[proofplan] We construct the maps $\tilde f_n:P_n \to Q_n$ recursively. At degree $0$, projectivity of $P_0$ lifts the map $f\circ \varepsilon_P:P_0\to N$ through the augmentation $\varepsilon_Q:Q_0\to N$. The degree-one step uses exactness at $Q_0$ to turn the vanishing of $\varepsilon_Q\tilde f_0 d_1^P$ into a lift through $d_1^Q$. The general induction step is the same argument one degree higher: the already constructed chain-map identity forces $\tilde f_n d_{n+1}^P$ to land in $\ker d_n^Q$, and exactness identifies this kernel with the image of $d_{n+1}^Q$, allowing projectivity of $P_{n+1}$ to provide the next lift. [/proofplan]

[step:Lift the degree-zero map through the augmentation of $Q_\bullet$]Since $Q_\bullet \to N$ is a projective resolution, the augmentation \begin{align*} \varepsilon_Q:Q_0 \to N \end{align*} is surjective. Define the $R$-module homomorphism \begin{align*} g_0:P_0 &\to N \\ x &\mapsto f(\varepsilon_P(x)). \end{align*} Because $P_0$ is projective and $\varepsilon_Q$ is surjective, there exists an $R$-module homomorphism \begin{align*} \tilde f_0:P_0 \to Q_0 \end{align*} such that \begin{align*} \varepsilon_Q \circ \tilde f_0 = g_0 = f \circ \varepsilon_P. \end{align*} This proves the required augmented identity at degree $0$.[/step]

[guided]We begin at the only place where the given map $f:M\to N$ appears directly. The source resolution ends in the augmentation $\varepsilon_P:P_0\to M$, so the composite map we must lift is \begin{align*} g_0:P_0 &\to N \\ x &\mapsto f(\varepsilon_P(x)). \end{align*} This is an $R$-module homomorphism because it is the composite of the $R$-module homomorphisms $\varepsilon_P$ and $f$. The target resolution is exact at $N$, so its augmentation \begin{align*} \varepsilon_Q:Q_0\to N \end{align*} is surjective. Since $P_0$ is projective, the defining lifting property of projective modules applies to the surjection $\varepsilon_Q:Q_0\to N$ and the map $g_0:P_0\to N$. Therefore there exists an $R$-module homomorphism \begin{align*} \tilde f_0:P_0\to Q_0 \end{align*} with \begin{align*} \varepsilon_Q\circ \tilde f_0 = g_0 = f\circ \varepsilon_P. \end{align*} This is exactly the commutative square at the augmented end of the resolutions.[/guided]

[step:Lift the first differential through $d_1^Q$]Define the $R$-module homomorphism \begin{align*} a_1:P_1 &\to Q_0 \\ x &\mapsto \tilde f_0(d_1^P(x)). \end{align*} Then \begin{align*} \varepsilon_Q(a_1(x)) &= \varepsilon_Q(\tilde f_0(d_1^P(x))) \\ &= f(\varepsilon_P(d_1^P(x))) \\ &= 0 \end{align*} for every $x\in P_1$, because $\varepsilon_P\circ d_1^P=0$. Hence $a_1(P_1)\subseteq \ker \varepsilon_Q$. Exactness of $Q_\bullet\to N$ at $Q_0$ gives \begin{align*} \ker \varepsilon_Q = \operatorname{im} d_1^Q. \end{align*} Let \begin{align*} \rho_1:Q_1 &\to \ker \varepsilon_Q \\ y &\mapsto d_1^Q(y) \end{align*} be the codomain-restricted map. The equality $\ker \varepsilon_Q=\operatorname{im} d_1^Q$ says precisely that $\rho_1$ is surjective. Let \begin{align*} \bar a_1:P_1 &\to \ker \varepsilon_Q \\ x &\mapsto a_1(x) \end{align*} be the same homomorphism with restricted codomain. Since $P_1$ is projective, there exists an $R$-module homomorphism \begin{align*} \tilde f_1:P_1\to Q_1 \end{align*} such that \begin{align*} \rho_1\circ \tilde f_1=\bar a_1. \end{align*} By the definition of $\rho_1$ and $\bar a_1$, this is \begin{align*} d_1^Q\circ \tilde f_1=\tilde f_0\circ d_1^P. \end{align*}[/step]

[guided]The next map $\tilde f_1:P_1\to Q_1$ must make the degree-one chain-map square commute. This means we want \begin{align*} d_1^Q\circ \tilde f_1=\tilde f_0\circ d_1^P. \end{align*} So define the right-hand side as a map \begin{align*} a_1:P_1 &\to Q_0 \\ x &\mapsto \tilde f_0(d_1^P(x)). \end{align*} To lift $a_1$ through $d_1^Q$, we first verify that its image lies in the image of $d_1^Q$. For every $x\in P_1$, the degree-zero identity already constructed gives \begin{align*} \varepsilon_Q(a_1(x)) &= \varepsilon_Q(\tilde f_0(d_1^P(x))) \\ &= f(\varepsilon_P(d_1^P(x))). \end{align*} Because $P_\bullet\to M$ is a chain complex, $\varepsilon_P\circ d_1^P=0$. Therefore \begin{align*} \varepsilon_Q(a_1(x))=0. \end{align*} Thus $a_1(P_1)\subseteq \ker\varepsilon_Q$. Exactness of the augmented complex $Q_\bullet\to N$ at $Q_0$ gives \begin{align*} \ker \varepsilon_Q=\operatorname{im} d_1^Q. \end{align*} Instead of treating $d_1^Q$ as a map merely into $Q_0$, restrict its codomain to its exact target: \begin{align*} \rho_1:Q_1 &\to \ker \varepsilon_Q \\ y &\mapsto d_1^Q(y). \end{align*} Exactness says that $\rho_1$ is surjective. Also define \begin{align*} \bar a_1:P_1 &\to \ker \varepsilon_Q \\ x &\mapsto a_1(x). \end{align*} This is well-defined because we just proved $a_1(P_1)\subseteq \ker\varepsilon_Q$. Now apply the defining lifting property of the projective module $P_1$ to the surjection $\rho_1:Q_1\to \ker\varepsilon_Q$ and the map $\bar a_1:P_1\to\ker\varepsilon_Q$. We obtain an $R$-module homomorphism \begin{align*} \tilde f_1:P_1\to Q_1 \end{align*} such that \begin{align*} \rho_1\circ \tilde f_1=\bar a_1. \end{align*} Unpacking the definitions of $\rho_1$ and $\bar a_1$, this becomes \begin{align*} d_1^Q\circ \tilde f_1=\tilde f_0\circ d_1^P. \end{align*} So the chain-map identity holds in degree $1$.[/guided]

[step:Inductively lift each higher differential through the next boundary map of $Q_\bullet$]Let $n\geq 1$, and suppose that $R$-module homomorphisms \begin{align*} \tilde f_i:P_i\to Q_i \qquad 0\leq i\leq n \end{align*} have been constructed such that \begin{align*} d_i^Q\circ \tilde f_i=\tilde f_{i-1}\circ d_i^P \end{align*} for every $1\leq i\leq n$. Define \begin{align*} a_{n+1}:P_{n+1} &\to Q_n \\ x &\mapsto \tilde f_n(d_{n+1}^P(x)). \end{align*} For every $x\in P_{n+1}$, \begin{align*} d_n^Q(a_{n+1}(x)) &= d_n^Q(\tilde f_n(d_{n+1}^P(x))) \\ &= \tilde f_{n-1}(d_n^P(d_{n+1}^P(x))) \\ &=0, \end{align*} because $d_n^P\circ d_{n+1}^P=0$. Hence $a_{n+1}(P_{n+1})\subseteq \ker d_n^Q$. Exactness of $Q_\bullet$ at $Q_n$ gives \begin{align*} \ker d_n^Q=\operatorname{im} d_{n+1}^Q. \end{align*} Let \begin{align*} \rho_{n+1}:Q_{n+1} &\to \ker d_n^Q \\ y &\mapsto d_{n+1}^Q(y) \end{align*} be the codomain-restricted boundary map. This map is surjective. Let \begin{align*} \bar a_{n+1}:P_{n+1} &\to \ker d_n^Q \\ x &\mapsto a_{n+1}(x) \end{align*} be the same homomorphism with restricted codomain. Since $P_{n+1}$ is projective, there exists an $R$-module homomorphism \begin{align*} \tilde f_{n+1}:P_{n+1}\to Q_{n+1} \end{align*} such that \begin{align*} \rho_{n+1}\circ \tilde f_{n+1}=\bar a_{n+1}. \end{align*} Equivalently, \begin{align*} d_{n+1}^Q\circ \tilde f_{n+1}=\tilde f_n\circ d_{n+1}^P. \end{align*}[/step]

[guided]Assume that the maps have already been constructed through degree $n$, where $n\geq 1$, and that the chain-map identities \begin{align*} d_i^Q\circ \tilde f_i=\tilde f_{i-1}\circ d_i^P \end{align*} hold for every $1\leq i\leq n$. We now construct the next map \begin{align*} \tilde f_{n+1}:P_{n+1}\to Q_{n+1}. \end{align*} The desired identity in degree $n+1$ is \begin{align*} d_{n+1}^Q\circ \tilde f_{n+1}=\tilde f_n\circ d_{n+1}^P. \end{align*} So define the right-hand side as \begin{align*} a_{n+1}:P_{n+1} &\to Q_n \\ x &\mapsto \tilde f_n(d_{n+1}^P(x)). \end{align*} To lift $a_{n+1}$ through $d_{n+1}^Q$, we must prove that $a_{n+1}$ lands in $\operatorname{im} d_{n+1}^Q$. Exactness will identify this image with $\ker d_n^Q$, so we check membership in that kernel. For every $x\in P_{n+1}$, the induction hypothesis in degree $n$ gives \begin{align*} d_n^Q(a_{n+1}(x)) &=d_n^Q(\tilde f_n(d_{n+1}^P(x))) \\ &=\tilde f_{n-1}(d_n^P(d_{n+1}^P(x))). \end{align*} Since $P_\bullet$ is a chain complex, \begin{align*} d_n^P\circ d_{n+1}^P=0. \end{align*} Therefore \begin{align*} d_n^Q(a_{n+1}(x))=0. \end{align*} Thus \begin{align*} a_{n+1}(P_{n+1})\subseteq \ker d_n^Q. \end{align*} Exactness of $Q_\bullet$ at $Q_n$ gives \begin{align*} \ker d_n^Q=\operatorname{im} d_{n+1}^Q. \end{align*} Now restrict the codomain of $d_{n+1}^Q$ to the kernel it surjects onto: \begin{align*} \rho_{n+1}:Q_{n+1} &\to \ker d_n^Q \\ y &\mapsto d_{n+1}^Q(y). \end{align*} This map is surjective by exactness. Also define \begin{align*} \bar a_{n+1}:P_{n+1} &\to \ker d_n^Q \\ x &\mapsto a_{n+1}(x). \end{align*} This is well-defined because the previous computation proved that $a_{n+1}$ lands in $\ker d_n^Q$. Since $P_{n+1}$ is projective, the lifting property applies to the surjection $\rho_{n+1}:Q_{n+1}\to\ker d_n^Q$ and the map $\bar a_{n+1}:P_{n+1}\to\ker d_n^Q$. Hence there exists an $R$-module homomorphism \begin{align*} \tilde f_{n+1}:P_{n+1}\to Q_{n+1} \end{align*} such that \begin{align*} \rho_{n+1}\circ \tilde f_{n+1}=\bar a_{n+1}. \end{align*} Unwinding the definitions gives \begin{align*} d_{n+1}^Q\circ \tilde f_{n+1}=\tilde f_n\circ d_{n+1}^P. \end{align*} This is exactly the next chain-map identity.[/guided]

[step:Assemble the recursively constructed maps into the required chain map] By the degree-zero construction, the identity \begin{align*} \varepsilon_Q\circ \tilde f_0=f\circ \varepsilon_P \end{align*} holds. By the degree-one construction and the induction step applied successively for $n=1,2,3,\dots$, there exist $R$-module homomorphisms \begin{align*} \tilde f_n:P_n\to Q_n \qquad n\geq 0 \end{align*} satisfying \begin{align*} d_n^Q\circ \tilde f_n=\tilde f_{n-1}\circ d_n^P \end{align*} for every $n\geq 1$. Therefore the family $\tilde f_\bullet=(\tilde f_n)_{n\geq 0}$ is a chain map $P_\bullet\to Q_\bullet$ lifting $f$ at the augmented end. This proves the theorem. [/step]