Let's suppose that 


$f:X\rightarrow X$ is a continuous map such that 


1. $H\_{\ast}(f): H\_{\ast}(X)\rightarrow H\_{\ast}(X)$ is a homology isomorphism (with integral coefficients)
2. $X$ is a finite connected CW-complex.
3. $\pi\_{1}(f): \pi\_{1}(X)\rightarrow \pi\_{1}(X)$ is an isomorphism of fundamental groups.
4. $\pi\_{1}(X)$ is a finitely presented group.
5. $\pi\_{n}(f)=0$ for $n>1$.
6. 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. 



When homology isomorphism implies homotopy isomorphism
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 $\simeq S^1$.


