since $\phi$ preserves the projections.
To show that the naturality square commutes for $\eta$ constructed this way, use the uniqueness of path-lifting for covering spaces.