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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.

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  • 1
    $\begingroup$ Are you aware of Whitehead’s theorem? And, what role does the limit object play or is ought to play? $\endgroup$ – user51223 Feb 10 at 3:35
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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 $\simeq S^1$.

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