[proofplan] We verify the chain map identities degree by degree. The inclusions $j_D$ and $j_C$ are compatible with the mapping cylinder differential because that differential restricts to $d_D$ on the displayed copy of $D$ and to $d_C$ on the displayed copy of $C$. For the projection $q$, the two terms involving the suspension coordinate cancel after applying $f$, and the remaining equality is exactly the chain map identity for $f$. The two factorisation identities then follow by evaluating the composites on homogeneous elements. [/proofplan]

[step:Check that the inclusion of $D$ commutes with the differentials] Fix $n \in \mathbb{Z}$ and let $y \in D_n$. The map $(j_D)_n: D_n \to \operatorname{Cyl}(f)_n$ is defined by $(j_D)_n(y) = (y,0,0)$. Applying the mapping cylinder differential gives \begin{align*} d_{\operatorname{Cyl}(f)}(j_D)_n(y) &= d_{\operatorname{Cyl}(f)}(y,0,0) \\ &= \bigl(d_D y - f_{n-1}(0),\, d_C 0 + 0,\, -d_C 0\bigr) \\ &= (d_D y,0,0) \\ &= (j_D)_{n-1}(d_D y). \end{align*} Thus \begin{align*} d_{\operatorname{Cyl}(f)} \circ (j_D)_n = (j_D)_{n-1} \circ d_D \end{align*} for every $n \in \mathbb{Z}$, so $j_D: D \to \operatorname{Cyl}(f)$ is a chain map. [/step]

[step:Check that the inclusion of $C$ commutes with the differentials] Fix $n \in \mathbb{Z}$ and let $x \in C_n$. The map $(j_C)_n: C_n \to \operatorname{Cyl}(f)_n$ is defined by $(j_C)_n(x) = (0,x,0)$. Applying the mapping cylinder differential gives \begin{align*} d_{\operatorname{Cyl}(f)}(j_C)_n(x) &= d_{\operatorname{Cyl}(f)}(0,x,0) \\ &= \bigl(d_D 0 - f_{n-1}(0),\, d_C x + 0,\, -d_C 0\bigr) \\ &= (0,d_C x,0) \\ &= (j_C)_{n-1}(d_C x). \end{align*} Thus \begin{align*} d_{\operatorname{Cyl}(f)} \circ (j_C)_n = (j_C)_{n-1} \circ d_C \end{align*} for every $n \in \mathbb{Z}$, so $j_C: C \to \operatorname{Cyl}(f)$ is a chain map. [/step]

[step:Use the chain map identity for $f$ to prove that $q$ commutes with the differentials]Fix $n \in \mathbb{Z}$ and let $(y,x,s) \in \operatorname{Cyl}(f)_n = D_n \oplus C_n \oplus C_{n-1}$. Thus $y \in D_n$, $x \in C_n$, and $s \in C_{n-1}$. By the definition of $q_{n-1}: \operatorname{Cyl}(f)_{n-1} \to D_{n-1}$ and of $d_{\operatorname{Cyl}(f)}$, we have \begin{align*} q_{n-1}\bigl(d_{\operatorname{Cyl}(f)}(y,x,s)\bigr) &= q_{n-1}\bigl(d_D y - f_{n-1}(s),\, d_C x + s,\, -d_C s\bigr) \\ &= d_D y - f_{n-1}(s) + f_{n-1}(d_C x + s) \\ &= d_D y - f_{n-1}(s) + f_{n-1}(d_C x) + f_{n-1}(s) \\ &= d_D y + f_{n-1}(d_C x). \end{align*} Since $f: C \to D$ is a chain map, $f_{n-1}(d_C x) = d_D(f_n(x))$. Therefore \begin{align*} q_{n-1}\bigl(d_{\operatorname{Cyl}(f)}(y,x,s)\bigr) &= d_D y + d_D(f_n(x)) \\ &= d_D\bigl(y + f_n(x)\bigr) \\ &= d_D(q_n(y,x,s)). \end{align*} Hence \begin{align*} q_{n-1} \circ d_{\operatorname{Cyl}(f)} = d_D \circ q_n \end{align*} for every $n \in \mathbb{Z}$, so $q: \operatorname{Cyl}(f) \to D$ is a chain map.[/step]

[guided]We must prove the chain map identity for $q$, namely that applying the mapping cylinder differential first and then $q$ gives the same result as applying $q$ first and then the differential of $D$. Fix $n \in \mathbb{Z}$ and take an arbitrary element \begin{align*} (y,x,s) \in \operatorname{Cyl}(f)_n = D_n \oplus C_n \oplus C_{n-1}. \end{align*} This means $y \in D_n$, $x \in C_n$, and $s \in C_{n-1}$. First apply the mapping cylinder differential: \begin{align*} d_{\operatorname{Cyl}(f)}(y,x,s) = \bigl(d_D y - f_{n-1}(s),\, d_C x + s,\, -d_C s\bigr). \end{align*} This element lies in \begin{align*} D_{n-1} \oplus C_{n-1} \oplus C_{n-2} = \operatorname{Cyl}(f)_{n-1}, \end{align*} so we may apply $q_{n-1}: \operatorname{Cyl}(f)_{n-1} \to D_{n-1}$. By definition, $q_{n-1}(a,b,t) = a + f_{n-1}(b)$ for $a \in D_{n-1}$, $b \in C_{n-1}$, and $t \in C_{n-2}$. Therefore \begin{align*} q_{n-1}\bigl(d_{\operatorname{Cyl}(f)}(y,x,s)\bigr) &= q_{n-1}\bigl(d_D y - f_{n-1}(s),\, d_C x + s,\, -d_C s\bigr) \\ &= d_D y - f_{n-1}(s) + f_{n-1}(d_C x + s). \end{align*} Now use $R$-linearity of $f_{n-1}: C_{n-1} \to D_{n-1}$: \begin{align*} f_{n-1}(d_C x + s) = f_{n-1}(d_C x) + f_{n-1}(s). \end{align*} Substituting this into the previous expression gives the cancellation of the two $s$-terms: \begin{align*} q_{n-1}\bigl(d_{\operatorname{Cyl}(f)}(y,x,s)\bigr) &= d_D y - f_{n-1}(s) + f_{n-1}(d_C x) + f_{n-1}(s) \\ &= d_D y + f_{n-1}(d_C x). \end{align*} The remaining term is where the hypothesis that $f$ is a chain map is used. The chain map identity says \begin{align*} d_D \circ f_n = f_{n-1} \circ d_C, \end{align*} so, evaluated at $x \in C_n$, \begin{align*} f_{n-1}(d_C x) = d_D(f_n(x)). \end{align*} Thus \begin{align*} q_{n-1}\bigl(d_{\operatorname{Cyl}(f)}(y,x,s)\bigr) &= d_D y + d_D(f_n(x)) \\ &= d_D\bigl(y + f_n(x)\bigr) \\ &= d_D(q_n(y,x,s)). \end{align*} Since this holds for every degree $n$ and every element $(y,x,s) \in \operatorname{Cyl}(f)_n$, the map $q: \operatorname{Cyl}(f) \to D$ is a chain map.[/guided]

[step:Evaluate the two composites on homogeneous elements] Fix $n \in \mathbb{Z}$. For $y \in D_n$, the composite $(q \circ j_D)_n: D_n \to D_n$ satisfies \begin{align*} (q \circ j_D)_n(y) &= q_n((j_D)_n(y)) \\ &= q_n(y,0,0) \\ &= y + f_n(0) \\ &= y. \end{align*} Hence $q \circ j_D = \operatorname{id}_D$. For $x \in C_n$, the composite $(q \circ j_C)_n: C_n \to D_n$ satisfies \begin{align*} (q \circ j_C)_n(x) &= q_n((j_C)_n(x)) \\ &= q_n(0,x,0) \\ &= 0 + f_n(x) \\ &= f_n(x). \end{align*} Hence $q \circ j_C = f$. Together with the previous steps, this proves the stated factorisation identities by chain maps. [/step]