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Let $X$ be a path-connected space and $H$ a subgroup of $\pi_1(x,X)$ and let $x\in X$ be a point. Is the following true?

There exists a covering map $\rho:C\rightarrow X$ and $c\in \rho^{-1}(x)$ such that $\pi_1(C,c)\cong H$.

The thing that is clear is that $\pi_1(C,c)$ is congruent to a subgroup of $\pi_1(X,x)$. Since $\rho^*$ is injective.

Asinomás
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  • If $X$ is simply connected, then $\pi_1(X,x)$ is trivial. Do you mean something else? – Ashwin Iyengar Dec 06 '16 at 21:24
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    Presumably the OP means "connected." The answer is yes. In fact as $c$ runs over all preimages of $x$, the fundamental group at $c$ runs over all conjugates of $H$. – Qiaochu Yuan Dec 06 '16 at 21:27
  • @QiaochuYuan What I meant to ask is if there exists a covering space for each $H$. – Asinomás Dec 06 '16 at 21:35
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    Yes, that's part of the classification of covering spaces (given some mild hypotheses on $X$; my preferred hypothesis is locally contractible). – Qiaochu Yuan Dec 06 '16 at 21:40

1 Answers1

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According to Proposition 1.3.6 of Hatcher's Algebraic Topology,

Suppose $X$ is path-connected, locally path-connected and semilocally simply-connected. Then for every subgroup $H \subseteq \pi_1(X,x_0)$, there is a covering space $p : X_H \rightarrow X$ such that $p_* \pi_1(X_H, \tilde x_0) = H$ for a suitably chosen basepoint $\tilde x_0 \in X_H.$

At least some sort of condition is necessary, as the well-known Hawaiian earring space (for example) has no universal cover and so this result fails. Of course the condition of being simply-connected is much too strong to be interesting here.

  • thank you very much. I recall that was the same condition we used in class to build the "universal covering". – Asinomás Dec 06 '16 at 21:29
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    The Hawaiian earring space is path-connected so it is a counterexample to the new question –  Dec 06 '16 at 21:32