1. ## covering spaces

Let X be a locally simply connected space. (X1, p1) is a cover for X. (X2,p2) is a covering space for X1. Then prove that (X2,p1*p2) is a covering space for X.

I thought of this. But I am quite sure that I am wrong.

Let x1 be an element X1 such that p(x1)=x. Then there is an admissible open set U1 is X1, such that p2^(-1) (U1) is a disjoint union of open sets each of which is homeomorphically mapped onto U1 by the map p2. I thought p1(U1) will be the required admissible open set about x. But I am not sure.

2. ## Re: covering spaces

What definition are you using for a covering?

3. ## Re: covering spaces

(X1,p) is a covering space for X if for each x in X, there exists an admissible open set U in X containing x, such that p^(-1) (U) is a disjoint union of open sets, each of which is homeomorphically mapped onto U by p. p should be surjective too.

4. ## Re: covering spaces

If $X$ is locally simply connected then for each $x\in X$ there exists a neighborhood $U$ such that the map $\pi_1(U,x)\rightarrow \pi_1(X,x)$ induced by the inclusion is trivial. Use this fact to first show that $p_1^{-1}(U)$ is a disjoint union of open sets $C_\alpha$ such that $p_1 |_{C_\alpha}$ is a homeomorphism for all $\alpha$.

By the way, semi-locally simply connected is sufficient for this result.