[proofplan] We verify the three defining properties of an [equivalence relation](/page/Equivalence%20Relation) directly from the chain homotopy equation. Reflexivity is witnessed by the zero family of maps, symmetry by negating a chosen homotopy, and transitivity by adding two homotopies degree by degree. Each construction is valid because each $\operatorname{Hom}_R(C_n, D_{n+1})$ is an abelian group under pointwise addition. [/proofplan] [step:Use the zero family to prove reflexivity] Let $f_\bullet \in \operatorname{Ch}_R(C_\bullet, D_\bullet)$. For each $n \in \mathbb{Z}$, define the zero $R$-[linear map](/page/Linear%20Map) \begin{align*} 0_n: C_n \to D_{n+1} \end{align*} by $0_n(x) = 0$ for all $x \in C_n$. Then for every $n \in \mathbb{Z}$, \begin{align*} d^D_{n+1} \circ 0_n + 0_{n-1} \circ d^C_n = 0 + 0 = 0 = f_n - f_n. \end{align*} Thus the family $(0_n)_{n \in \mathbb{Z}}$ is a chain homotopy from $f_\bullet$ to $f_\bullet$, so $f_\bullet \sim f_\bullet$. [/step] [step:Negate a homotopy to prove symmetry] Let $f_\bullet, g_\bullet \in \operatorname{Ch}_R(C_\bullet, D_\bullet)$ and suppose $f_\bullet \sim g_\bullet$. Choose a chain homotopy from $f_\bullet$ to $g_\bullet$, meaning a family of $R$-linear maps \begin{align*} s_n: C_n \to D_{n+1} \end{align*} such that, for every $n \in \mathbb{Z}$, \begin{align*} f_n - g_n = d^D_{n+1} \circ s_n + s_{n-1} \circ d^C_n. \end{align*} For each $n \in \mathbb{Z}$, define the $R$-linear map \begin{align*} (-s)_n: C_n \to D_{n+1} \end{align*} by $(-s)_n(x) = -s_n(x)$ for all $x \in C_n$. Since composition of $R$-linear maps respects addition and additive inverses, \begin{align*} d^D_{n+1} \circ (-s)_n + (-s)_{n-1} \circ d^C_n &= -d^D_{n+1} \circ s_n - s_{n-1} \circ d^C_n \\ &= -(f_n - g_n) \\ &= g_n - f_n. \end{align*} Thus $(-s)_\bullet$ is a chain homotopy from $g_\bullet$ to $f_\bullet$, so $g_\bullet \sim f_\bullet$. [/step] [step:Add two homotopies to prove transitivity] Let $f_\bullet, g_\bullet, h_\bullet \in \operatorname{Ch}_R(C_\bullet, D_\bullet)$ and suppose $f_\bullet \sim g_\bullet$ and $g_\bullet \sim h_\bullet$. Choose a chain homotopy $(s_n)_{n \in \mathbb{Z}}$ from $f_\bullet$ to $g_\bullet$ and a chain homotopy $(t_n)_{n \in \mathbb{Z}}$ from $g_\bullet$ to $h_\bullet$. Thus, for every $n \in \mathbb{Z}$, \begin{align*} f_n - g_n &= d^D_{n+1} \circ s_n + s_{n-1} \circ d^C_n, \\ g_n - h_n &= d^D_{n+1} \circ t_n + t_{n-1} \circ d^C_n. \end{align*} For each $n \in \mathbb{Z}$, define the $R$-linear map \begin{align*} (s+t)_n: C_n \to D_{n+1} \end{align*} by $(s+t)_n(x) = s_n(x) + t_n(x)$ for all $x \in C_n$. Adding the two displayed homotopy equations and using distributivity of composition over addition gives \begin{align*} f_n - h_n &= (f_n - g_n) + (g_n - h_n) \\ &= d^D_{n+1} \circ s_n + s_{n-1} \circ d^C_n + d^D_{n+1} \circ t_n + t_{n-1} \circ d^C_n \\ &= d^D_{n+1} \circ (s+t)_n + (s+t)_{n-1} \circ d^C_n. \end{align*} Therefore $(s+t)_\bullet$ is a chain homotopy from $f_\bullet$ to $h_\bullet$, so $f_\bullet \sim h_\bullet$. [/step] [step:Conclude that chain homotopy is an equivalence relation] We have shown that $\sim$ is reflexive, symmetric, and transitive on $\operatorname{Ch}_R(C_\bullet, D_\bullet)$. Hence chain [homotopy is an equivalence relation](/theorems/1872) on the set of chain maps $C_\bullet \to D_\bullet$. [/step]