Does a map of spaces inducing an isomorphism on homology induce an isomorphism between the homologies of the  loop spaces? 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? 
 A: 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.
A: 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. 
