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Let $X$ be a connected and locally path-connected space. In Spanier's book, the universal cover of $X$ is defined to be a connected covering space $p\colon\tilde{X}\to X$ such that for any covering $q\colon Y\to X$ there is a fiber-preserving map $f\colon \tilde{X}\to Y$.

The author stated without proof that any two universal covers are equivalent meaning that there is a fiber-preserving homeomorphism between two universal covering spaces.

Of course, for simply-connected universal covers, this is not a problem. Besides, if the base space is also semi-locally simply-connected, then every universal covering is simply-connected.

My Question is: Is Spanier's statement true, or is it just a mistake in the book?

I've asked a somewhat related question here.

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    $\begingroup$ Does the given definition of universal covering require that $f : \tilde{X} \to Y$ is unique as a fiber-preserving map? $\endgroup$
    – Anon
    Aug 5, 2016 at 19:55
  • $\begingroup$ @QiaochuYuan I don't get it. I didn't mean universal covers are unique up to a unique homeomorphism. Only one fiber-preserving homeomorphism suffices. In nice cases, fundamental groups of universal covers can be mapped to conjugate subgroups, so there is a homeomorphism. Did I miss something really important here? Could you elaborate a little bit? Thanks. $\endgroup$
    – YYF
    Aug 6, 2016 at 23:10
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    $\begingroup$ When you say "the initial object in the category of connected covering spaces" you are implying that universal covers are unique up to unique homeomorphism (because initial objects are unique up to unique isomorphism). $\endgroup$
    – Qiaochu Yuan
    Aug 6, 2016 at 23:18
  • $\begingroup$ @QiaochuYuan oh, you are definitely right. I just realized you were referring to the comment! But that's not the problem I am asking though. $\endgroup$
    – YYF
    Aug 6, 2016 at 23:53

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An important fact I missed is that every universal cover, in the sense of Spanier, is regular. So if $p\colon\tilde{X}\to X$ is a universal cover, then $p_{*}\pi(\tilde{X},\tilde{x}_0)=p_{*}\pi(\tilde{X},\tilde{x}_0')$ provided that $\tilde{x}_0,\tilde{x}_0'$ are in the same fiber of $p$. This immediately shows that there is always a fiber-preserving homeomorphism between two universal covers.

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