Composition of covering map map if X has universal cover

Question

Let X be topological space which has a universal cover. $(Y,p)$ is covering of X and $(Z,q)$ is covering space of Y. Then I know that $(Z,p\circ q) $ is covering of X.

I know that in general composition of covering map need not covering. But Here along with existence of universal cover assumption, we are having that property. I do not know how to start this problem.

Any Help Hint will be appreciated.

Answer 1

There is a Galois correspondence for covering spaces relating subgroups of the fundamental group under inclusion and path connected covering spaces of the space. Any universal cover of X is in particular also a universal cover of Y for Y path connected. Try using that?

Answer 2

We need a lemma.

Lemma If we have maps $q : X \to Y$, $r : Y \to Z$, and $p = r\circ q : X \to Z$, then $$ p \mbox{ and } q \mbox{ are coverings} \Rightarrow r \mbox{ is covering}$$ $$ p \mbox{ and } r \mbox{ are coverings} \Rightarrow q \mbox{ is covering}$$

Now suppose that we have coverings $q : X \to Y$ and $r : Y \to Z$ with $p = r \circ q$ and $u : \tilde{Z} \to Z$ be an universal covering.

Since $\tilde{Z}$ is universal cover, it is simply connected so we can lift $u$ with respect to $r$ and obtain $\tilde{r} : \tilde{Z} \to Y$. Again, we can lift $\tilde{r}$ with respect to $q$ and thus we get $\tilde{q} : \tilde{Z} \to X$.

Now, applying the lemma to $u$, $r$ and $\tilde{r}$, we get a covering $\tilde{r} : \tilde{Z} \to Y$. Again, applying the lemma to $\tilde{r}$, $q$ and $\tilde{q}$, $\tilde{q}$ is verified to be a covering.

Finally applying the lemma to $\tilde{q}$, $p$ and $u$ concludes that $p$ is a covering.