[proofplan] We construct the comparison morphism degree by degree. The degree $0$ map is obtained by extending the composite $M \to N \to J^0$ across the monomorphism $M \to I^0$, using injectivity of $J^0$. For the induction step, the obstruction to defining $\alpha^{q+1}$ is the composite $d_J^q \circ \alpha^q$; exactness of the source resolution shows that this composite vanishes on $\ker d_I^q$, so it factors through $\operatorname{im} d_I^q \subset I^{q+1}$. Injectivity of $J^{q+1}$ then extends the factored map to all of $I^{q+1}$, giving the next cochain identity. [/proofplan] [step:Extend the degree zero composite across the first monomorphism] Define the $R$-module homomorphism \begin{align*} \theta^0: M \to J^0 \end{align*} by \begin{align*} \theta^0 = \varepsilon_J \circ f. \end{align*} Because \begin{align*} 0 \longrightarrow M \xrightarrow{\varepsilon_I} I^0 \end{align*} is exact, $\varepsilon_I$ is a monomorphism. Since $J^0$ is injective and $\theta^0: M \to J^0$ is an $R$-module homomorphism, $\theta^0$ extends across $\varepsilon_I$. Thus there exists an $R$-module homomorphism \begin{align*} \alpha^0: I^0 \to J^0 \end{align*} such that \begin{align*} \alpha^0 \circ \varepsilon_I = \theta^0 = \varepsilon_J \circ f. \end{align*} [guided] The first square of the augmented complexes requires \begin{align*} \alpha^0 \circ \varepsilon_I = \varepsilon_J \circ f. \end{align*} So the map that must be extended from $M$ to $I^0$ is the composite \begin{align*} \theta^0: M \to J^0, \qquad \theta^0 = \varepsilon_J \circ f. \end{align*} The map $\varepsilon_I: M \to I^0$ is a monomorphism because the augmented sequence \begin{align*} 0 \longrightarrow M \xrightarrow{\varepsilon_I} I^0 \end{align*} is exact. The module $J^0$ is injective by hypothesis, since $J^\bullet$ is an injective resolution. Therefore the defining extension property of injective modules applies to the monomorphism $\varepsilon_I: M \to I^0$ and the map $\theta^0: M \to J^0$. It gives an $R$-module homomorphism \begin{align*} \alpha^0: I^0 \to J^0 \end{align*} satisfying \begin{align*} \alpha^0 \circ \varepsilon_I = \theta^0 = \varepsilon_J \circ f. \end{align*} This constructs the degree $0$ part of the comparison morphism and verifies the required compatibility with the augmentations. [/guided] [/step] [step:Construct the next map from the obstruction composite] Assume that, for some integer $q \geq 0$, the maps \begin{align*} \alpha^k: I^k \to J^k \qquad 0 \leq k \leq q \end{align*} have been constructed and satisfy \begin{align*} d_J^k \circ \alpha^k = \alpha^{k+1} \circ d_I^k \qquad 0 \leq k \leq q-1, \end{align*} with the augmentation identity already established when $q=0$. Define the $R$-module homomorphism \begin{align*} \beta^q: I^q \to J^{q+1} \end{align*} by \begin{align*} \beta^q = d_J^q \circ \alpha^q. \end{align*} We show that $\beta^q$ vanishes on $\ker d_I^q$. If $q=0$ and $x \in \ker d_I^0$, exactness of the augmented resolution $I^\bullet$ gives $x \in \operatorname{im}\varepsilon_I$, so $x=\varepsilon_I(m)$ for some $m \in M$. Then \begin{align*} \beta^0(x) &= d_J^0\alpha^0\varepsilon_I(m) \\ &= d_J^0\varepsilon_J f(m) \\ &= 0, \end{align*} because $d_J^0 \circ \varepsilon_J = 0$ in the augmented cochain complex $J^\bullet$. If $q \geq 1$ and $x \in \ker d_I^q$, exactness of $I^\bullet$ gives $x \in \operatorname{im} d_I^{q-1}$, so $x=d_I^{q-1}(y)$ for some $y \in I^{q-1}$. Using the induction hypothesis in degree $q-1$ and the cochain identity $d_J^q \circ d_J^{q-1}=0$, we obtain \begin{align*} \beta^q(x) &= d_J^q\alpha^q d_I^{q-1}(y) \\ &= d_J^q d_J^{q-1}\alpha^{q-1}(y) \\ &= 0. \end{align*} Thus $\ker d_I^q \subseteq \ker \beta^q$. Consequently $\beta^q$ factors through the image of $d_I^q$. Define \begin{align*} \gamma^q: \operatorname{im} d_I^q \to J^{q+1} \end{align*} by \begin{align*} \gamma^q(d_I^q(x)) = \beta^q(x) \qquad \text{for } x \in I^q. \end{align*} This is well-defined because if $d_I^q(x)=d_I^q(x')$, then $x-x' \in \ker d_I^q$, hence \begin{align*} \beta^q(x)-\beta^q(x')=\beta^q(x-x')=0. \end{align*} [guided] Suppose the comparison maps have already been constructed through degree $q$. The next cochain identity we want is \begin{align*} \alpha^{q+1} \circ d_I^q = d_J^q \circ \alpha^q. \end{align*} The right-hand side is already known, so define the obstruction composite \begin{align*} \beta^q: I^q \to J^{q+1}, \qquad \beta^q = d_J^q \circ \alpha^q. \end{align*} To make $\beta^q$ equal to $\alpha^{q+1}\circ d_I^q$, the map $\beta^q$ must depend only on $d_I^q(x)$, not on the representative $x \in I^q$. This is exactly the condition that $\beta^q$ vanish on $\ker d_I^q$. First consider $q=0$. If $x \in \ker d_I^0$, exactness of \begin{align*} 0 \longrightarrow M \xrightarrow{\varepsilon_I} I^0 \xrightarrow{d_I^0} I^1 \end{align*} gives $x=\varepsilon_I(m)$ for some $m \in M$. Then \begin{align*} \beta^0(x) &= d_J^0\alpha^0\varepsilon_I(m) \\ &= d_J^0\varepsilon_J f(m) \\ &= 0, \end{align*} because $\alpha^0\circ\varepsilon_I=\varepsilon_J\circ f$ and because $d_J^0\circ\varepsilon_J=0$ in the augmented complex resolving $N$. Now let $q \geq 1$. If $x \in \ker d_I^q$, exactness at $I^q$ gives $x=d_I^{q-1}(y)$ for some $y \in I^{q-1}$. The induction hypothesis says \begin{align*} \alpha^q \circ d_I^{q-1} = d_J^{q-1} \circ \alpha^{q-1}. \end{align*} Therefore \begin{align*} \beta^q(x) &= d_J^q\alpha^q d_I^{q-1}(y) \\ &= d_J^q d_J^{q-1}\alpha^{q-1}(y) \\ &= 0, \end{align*} because consecutive differentials in the cochain complex $J^\bullet$ compose to zero. Hence $\ker d_I^q \subseteq \ker\beta^q$ in every case. This containment is precisely what is needed to factor $\beta^q$ through the quotient of $I^q$ by $\ker d_I^q$, equivalently through the image $\operatorname{im}d_I^q$. Define \begin{align*} \gamma^q: \operatorname{im} d_I^q \to J^{q+1} \end{align*} by \begin{align*} \gamma^q(d_I^q(x)) = \beta^q(x). \end{align*} The definition is independent of the choice of $x$: if $d_I^q(x)=d_I^q(x')$, then $x-x'\in\ker d_I^q$, and therefore \begin{align*} \beta^q(x)-\beta^q(x')=\beta^q(x-x')=0. \end{align*} So $\gamma^q$ is a well-defined $R$-module homomorphism on the submodule $\operatorname{im}d_I^q \subseteq I^{q+1}$. [/guided] [/step] [step:Extend the factored map to all of the next injective term] Let \begin{align*} \iota^q: \operatorname{im} d_I^q \hookrightarrow I^{q+1} \end{align*} denote the inclusion homomorphism. Since $J^{q+1}$ is injective and $\iota^q$ is a monomorphism, the homomorphism \begin{align*} \gamma^q: \operatorname{im} d_I^q \to J^{q+1} \end{align*} extends across $\iota^q$. Hence there exists an $R$-module homomorphism \begin{align*} \alpha^{q+1}: I^{q+1} \to J^{q+1} \end{align*} such that \begin{align*} \alpha^{q+1}\circ \iota^q = \gamma^q. \end{align*} For every $x \in I^q$, this gives \begin{align*} \alpha^{q+1}d_I^q(x) &= \gamma^q(d_I^q(x)) \\ &= \beta^q(x) \\ &= d_J^q\alpha^q(x). \end{align*} Therefore \begin{align*} \alpha^{q+1}\circ d_I^q = d_J^q\circ\alpha^q. \end{align*} [guided] At this point we have constructed a map only on the submodule $\operatorname{im}d_I^q$ of $I^{q+1}$. Let \begin{align*} \iota^q: \operatorname{im} d_I^q \hookrightarrow I^{q+1} \end{align*} be the inclusion homomorphism. This map is a monomorphism because it is the inclusion of a submodule. The target module $J^{q+1}$ is injective by hypothesis, since $J^\bullet$ is an injective resolution. Applying the defining extension property of injective modules to the monomorphism $\iota^q$ and the homomorphism \begin{align*} \gamma^q: \operatorname{im}d_I^q \to J^{q+1}, \end{align*} we obtain an $R$-module homomorphism \begin{align*} \alpha^{q+1}: I^{q+1} \to J^{q+1} \end{align*} such that \begin{align*} \alpha^{q+1}\circ \iota^q = \gamma^q. \end{align*} Now verify that this extension has the required cochain property. For every $x \in I^q$, the element $d_I^q(x)$ lies in $\operatorname{im}d_I^q$, so \begin{align*} \alpha^{q+1}d_I^q(x) &= \gamma^q(d_I^q(x)) \\ &= \beta^q(x) \\ &= d_J^q\alpha^q(x). \end{align*} Since this equality holds for every $x \in I^q$, it is an equality of homomorphisms: \begin{align*} \alpha^{q+1}\circ d_I^q = d_J^q\circ\alpha^q. \end{align*} This completes the induction step. [/guided] [/step] [step:Assemble the degreewise maps into a morphism of augmented cochain complexes] Starting from $\alpha^0$ and iterating the induction step for $q=0,1,2,\dots$, we obtain $R$-module homomorphisms \begin{align*} \alpha^q: I^q \to J^q \qquad q \geq 0 \end{align*} such that \begin{align*} \alpha^0 \circ \varepsilon_I &= \varepsilon_J \circ f, \\ d_J^q \circ \alpha^q &= \alpha^{q+1} \circ d_I^q \qquad \text{for every } q \geq 0. \end{align*} Thus $f$ together with the family $(\alpha^q)_{q\geq 0}$ is a morphism of augmented cochain complexes from the injective resolution of $M$ to the injective resolution of $N$. This proves the theorem. [/step]