Are all covering maps of spaces with the same fundamental group homeomorphisms?

Question

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

Answer 1

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 $x=y$.

Answer 2

No, because the homomorphism need not be surjective. For example, $z\mapsto z^2$ is a two sheeted covering space of a circle by itself, which is clearly not a homeomorphism. The image in $\pi_1(S^1)\simeq \mathbb Z$ of the induced homomorphism of fundamental groups is $2\mathbb Z$.

Interestingly, the result does hold for finite fundamental groups, which generalizes the simply connected case.

Answer 3

If the covering map induces an isomorphism $p_*\colon \pi_1(X) \stackrel{\cong}{\to} \pi_1(Y)$ of fundamental groups then the answer is yes. In general a covering map induces an injection on fundamental groups, and the cardinality of the quotient set $\pi_1(Y)/im(\pi_1(X))$ gives the number of sheets.