Suppose that $\pi_1(X)$ is a finite group. Show that any map $f:X \to S^1$ is nullhomotopic.

My attempt: Since $\pi_1(X)$ is finite and $\pi_1(S^1)=\mathbb{Z}$ torsion-free, then the induced homomorphism $f_*: \pi_1(X) \to \pi_1(S^1)$ has to be trivial. Therefore it is homotopic to a constant map and hence by definition nullhomotopic.

Is my reasoning correct? I have seen the solution to this problem using the covering spaces, lifting $f$ to $\mathbb{R}$ and then using the fact that $\mathbb{R}$ is contractible, but is this additional machinery really needed?

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    $\begingroup$ You need a little more than the induced map on $\pi_1$ being trivial to conclude that a map is nullhomotopic. For instance, the identity map $S^2\rightarrow S^2$ induces the trivial map on $\pi_1$, but is not nullhomotopic. $\endgroup$ – Joe Johnson 126 May 3 '13 at 15:49
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    $\begingroup$ Why do you know that if $f_*$ is trivial, it is homotopic to a constant? $\endgroup$ – Thomas Andrews May 3 '13 at 15:49
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    $\begingroup$ Think about the simply connected cover of $S^1$. $\endgroup$ – rondo9 May 3 '13 at 15:54

Fancy: $\pi_1(X)$ finite implies $H_1(X)$ finite implies $0 = Hom(H_1(X), \mathbb Z) = H^1(X, \mathbb Z) = [X, K(\mathbb Z, 1)] = [X, S^1]$.

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    $\begingroup$ Which requires that $X$ has the homotopy type of a cw complex. The covering space argument however works at least for connected and locally path connected spaces. $\endgroup$ – Alexander Thumm May 3 '13 at 16:02
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    $\begingroup$ Ok, but no hypotheses on X were stated, and I thought this was amusing. $\endgroup$ – Justin Young May 3 '13 at 21:16

The stated proof is not sufficient. Consider a case where the identity map $S^2 \to S^2$ induces the trivial map on $\pi_1$, but is not nullhomotopic.


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