All maps from a CW complex to S^1 null-homotopic implies finite first homology 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?
Thanks!
 A: 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.  
A: 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!
