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 $[\beta] \bullet f(y_0) = f(y)$.