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Classifying spaces are obviously unique up to homotopy type. I am wondering, whether under stronger conditions, one can also say that they are unique up to homeomorphism. In particular, suppose $\Gamma$ is a group and there exist a model $X$ for $B\Gamma$, which is closed (compact without boundary). Suppose $Y$ is also a model for $B\Gamma$ and $Y$ is also closed. In my baby examples it seems reasonable that $X\cong Y$. Is this always true?

Furthermore, if $X$ and $Y$ are models for $B\Gamma$ and $X$ is a closed $n$-dimensional manifold and $Y$ is also an $n$-dimensional manifold. Is it true that $Y$ is closed as well?

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    Re: the first sentence, that depends on how you define classifying spaces. If you define them in terms of the functor they represent then of course they are unique up to homotopy type. But if you define them as quotients of weakly contractible spaces by free actions then this seems less clear to me (but then again I have not really studied this).2012-06-21
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    Thank you. I only want to consider discrete groups.2012-06-21
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    But even then I do not think it is obvious that Eilenberg-MacLane spaces are unique up to homotopy type (again unless one defines them via the functor they represent). This was first proven by Hopf, I think.2012-06-21
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    Hm, I see. Do you have a reference for this?2012-06-21
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    For CW-complexes this is Theorem 1B.8 in Hatcher.2012-06-21
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    Anyway, a positive result: if I'm not horribly mistaken this follows for compact hyperbolic $n$-manifolds, $n \ge 3$ by Mostow rigidity (http://en.wikipedia.org/wiki/Mostow_rigidity_theorem).2012-06-21

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