[proofplan] We prove the homotopy identities by induction on the cochain degree. The difference $\delta^q := \alpha^q - \beta^q$ is itself compatible with the differentials and vanishes on the augmentation in degree $0$. In each degree, exactness of the source resolution makes the remaining error factor through an image submodule, and injectivity of the corresponding $J^q$ extends that partially defined map to all of the next term $I^{q+1}$. [/proofplan] [step:Define the difference cochain map] For each integer $q \geq 0$, define the $R$-module homomorphism \begin{align*} \delta^q: I^q &\to J^q \\ x &\mapsto \alpha^q(x) - \beta^q(x). \end{align*} Since $\alpha^\bullet$ and $\beta^\bullet$ are cochain morphisms, for every $q \geq 0$ we have \begin{align*} d_J^q \circ \delta^q &= d_J^q \circ \alpha^q - d_J^q \circ \beta^q \\ &= \alpha^{q+1} \circ d_I^q - \beta^{q+1} \circ d_I^q \\ &= \delta^{q+1} \circ d_I^q. \end{align*} Thus $\delta^\bullet$ is a cochain morphism from $I^\bullet$ to $J^\bullet$. Moreover, \begin{align*} \delta^0 \circ \varepsilon_I &= \alpha^0 \circ \varepsilon_I - \beta^0 \circ \varepsilon_I \\ &= \varepsilon_J \circ f - \varepsilon_J \circ f \\ &= 0. \end{align*} [/step] [step:Construct the first homotopy map from vanishing on $\ker d_I^0$] Exactness of the source resolution at $I^0$ gives \begin{align*} \operatorname{im}\varepsilon_I = \ker d_I^0. \end{align*} Since $\delta^0 \circ \varepsilon_I = 0$, the map $\delta^0$ vanishes on $\ker d_I^0$. Define an $R$-module homomorphism \begin{align*} \varphi_0: \operatorname{im} d_I^0 &\to J^0 \end{align*} by the rule \begin{align*} \varphi_0(d_I^0(x)) := \delta^0(x) \end{align*} for $x \in I^0$. This is well-defined: if $d_I^0(x) = d_I^0(x')$, then $x - x' \in \ker d_I^0$, so \begin{align*} \delta^0(x) - \delta^0(x') = \delta^0(x - x') = 0. \end{align*} The submodule $\operatorname{im} d_I^0$ is contained in $I^1$, and $J^0$ is injective. Therefore $\varphi_0$ extends to an $R$-module homomorphism \begin{align*} s^1: I^1 \to J^0 \end{align*} such that \begin{align*} s^1|_{\operatorname{im} d_I^0} = \varphi_0. \end{align*} For every $x \in I^0$, \begin{align*} (s^1 \circ d_I^0)(x) = \varphi_0(d_I^0(x)) = \delta^0(x). \end{align*} Hence \begin{align*} \delta^0 = s^1 \circ d_I^0. \end{align*} This is the required homotopy identity in degree $0$. [guided] The first degree is the model for the whole induction. We want a map $s^1: I^1 \to J^0$ satisfying \begin{align*} \delta^0 = s^1 \circ d_I^0. \end{align*} Such a formula says that $\delta^0$ depends only on the image $d_I^0(x)$, not on the chosen representative $x \in I^0$. Exactness gives \begin{align*} \operatorname{im}\varepsilon_I = \ker d_I^0. \end{align*} The two morphisms $\alpha^\bullet$ and $\beta^\bullet$ both lie over the same map $f$, so \begin{align*} \delta^0 \circ \varepsilon_I = \alpha^0 \circ \varepsilon_I - \beta^0 \circ \varepsilon_I = \varepsilon_J \circ f - \varepsilon_J \circ f = 0. \end{align*} Therefore $\delta^0$ vanishes on $\operatorname{im}\varepsilon_I$, hence on $\ker d_I^0$. Now define \begin{align*} \varphi_0: \operatorname{im} d_I^0 &\to J^0 \end{align*} by \begin{align*} \varphi_0(d_I^0(x)) := \delta^0(x). \end{align*} This definition uses representatives, so we check well-definedness. If $d_I^0(x) = d_I^0(x')$, then $x - x' \in \ker d_I^0$. Since $\delta^0$ vanishes on $\ker d_I^0$, \begin{align*} \delta^0(x) - \delta^0(x') = \delta^0(x - x') = 0. \end{align*} Thus $\varphi_0$ is a well-defined $R$-module homomorphism. The point of injectivity is now exactly this extension problem: $\varphi_0$ is only defined on the submodule $\operatorname{im} d_I^0 \subset I^1$, and $J^0$ is injective. Hence $\varphi_0$ extends to an $R$-module homomorphism \begin{align*} s^1: I^1 \to J^0 \end{align*} with $s^1|_{\operatorname{im} d_I^0} = \varphi_0$. Therefore, for every $x \in I^0$, \begin{align*} (s^1 \circ d_I^0)(x) = \varphi_0(d_I^0(x)) = \delta^0(x), \end{align*} which proves \begin{align*} \delta^0 = s^1 \circ d_I^0. \end{align*} [/guided] [/step] [step:Extend the homotopy one degree at a time] Assume that $q \geq 1$ and that $R$-module homomorphisms \begin{align*} s^1: I^1 \to J^0,\quad s^2: I^2 \to J^1,\quad \dots,\quad s^q: I^q \to J^{q-1} \end{align*} have been constructed so that \begin{align*} \delta^k = d_J^{k-1} \circ s^k + s^{k+1} \circ d_I^k \end{align*} for every integer $k$ with $0 \leq k \leq q-1$, again omitting the first term when $k=0$. Define the remaining error in degree $q$ by \begin{align*} \gamma^q: I^q &\to J^q \\ x &\mapsto \delta^q(x) - (d_J^{q-1} \circ s^q)(x). \end{align*} We show that $\gamma^q$ vanishes on $\ker d_I^q$. Exactness of the source resolution gives \begin{align*} \ker d_I^q = \operatorname{im} d_I^{q-1}. \end{align*} Let $x \in \ker d_I^q$. Then there exists $y \in I^{q-1}$ such that $x = d_I^{q-1}(y)$. Using the cochain identity for $\delta^\bullet$, the induction hypothesis in degree $q-1$, and the relation $d_J^{q-1}\circ d_J^{q-2}=0$, we obtain \begin{align*} \gamma^q(x) &= \gamma^q(d_I^{q-1}(y)) \\ &= \delta^q(d_I^{q-1}(y)) - d_J^{q-1}(s^q(d_I^{q-1}(y))) \\ &= d_J^{q-1}(\delta^{q-1}(y)) - d_J^{q-1}(s^q(d_I^{q-1}(y))) \\ &= d_J^{q-1}\bigl(d_J^{q-2}(s^{q-1}(y)) + s^q(d_I^{q-1}(y))\bigr) - d_J^{q-1}(s^q(d_I^{q-1}(y))) \\ &= 0. \end{align*} For $q=1$, the displayed computation is interpreted using the already proved identity $\delta^0 = s^1 \circ d_I^0$, so no term involving $d_J^{-1}$ appears. Define \begin{align*} \varphi_q: \operatorname{im} d_I^q &\to J^q \end{align*} by \begin{align*} \varphi_q(d_I^q(x)) := \gamma^q(x) \end{align*} for $x \in I^q$. This is well-defined because $\gamma^q$ vanishes on $\ker d_I^q$. Since $\operatorname{im} d_I^q \subset I^{q+1}$ and $J^q$ is injective, $\varphi_q$ extends to an $R$-module homomorphism \begin{align*} s^{q+1}: I^{q+1} \to J^q \end{align*} such that \begin{align*} s^{q+1}|_{\operatorname{im} d_I^q} = \varphi_q. \end{align*} Then for every $x \in I^q$, \begin{align*} (s^{q+1} \circ d_I^q)(x) = \varphi_q(d_I^q(x)) = \gamma^q(x). \end{align*} Hence \begin{align*} \delta^q = d_J^{q-1} \circ s^q + s^{q+1} \circ d_I^q. \end{align*} This completes the induction step. [guided] Assume the homotopy identity has already been constructed through degree $q-1$. We now correct the error in degree $q$. The map $s^q: I^q \to J^{q-1}$ already exists, so the term $d_J^{q-1}\circ s^q$ is already determined. Define the part of $\delta^q$ not yet accounted for by \begin{align*} \gamma^q: I^q &\to J^q \\ x &\mapsto \delta^q(x) - (d_J^{q-1} \circ s^q)(x). \end{align*} To construct $s^{q+1}: I^{q+1} \to J^q$, we want $\gamma^q$ to factor through $d_I^q$. That requires $\gamma^q$ to vanish on $\ker d_I^q$. Exactness of the source resolution at $I^q$ gives \begin{align*} \ker d_I^q = \operatorname{im} d_I^{q-1}. \end{align*} Take $x \in \ker d_I^q$. Then $x = d_I^{q-1}(y)$ for some $y \in I^{q-1}$. Since $\delta^\bullet$ is a cochain morphism, \begin{align*} \delta^q \circ d_I^{q-1} = d_J^{q-1} \circ \delta^{q-1}. \end{align*} By the induction hypothesis in degree $q-1$, \begin{align*} \delta^{q-1} = d_J^{q-2} \circ s^{q-1} + s^q \circ d_I^{q-1}, \end{align*} with the first term absent when $q=1$. Therefore \begin{align*} \gamma^q(x) &= \gamma^q(d_I^{q-1}(y)) \\ &= \delta^q(d_I^{q-1}(y)) - d_J^{q-1}(s^q(d_I^{q-1}(y))) \\ &= d_J^{q-1}(\delta^{q-1}(y)) - d_J^{q-1}(s^q(d_I^{q-1}(y))) \\ &= d_J^{q-1}\bigl(d_J^{q-2}(s^{q-1}(y)) + s^q(d_I^{q-1}(y))\bigr) - d_J^{q-1}(s^q(d_I^{q-1}(y))) \\ &= 0, \end{align*} because consecutive differentials in the cochain complex $J^\bullet$ compose to zero: \begin{align*} d_J^{q-1}\circ d_J^{q-2}=0. \end{align*} Now $\gamma^q$ factors through the quotient of $I^q$ by $\ker d_I^q$, equivalently through the image $\operatorname{im} d_I^q$. Define \begin{align*} \varphi_q: \operatorname{im} d_I^q &\to J^q \end{align*} by \begin{align*} \varphi_q(d_I^q(x)) := \gamma^q(x). \end{align*} This is well-defined: if $d_I^q(x)=d_I^q(x')$, then $x-x' \in \ker d_I^q$, and hence \begin{align*} \gamma^q(x)-\gamma^q(x')=\gamma^q(x-x')=0. \end{align*} The submodule $\operatorname{im} d_I^q$ lies in $I^{q+1}$, and $J^q$ is injective. Hence $\varphi_q$ extends to an $R$-module homomorphism \begin{align*} s^{q+1}: I^{q+1} \to J^q. \end{align*} For every $x \in I^q$, \begin{align*} (s^{q+1}\circ d_I^q)(x)=\varphi_q(d_I^q(x))=\gamma^q(x). \end{align*} Substituting the definition of $\gamma^q$ gives \begin{align*} \delta^q = d_J^{q-1}\circ s^q + s^{q+1}\circ d_I^q. \end{align*} This is exactly the homotopy identity in degree $q$. [/guided] [/step] [step:Assemble the maps into a cochain homotopy] Starting from $s^1$ and repeatedly applying the induction step, we obtain $R$-module homomorphisms \begin{align*} s^{q+1}: I^{q+1} \to J^q \end{align*} for every $q \geq 0$ such that \begin{align*} \delta^q = d_J^{q-1} \circ s^q + s^{q+1} \circ d_I^q \end{align*} for every $q \geq 0$, with the term $d_J^{-1}\circ s^0$ omitted in degree $0$. Since $\delta^q = \alpha^q - \beta^q$ for every $q \geq 0$, this says precisely that \begin{align*} \alpha^q - \beta^q = d_J^{q-1} \circ s^q + s^{q+1} \circ d_I^q \end{align*} for all $q \geq 0$. Therefore $\alpha^\bullet$ and $\beta^\bullet$ are cochain homotopic. [/step]