I report here the problem: Let $X$ be a space which has a universal covering space. If $(X_1,p_1)$ is a covering space of $X$ and $(X_2,p_2)$ is a covering space of $X_1$, then $(X_2,p_1 \circ p_2)$ is a covering space of $X$.
We see easily that $p_1 \circ p_2$ are
continuous and surjective. Let $(\tilde X,r)$ be the universal covering space. By the theory we know that $\tilde X$ is also covering space of both $X_1$ and $X_2$, we call them $(\tilde X,r_1)$ and $(\tilde X,r_2)$
respectively. Thus, $(p_1 \circ p_2) \circ r_2=r$. Now, I have no idea how to procede. I know that, if $x \in X$ I have to find an elementary neighbourhood of $x$ and I suppose I can build it using elementary neighbourhoods of $x$ seen with the map $r$ and $p_1$ but I don't know how.
Best Answer
You have to use the universal property of the universal covering space.
Let $(X, x)$ be the base space, $p_1 \colon (X_1, x_1) \to (X, x)$ be the first covering map and $p_2 \colon (X_2, x_2) \to (X_1, x_1)$ be the second one. Also, let $u \colon (U, u) \to (X, x)$ be the universal covering map.
Then there exists a unique covering map $q_1 \colon (U, u) \to (X_1, x_1)$ such that $p_1 q_1 = u$, and there exists a unique covering map $q_2 \colon (U, u) \to (X_2, x_2)$ such that $p_2 q_2 = q_1$.
This means that $u = (p_1 p_2) q_2$, $u$ and $q_2$ are covering maps, hence $p_1 p_2$ is a covering map by the following:
Lemma (Topology - Munkres, 80.2) Assume that your spaces are path connected and locally path connected. Let $f = gh$ be continuous maps, and assume that $f$ is a covering map. Then $g$ is a covering map if and only if $h$ is a covering map.