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So given a connected compact CW-complex $X$, a quick covering space argument shows that if $H_1(X)$ is finite, then every map $X \to S^1$ is null-homotopic. I was curious if the converse was true: given such a CW-complex $X$, if every map $X \to S^1$ is null-homotopic, does that imply that $H_1(X)$ is finite?


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up vote 2 down vote accepted

The space $S^1$ is a $K(\mathbb Z,1)$, meaning that $H^1(X) = [X,S^1]$ (homotopy classes of maps from $X$ to $S^1$), for a reasonable space $X$. (The isomorphism from right to left is given by sending a map $f$ to $f^*$ of the fundamental class in $H^1(S^1)$.)

Thus if $[X,S^1]$ consists of just a single point (the class of null-homotopic maps), then $H^1(X) = 0$. By universal coefficients, we have $H^1(X) = Hom(H_1(X),\mathbb Z)$, and so if $H_1(X)$ is finitely generated, then we see that $H^1(X) = 0$ iff $H_1(X)$ is finite.

So the answer to your question seems to be yes.

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Is there a way to see this without using this much machinery? – msteve Jul 24 '14 at 17:25

In general 'no' but for your example 'yes'.

It turns out that homotopy classes of maps to $S^1$ are in correspondence with first cohomology, $H^1(X, \mathbb{Z})$. If this is zero then, by the universal coefficient theorem we must have an exact sequence

$$ 0 \rightarrow \text{Ext}(H_0, \mathbb{Z}) \rightarrow H^1(X, \mathbb{Z}) \rightarrow \text{Hom}(H_1(X), \mathbb{Z}) \rightarrow 0 $$ If the middle term is zero, then so is the right hand term, and vice-versa (since $H_0$ is always free).

Now, if $H_1(X)$ is finitely generated, then it's clear that it must be finite in order for the right hand side to vanish. In your example $H_1(X)$ is finitely generated, so we're good.

On the other hand, if $H_1(X) = \mathbb{Q}$, for example, the right hand side would also vanish- and $\mathbb{Q}$ is definitely not finite!

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