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
Best Answer
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.