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Let $X$ be a compact, connected, locally path-connected, and semilocally simply connected topological group with $\pi_1(X) \cong \mathbb{Z}$.

Let $u\colon \widetilde{X}\to X$ be its topological universal cover, which also has the structure of a topological group.

Let $Y$ be a locally path-connected topological space.

Suppose there exists a surjective, local homeomorphism $f\colon \widetilde{X}\to Y$ and a (surjective) morphism $a\colon Y \to X$ of topological spaces with connected fibers such that $u = a\circ f$.

Thus, one could say that the universal cover $u$ ''factors'' through $f$ and $a$.

Question. With these above conditions, can one deduce that $Y$ is a topological group?

Thanks in advance for the help!

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  • $\begingroup$ What do you mean by "morphism $Y\to X$"? morphism of what? $\endgroup$
    – YCor
    Commented Oct 30, 2019 at 21:50
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    $\begingroup$ @KevinCasto $X$ is a group by assumption (I agree the notation is misleading). $\endgroup$
    – YCor
    Commented Oct 30, 2019 at 21:53
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    $\begingroup$ It easily follows from the assumptions that $a$ is bijective. $\endgroup$
    – YCor
    Commented Oct 30, 2019 at 22:09
  • $\begingroup$ @YCor Sorry about that! I meant a morphism of topological spaces $\endgroup$
    – Jackson Morrow
    Commented Oct 30, 2019 at 22:37
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    $\begingroup$ @user44191 what you prove is what I just said, namely that $a$ is bijective. A little more step is needed to obtain that $a$ is a homeomorphism. $\endgroup$
    – YCor
    Commented Oct 30, 2019 at 22:45

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