[proofplan] We compare the two induced maps on an arbitrary homology class represented by a cycle $z \in C_n$. The chain homotopy identity expresses $f_n(z)-g_n(z)$ as the sum of a boundary in $D_n$ and a term involving $d_n^C(z)$. Since $z$ is a cycle, the second term vanishes, so $f_n(z)$ and $g_n(z)$ differ by a boundary and therefore determine the same class in $H_n(D_\bullet)$. [/proofplan] [step:Fix a homology class and choose a cycle representative] Fix $n \in \mathbb Z$. Define the cycle and boundary submodules \begin{align*} Z_n(C_\bullet) &:= \ker(d_n^C: C_n \to C_{n-1}), \\ B_n(C_\bullet) &:= \operatorname{im}(d_{n+1}^C: C_{n+1} \to C_n), \end{align*} and similarly define $Z_n(D_\bullet)$ and $B_n(D_\bullet)$ for $D_\bullet$. Let $[z] \in H_n(C_\bullet)$ be an arbitrary homology class, represented by a cycle $z \in Z_n(C_\bullet)$. Thus $z \in C_n$ and \begin{align*} d_n^C(z) = 0. \end{align*} [/step] [step:Verify that the images of the cycle are cycles in $D_\bullet$] Since $f: C_\bullet \to D_\bullet$ is a chain map, its components satisfy \begin{align*} d_n^D \circ f_n = f_{n-1} \circ d_n^C. \end{align*} Applying this identity to $z$ gives \begin{align*} d_n^D(f_n(z)) = f_{n-1}(d_n^C(z)) = f_{n-1}(0) = 0. \end{align*} Hence $f_n(z) \in Z_n(D_\bullet)$. The same argument applied to the chain map $g: C_\bullet \to D_\bullet$ gives \begin{align*} d_n^D(g_n(z)) = g_{n-1}(d_n^C(z)) = g_{n-1}(0) = 0, \end{align*} so $g_n(z) \in Z_n(D_\bullet)$. Therefore both homology classes $[f_n(z)]$ and $[g_n(z)]$ are defined in $H_n(D_\bullet)$. [/step] [step:Use the chain homotopy identity to show the two images differ by a boundary] By the chain homotopy identity in degree $n$, \begin{align*} f_n(z) - g_n(z) &= d_{n+1}^D(h_n(z)) + h_{n-1}(d_n^C(z)). \end{align*} Since $z$ is a cycle, $d_n^C(z) = 0$. Since $h_{n-1}: C_{n-1} \to D_n$ is $R$-linear, $h_{n-1}(0) = 0$. Therefore \begin{align*} f_n(z) - g_n(z) &= d_{n+1}^D(h_n(z)). \end{align*} Here $h_n(z) \in D_{n+1}$, so $d_{n+1}^D(h_n(z)) \in \operatorname{im}(d_{n+1}^D) = B_n(D_\bullet)$. Hence \begin{align*} f_n(z) - g_n(z) \in B_n(D_\bullet). \end{align*} [/step] [step:Conclude equality of the induced homomorphisms on homology] Since $f_n(z)$ and $g_n(z)$ are cycles in $D_n$ and their difference lies in $B_n(D_\bullet)$, they represent the same coset in \begin{align*} H_n(D_\bullet) = Z_n(D_\bullet) / B_n(D_\bullet). \end{align*} Thus \begin{align*} H_n(f)([z]) = [f_n(z)] = [g_n(z)] = H_n(g)([z]). \end{align*} The class $[z] \in H_n(C_\bullet)$ was arbitrary, so \begin{align*} H_n(f) = H_n(g): H_n(C_\bullet) \to H_n(D_\bullet). \end{align*} Since $n \in \mathbb Z$ was arbitrary, the equality holds in every degree. [/step]