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,



1 Answer 1


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.


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