Theorem: Chain homotopy of chain maps is an equivalence relation.

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Proof: Suppose we have chain maps $f_\sharp, g_\sharp, h_\sharp : C_n(X) \rightarrow C_n(Y)$. For $f_\sharp$ and $g_\sharp$ to be chain-homotopic, which we will denote for the duration of this proof by $f_\sharp \sim g_\sharp$, means that there exists a map $P : C_n(X) \rightarrow C_{n+1}(Y)$ such that $\delta_{n+2} P_{n+1} + P_n \delta_{n+1} = g_\sharp - f_\sharp$ (for all $n$). We will abuse notation and and just write the left side as $\delta P + P \delta$, leaving the particular instance (subscripts) of $\delta$ and $P$ to be inferred from the context.

Clearly, $\sim$ is reflexive. Simply take $P_0$ to be the zero map, and we have $\delta P_0 + P_0 \delta = 0 = f_\sharp - f_\sharp$. So, $f_\sharp \sim f_\sharp$.

It's also easy to see that $\sim$ is symmetric. Suppose $f_\sharp \sim g_\sharp$ so that $\delta P + P \delta = f_\sharp - g_\sharp$ for some $P$. Then, we have $g_\sharp - f_\sharp = - (\delta P + P \delta) = \delta (-P) + (-P) \delta$. So $g_\sharp \sim f_\sharp$ via the map $-P$.

Finally, suppose $f_\sharp \sim g_\sharp$ and $g_\sharp \sim h_\sharp$. Then, we have $\delta Q + Q \delta = f_\sharp - g_\sharp$ and $\delta R + R \delta = g_\sharp - h_\sharp$ for maps $Q$ and $R$. Then,

$$f_\sharp - h_\sharp = (f_\sharp - g_\sharp) + (g_\sharp - h_\sharp) = \delta Q + Q \delta + \delta R + R \delta = \delta (Q + R) + (Q + R) \delta$$

This last line perhaps merits a bit of explanation -- since these are all maps between abelian groups, addition of maps (defined in the obvious way) is clearly commutative, and the fact that $\delta$ can factor out of sums follows from the fact that $\delta$ is a homomorphism. So, $f_\sharp \sim h_\sharp$ via the map $Q+R$.

So, chain-homotopy is an equivalence relation.