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Let X be a path connected and locally path connected topological space. Suppose there exists a continuous $f: X \rightarrow S^1$ inducing the trivial map between the fundamental groups of X and $S^1$.

How do I show that this implies the null-homotopy of f? Thanks in advance!

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1 Answer 1

up vote 6 down vote accepted

You can use a bit of covering space theory. There is a theorem that says if $f:Y\rightarrow Z$ induces the trivial map on the fundamental group, for $Y$ path connected, locally path connected then there is a lift of $f$, $\tilde{f}:Y\rightarrow \tilde{Z}$ where $\tilde{Z}$ is any covering space, i.e. $p\circ\tilde{f}=f$, where $p:\tilde{Z}\rightarrow Z$ is the covering map. See, for instance, Proposition 1.33 in Allen Hatcher's $\textit{Algebraic Topology}$. Since $S^1$ has a contractible universal cover, $\mathbb{R}$, the map $f$ factors through a contractible space and is therefore null-homotopic.

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Very nice solution. Thank you! –  user18679 Nov 1 '11 at 17:00

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