If two chain maps over a PID induce the same homomorphism, then they are homotopic

Question

If two chain maps $f,g:\mathcal{X} \rightarrow \mathcal{Y}$, where $\mathcal{X},\mathcal{Y}$ are chain complexes with free modules $X_p$ and $Y_p$ over a PID, $R$, induce the same homomorphism in the homology, then how to prove that they are homotopic?

Answer 1

I was asking myself the same question, and actually in general the answer is no.

Consider the chain complex $\mathcal{X} = 0 \to \mathbb{Z}\oplus\mathbb{Z} \to \mathbb{Z} \to 0$ with differential $(x,y) \mapsto 2y$. Its homology is therefore $H_0(\mathcal{X}) = \mathbb{Z}/2\mathbb{Z}$ and $H_1(\mathcal{X}) = \mathbb{Z}$. Moreover consider the chain map $f\colon \mathcal{X}\to \mathcal{X}$ given by $f_1\colon (x,y)\mapsto (x+y, y)$ and $f_0 = \textrm{Id}$.

You can verify easily that $f$ is indeed a chain map, and that it induces the identity morphism on homology. However, there exists no homotopy between $f$ and the identity map (such a homotopy $h_0\colon \mathbb{Z} \to \mathbb{Z}\oplus\mathbb{Z}$ would have to send $2\in \mathbb Z$ to $(1,0)\in \mathbb{Z}\oplus\mathbb{Z}$). A sufficient condition for the statement to be true is that not only the modules $X_p$ are free, but also all the homology modules $H_p(\mathcal{X})$ are free as well (with no condition on $\mathcal{Y}$).

Nevertheless Tyler Lawson's answer is perfectly correct, but answers a different question: even though the chain map $f$ is not homotopic to the identity map in this example, it does admit an inverse (in general up to homotopy, but in this example it even admits a strict inverse).

Answer 2

Yes, see Tyler Lawson's answer here https://mathoverflow.net/a/10983 This is for free abelian groups but the same argument works in your case.