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Suppose $X$ is path connected, let $F:(X,x_0)\to (X,x_0)$ be a map such that $F_*: \Pi_1(X,x_0)\to \Pi_1(X,x_0)$ is identity, does it imply that $F$ is homotopic to identity?

Let $y_0$ be arbitrary, and choose any loop $\gamma:I\to X$ around $x_0$ such that $\gamma(1/2) = y_0$. Then there is a homotopy between $\gamma$ and $F\circ\gamma$, and when we restrict this homotopy to the point $1/2$ we will get a path from $y_0$ to $F(y_0)$. My idea is to show that this can be extended to a homotopy of $F$ to the identity map. But I don't know how to make this precise.

Why is it true if we assume $X$ to be $K(G,1)$?

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No, this is not true. For a silly counterexample, take $X=S^2$ and $F$ the antipodal map. Since $S^2$ is simply connected, $F_*$ is trivially the identity. However, it is well-known that the antipodal map in $S^2$ is not homotopic to the identity (for instance, since $deg(F) = -1$).

PD: Now I see that you also ask $F(x_0)=x_0$. In this case your claim is also false. In order to see it, pick $X=S^2$ and $F$ any map with $deg(F) \neq 1,-1$. Since $deg(F) \neq 1$, $F$ is not homotopic to the identity. Since $deg(F) \neq -1$, $F$ has some fixed point, so it suffices to choose $x_0$ as some fixed point of $F$ to obtain a counterexample.

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The statement is correct if $X$ is a $K(G,1)$, i.e. the fundamental group is the only non trivial homotopy group.

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  • $\begingroup$ Yes, but why so? $\endgroup$ – mez Sep 29 '14 at 0:20

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