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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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    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.2014-12-05

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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.

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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.