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That is, let $f:X \rightarrow Y$ be a map of spaces such that $f_*: H_*(X) \rightarrow H_*(Y)$ induces an isomorphism on homology. We get an induced map $\tilde{f}: \Omega X \rightarrow \Omega Y$, where $\Omega X$ is the loop space of $x$. Does $\tilde{f}$ also induce an isomorphism on homology?

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This is at least true if $X, Y$ are simply connected (or if the fundamental group acts trivially on the fibers); see Proposition 1.12 in Hatcher's book on spectral sequences. –  Akhil Mathew Mar 23 '11 at 2:56
    
You mean in the setting where $X$ and $Y$ are fibers of fibrations, and the fundamental group of the base acts trivially on the homology of the fibers? I want to use this to prove the Spectral Sequence theorem :) –  Tony Mar 23 '11 at 3:53
    
@Tony: Whoops, yes, I meant that $\pi_1$ acts trivially on the homologies, not on the fibers themselves. This follows directly from the spectral sequence comparison theorem if I am not mistaken. (I'm not sure why you wouldn't prove the comparison theorem purely algebraically though.) –  Akhil Mathew Mar 23 '11 at 4:08
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I'm not familiar enough with this comparison theorem to know if you guys are somehow implicitly referring to this, but if $X$ and $Y$ are simply-connected (or if $f_\#:\pi_1(X)\rightarrow \pi_1(Y)$ is an isomorphism) then $f$ is already a homotopy equivalence by the relative Hurewicz theorem and Whitehead's theorem. –  Aaron Mazel-Gee Mar 24 '11 at 8:26
    
@Aaron: Wait, doesn't this only work if $X,Y$ are CW complexes? (Or is it true that the loop space functor preserves weak equivalences?) –  Akhil Mathew Apr 10 '11 at 19:55

1 Answer 1

It's not true in general.

Say, take a ring $R$ and consider the map $BGL(R)\to BGL(R)^+$. It always induces an isomorphism on homology, but $$H_1(\Omega BGL(R))=H_1(GL(R))=0$$ ($GL(R)$ has discrete topology) and $$H_1(\Omega BGL(R)^+)=\pi_1(\Omega BGL(R)^+)=\pi_2(BGL(R)^+)=K_2(R)$$ is often non-trivial.

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