A detail about reconstructing covering space from the action $\pi_1(X,x_0)\to S_{p^{-1}(x_0)}$ in Hatcher's book

Question

I'm struggling understanding a small sentece from Hatcher's Algebraic topology book (available online for free).

In page 70 Hatcher wants to reconstruct the covering $p:\tilde X\to X$ from the associated action $\pi_1(X,x_0)\to F=p^{-1}(x_0)$ assuming that $X$ is path-connected, locally path-connected, and semilocally path connected. So he takes the universal covering space $\tilde X_0$ constructed some pages before that consists of the homotopy classes of paths starting at $x_0$. He then defines a map $h:\tilde X_0\times F\to \tilde X$ as $h([\gamma],\tilde x_0)=\tilde \gamma_{\tilde x_0}(1)$ where $\tilde \gamma_{\tilde x_0}$ is the unique lift of $\gamma$ starting at $\tilde x_0$. Then he proves that $h$ is continuous, and in fact, a local homeomorphism. Details about the covering space $\tilde X_0$ are given in page 64.

After proving this he says: It is obvious that $h$ is surjective since $X$ is path-connected. This is the sentence that I don't understand. So for $\tilde a\in \tilde X$ one has to prove that there is $\gamma$ path in $X$ starting at $x_0$ and $\tilde x_0\in F$ such that $h([\gamma],\tilde x_0)=\tilde \gamma_{\tilde x_0}(1)=\tilde a$, this is the same as saying that given $\tilde a$ there is $\tilde x_0\in F$ and a path $\tilde \gamma$ from $\tilde x_0$ to $\tilde a$ because then we can simply take $\gamma=p\tilde \gamma$. I don't get how $X$ being path-connected comes into play.

Answer 1

Hatcher assumes covering spaces are connected, and since $X$ is locally path connected any covering space has this property. This means every covering space considered is connected and locally path connected, so it is path connected. Thus $h$ is easily seen to be onto.

Answer 2

After some months, I've finally arrived to a satisfactory answer to this question (I believe the answer given by Pedro Tamaroff is wrong...).

First we are gonna need this lemma which is proven here surjective covering space

If $p:\tilde X\to X$ is a covering space with $\tilde X$ nonempty and $X$ connected then $p$ is surjective.

Note that Hatcher says explicitly that covering spaces need not be surjective but I think that's a misleading statement because every single space $X$ that he considers is connected, $\tilde X$ may not be connected (although in most cases it is).

Now we are gonna prove the following lemma:

If $p:\tilde X\to X$ is a covering space with $\tilde X$ nonempty, $X$ locally path connected and $\tilde A$ is a path component of $\tilde X$ then $p|\tilde A:\tilde A\to X$ is also a covering space.

Proof : Let $U$ be a nonempty open set of $X$ and $p^{-1}(U)=\coprod_{i\in I} \tilde U_i$ (disjoint union of sets) with $p|\tilde U_i:\tilde U_i\to U$ a homeomorphism for each $i\in I$. Note that nothing guarrantees that $U$ is path connected. Using the fact that $X$ is locally path connected we find a path connected open set $V\subseteq U$ and then $p^{-1}(V)=\coprod_{i\in I}\tilde U_i\cap p^{-1}(V)=\coprod_{i\in I}(p|\tilde U_i)^{-1}(V)$ and with $p|(p|\tilde U_i)^{-1}(V):(p|\tilde U_i)^{-1}(V)\to V$ a homeomorphism for each $i\in I$. Now since $\tilde A$ is a path component and each $(p|\tilde U_i)^{-1}(V)$ is path connected, we have that for each $i\in I$ either $(p|\tilde U_i)^{-1}(V)\subseteq \tilde A$ or $(p|\tilde U_i)^{-1}(V)\cap \tilde A=\emptyset$. Consider the set $I_0=\{i\in I:(p|\tilde U_i)^{-1}(V)\subseteq \tilde A\}$. We then have $(p|\tilde A)^{-1}(V)=\coprod_{i\in I_0}(p|\tilde U_i)^{-1}(V)$ and for each $i\in I_0$, $(p|\tilde A)|(p|\tilde U_i)^{-1}(V):(p|\tilde U_i)^{-1}(V)\to V$ is a homeomorphism. Finally $p|\tilde A$ is a covering space.

Now let's return to our original problem:

Let $\tilde x\in \tilde X$ and $\tilde A$ be the path component containing $\tilde x$. Then our last lemmas tell us that $p|\tilde A:\tilde A\to X$ is a surjective covering space. This means that there is $\tilde x_0\in \tilde A$ such that $p(\tilde x_0)=x_0$, i.e. such that $\tilde x_0\in p^{-1}(x_0)$. Now let $\tilde \gamma$ be a path in $\tilde A$ from $\tilde x_0$ to $\tilde x$. We then have $h([p\tilde \gamma],\tilde x_0)=\widetilde{(p\tilde \gamma)}_{\tilde x_0}(1)=\tilde \gamma(1)=\tilde x$, so $h$ is indeed surjective.

Answer 3

I don't think the steps provided by Zero above are necessary. I'm almost seven years too late, but for anyone who's still interested, I think the following works.

Given $\tilde a \in \tilde X$, let $a = p(\tilde a)$. Then there is a path $\gamma$ in $X$ from $a$ to $x_0$. Now, this has a unique lift $\tilde \gamma$ in $\tilde X$ starting at $\tilde a$. Let $\tilde x = \tilde \gamma (1)$. Then $\bar{ \tilde \gamma}$ is the unique lift of $\bar \gamma$ starting at $\tilde x$. Hence $h([\bar\gamma],\tilde x)=\bar{\tilde\gamma}(1)=\tilde\gamma(0)=\tilde a$.