Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?


share|cite|improve this question

2 Answers 2

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.

share|cite|improve this answer
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!

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.