# When homology isomorphism implies homotopy isomorphism

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.

• Are you aware of Whitehead’s theorem? And, what role does the limit object play or is ought to play? – user51223 Feb 10 at 3:35

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 $$\simeq S^1$$.