# Equivalence between Category of Covers and $\pi_1(X)$ Sets

I have a question about an argument used in Szamuely's "Galois Groups and Fundamental Groups" in the excerpt below (or look up at page 38):

In order to show the category equivalence claimed in Thm 2.3.4 to verify the fully flatness of the functor $$Y \mapsto Fib_x(Y)$$ we have to show that for given two covers $$p:Y \to X, q: Z \to X$$ each map $$\varphi : Fib_x (Y ) → Fib_x (Z)$$ of $$π_1(X, x)$$-sets comes from a unique map $$Y → Z$$ of covers of $$X$$.

The author argues as follows: We take a $$y \in Fib_x(Y)$$ and identifies it with corresponding map $$\pi_y: \widetilde{X}_x \to Y$$ (remark: $$\widetilde{X}_x$$ is the universal cover) as claimed in 2.3.5:

For the construction of $$\pi_y$$ see here:

Indeed this works since by Thm 2.3.5 the functor $$Fib_x$$ is representable by $$\widetilde{X}_x$$.

Then it is shown in the proof that the map

$$\pi_{\phi(y)}: \widetilde{X}_x \to Z$$ factorizes as $$\widetilde{X}_x \xrightarrow{\pi_y} Y \xrightarrow{\psi_y} U_y \backslash \widetilde{X}_x \xrightarrow{p_y} Z$$

The PROBLEM is why is $$p_y$$ independent of $$y$$?

Indeed following the proof and using the naturality of diagram from 2.3.5 to show the claim that $$\phi$$ comes from a map of covers $$m: Y \to Z$$ it suffice to show that the map $$Hom_X(\widetilde{X}_x , Y) \to Hom_X(\widetilde{X}_x , Z)$$ is induced by a map $$m:Y \to Z$$ of covers.

The author showed that by construction $$\pi_{\phi(y)}$$ is the image of $$\pi_y$$ by composing with $$p_y$$.

The goal is to show that for every $$y' \in Fib_x(Y)$$ the morphism $$\pi_{\phi(y')}:\widetilde{X}_x \to Z$$ is the composition $$\widetilde{X}_x \xrightarrow{\pi_{y'}} Y \xrightarrow{p_y} Z$$ (therefore $$p_y$$ is independent of $$y$$)) but this isn't clear to me why this holds.

If I understand correctly, you see why there exists a map $$f:Y \to Z$$ of covers whose restriction to fibres is $$\phi$$, and your question is why that $$f$$ is unique.

Assume that $$g:Y \to Z$$ was also a map of covers whose restriction to fibres was $$\phi$$, then we can think of $$f,g$$ as lifts of $$\pi_y$$ to the covering $$Z \to X$$. Now as $$Y$$ is connected, if they agree at a point, $$f = g$$ (Proposition 2.2.2 I think). But they both agree on the fibres $$Fib_x(Y)$$, thus they are the same.

Edit-

We have a map $$f:Y \to Z$$ of covers, such that for a point $$y_0 \in Fib_x(Y)$$, $$f(y_0) = \phi(y_0) \in Fib_x(Z)$$. Now we must show that for every $$y \in Fib_x(Y)$$, $$f(y) = \phi(y)$$.

We use that $$Y,Z$$ can be chosen path-connected. Then $$\pi_1(X,x)$$ acts transitively on $$Fib_x(Y)$$. To see this, for any $$y \in Fib_x(Y)$$, consider a path $$\alpha$$ from $$y_0$$ to $$y$$. Then $$[\beta] = [p \circ \alpha] \in \pi_1(X,x)$$ takes $$y_0$$ to $$y$$. Then $$\phi(y) = \phi([\beta] \bullet y_0) = [\beta]\bullet f(y_0)$$ where $$\bullet$$ denotes the action of $$\pi_1(X,x)$$.

It remains to show $$f(y) = [\beta]\bullet f(y_0)$$. To see this note that as $$qf = p$$, we have that $$[\beta] = [q \circ \gamma]$$, for the path $$\gamma = f \circ \alpha$$ from $$f(y_0)$$ to $$f(y)$$. Thus $$[\beta] \bullet f(y_0) = f(y)$$.

• No, the problem is the existence. In the construction the author picks a $y \in Fib_x(Y)$ and showed that there exist a $f_y:Y \to Z$ with $f_y(y)= \phi(y)$. The problem is that I don't see why this $f=f_y$ is not dependent of $y$. Namely it's not clear that the whole $\phi$ arises from restricting of $f$ ($=f_y$). The author shows that $f_y$ has only the same image as $\phi$ only for $y$ not for the whole fiber $Fib_x(Y)$. – KarlPeter Apr 18 at 10:35
• Ah, I see the issue. This follows from the fact that we can choose $Y,Z$ to be path connected, then the action of the fundamental group is transitive. I have updated the answer to reflect this. – Tim The Enchanter Apr 18 at 11:04
• ah yes of couse the $\pi_1(X,x)$ action we have take into account. One remark: Why we can assume that $Y,Z$ are path connected? Or equivlently why we can assume that $Y,Z$ are connected? (indeed here all spaces are locally simply connected and therefore connected and path connected are equivalent) – KarlPeter Apr 18 at 16:49
• As you mentioned, all spaces are locally simply connected, hence locally path connected and hence covers $Y$ and $Z$ are homeomorphic to disjoint unions of their path components. Thus to define a map of covers it suffices to define it on each path component of $Y$, and continuity ensures that each path-component of $Y$ maps into a path-component of $Z$. Now the above argument shows that restricted to each path component of $Y$ the map is uniquely defined, from which we recover the uniqueness of the map $Y \to Z$. – Tim The Enchanter Apr 18 at 17:51
• For the sake of completeness, the above comment also (somewhat subtly) uses the fact that here, the restriction of a covering map to a connected component is still a covering map. The local homeomorphism condition follows from local simple connectedness of $X$, as we can restrict our evenly covered neighbourhoods to be connected. To see surjectivity use path lifting and (path-)connectivity of $X$. – Tim The Enchanter Apr 18 at 18:07