If every map $X\to S^1$ is nullhomotopic, then $H_1 (X)$ is finite, without using cohomology?

Question

If $X$ is a finite connected CW complex, then $H_1 (X, \mathbb{Z})$ is finite iff every map $X \to S^1$ is nullhomotopic.

This question has already been asked and answered here. However, the problem appeared on a topology qualifying exam at University of Michigan, for which cohomology is not part of the syllabus. I am wondering if there is a way to show the direction $$\text{all maps $X\to S^1$ nullhomotopic $\implies$ $H_1 (X, \mathbb{Z})$ is finite}$$ very explicitly, without reference to cohomology, that $S^1$ is a $K(\mathbb{Z}, 1)$, universal coefficient theorem, etc.

Answer 1

Well, if you allow the fact that $H_1$ is the abelianization of $\pi_1$ (Hatcher 2A.1), then

$$\mathrm{Hom}(\pi_1(X),\mathbb Z) \cong \mathrm{Hom}(H_1(X), \mathbb Z).$$

If $H_1(X)$ is finitely generated and if $\mathrm{Hom}(H_1(X), \mathbb Z)$ is zero, then $H_1(X)$ is finite. Since $X$ is a finite $CW$ complex, $H_1$ is definitely finitely generated.

We then want to show that $\mathrm{Hom}(\pi_1(X), \pi_1(S^1))$ is trivial.

Proposition 1B.9 in hatcher's text proves that since $S^1$ has a contractible universal cover,$$\mathrm{Hom}_{\mathrm{Grp}}(\pi_1(X),\pi_1(S^1)) \cong [X,S^1] =0$$

where the $[X,S^1]$ are based maps.