My topology book stated the following:
If $X$ is a simply connected topological space, then every covering map $p\colon E \rightarrow X$ is trivial (in other words, it's a homeomorphism)
Is this true for spaces with more general fundamental groups? That is, let $p\colon E \rightarrow X$ be a covering map such that $\pi_1(E)$ and $\pi_1(X)$ are abstractly-isomorphic groups. Does this imply that $p$ is a homeomorphism?
Edit: made statement clearer
Best Answer
It depends on what you mean by $\pi_1(E) = \pi_1(X)$ : this $=$ can be ambiguous (the spaces for which there is a literal equal sign there are quite rare in algebraic topology).
If you simply mean an abstract isomorphism (i.e. "there exists an isomorphism $\pi_1(E)\to \pi_1(X)$") then the answer is no, as has been pointed out earlier : there are many non-homeomorphic coverings $S^1\to S^1$, in fact coverings in general correspond exactly to nontrivial subgroups of $\mathbb Z$, so they correspond to nonzero natural numbers.
However, if by this $=$ you mean "the map induced by $p$ is an isomorphism $\pi_1(E)\to \pi_1(X)$", which is a much stronger statement, then the answer is yes (assuming of course $E,X$ are path-connected). Indeed, path-connectedness of $X$ implies the surjectivity of $p$, $p$ is a local homeomorphism, so it suffices to show that it's injective.
But if $p(x) = p(y)$ , then taking a path $\gamma:x\to y$ we get that $p_*\gamma$ is a loop based at $p(x)$, so it lifts up to pointed homotopy (which lifts as well) to a loop $\delta$ based at $x$ (because $p_*$ is surjective), which is thus pointed homotopic to $\gamma$, from which it follows that $\gamma$ is a loop, hence $x=y$.