TITLE: When homology isomorphism implies homotopy isomorphism
QUESTION [8 upvotes]: Let's suppose that 
$f:X\rightarrow X$ is a continuous map such that 

$H_{\ast}(f): H_{\ast}(X)\rightarrow H_{\ast}(X)$ is a homology isomorphism (with integral coefficients)
$X$ is a finite connected CW-complex.
$\pi_{1}(f): \pi_{1}(X)\rightarrow \pi_{1}(X)$ is an isomorphism of fundamental groups.
$\pi_{1}(X)$ is a finitely presented group.
$\pi_{n}(f)=0$ for $n>1$.
the homotopy colimit $$hocolim(X\rightarrow_{f} X\rightarrow_{f} X\dots)$$ is homotopy equivalent to a finite CW-complex. 


Does it imply that $f$ has to be a weak homotopy equivalence ?

My guess is that the answer should be no but I don't have a counterexample.

REPLY [15 votes]: Here's a counterexample. 
Set $X'=S^1\vee S^2$. 
Consider the following map $F':S^2\vee S^2\vee S^2\rightarrow X'$: It maps the first $S^2$ summand to the $S^2$ summand of $X'$ via a map that represents $2\in\pi_{2}S^2$; it maps the second summand once around the $S^1$ factor of $X'$, and maps the third $S^2$ summand to the $S^2$ summand in $X'$ by a map that represents $-1\in\pi_{2}S^2$.
Let $F$ be the composition of $F'$ with the map $S^2\rightarrow S^2\vee S^2\vee S^2$ which collapses 2 different latitudinal circles.
Form $X$ by attaching a 3-cell to $X'$ by the map $F$. Note that $\pi_{1}X=\pi_{1}S^1$, and the inclusion $S^1\hookrightarrow X$ is a homology isomorphism.
The map $f:X\rightarrow X$ which collapses $X$ to its $S^1$ summand satisfies all the requirements. In this case the hocolim in requirement 6 is $\simeq S^1$.