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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?

Grigory M
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Tony
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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 02:56
  • 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 03:53
  • @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 04:08
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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 08:26
  • @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
  • @Akhil: Yes I believe it is true, just from the fact that $\pi_n(\Omega f) = \pi_{n+1}(f)$ (and using Whitehead's theorem again). – Aaron Mazel-Gee Apr 13 '11 at 07:20
  • @Aaron: Whoops! You're right, of course... – Akhil Mathew Apr 13 '11 at 16:28
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    If the spaces are simply connected and CW spaces, then $f$ is actually a homotopical equivalence. If the spaces are not simply connected but $f$ induces an isomorphism on $\pi_1$ and an isomorphism on homology with local coefficients, the same holds. – Mariano Suárez-Álvarez Dec 05 '14 at 07:59

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Here is a more explicit family of counterexamples. Let $X$ be a homology sphere of dimension at least $3$ with a point removed. Then $X$ is an acyclic space with the same fundamental group as the original homology sphere. Hence the constant map $f : X \to \text{pt}$ induces an isomorphism on homology. In order for the induced map

$$\Omega f : \Omega X \to \text{pt}$$

on loop spaces (with some choice of basepoint) to induce an isomorphism on homology, it must induce an isomorphism

$$H_0(\Omega f) : H_0(\Omega X) \cong \mathbb{Z}[\pi_1(X)] \to H_0(\text{pt}) \cong \mathbb{Z}$$

which is equivalent to $\pi_1(X)$ being trivial. But of course a homology sphere need not have trivial fundamental group, so for example we can take $X$ to be Poincare dodecahedral space minus a point.

Qiaochu Yuan
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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.

Grigory M
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