[proofplan] Subtract the two lifted chain maps and prove that the difference is null-homotopic. The equality of the lifted maps on $M$ forces the degree-zero component of the difference to land in $\ker \varepsilon_Q$, and exactness of the resolution $Q_\bullet \to N$ identifies this kernel with $\operatorname{im} d_1^Q$. Projectivity of each $P_n$ then lets us lift the relevant defect map through the surjection $d_{n+1}^Q:Q_{n+1}\to \ker d_n^Q$. An induction constructs the homotopy maps $s_n:P_n\to Q_{n+1}$ and proves the homotopy identity in every degree. [/proofplan] [step:Subtract the two lifts and reduce to a null-homotopy problem] Define, for each $n \geq 0$, the $R$-module homomorphism \begin{align*} a_n:P_n &\to Q_n \\ x &\mapsto g_n(x)-h_n(x). \end{align*} Since $g_\bullet$ and $h_\bullet$ are chain maps, for every $n \geq 1$ we have \begin{align*} d_n^Q \circ a_n &= d_n^Q \circ g_n - d_n^Q \circ h_n \\ &= g_{n-1}\circ d_n^P - h_{n-1}\circ d_n^P \\ &= a_{n-1}\circ d_n^P. \end{align*} Thus $a_\bullet:P_\bullet\to Q_\bullet$ is a chain map. Moreover, since $g_\bullet$ and $h_\bullet$ lift the same map $f:M\to N$, \begin{align*} \varepsilon_Q \circ a_0 &= \varepsilon_Q \circ g_0 - \varepsilon_Q \circ h_0 \\ &= f\circ \varepsilon_P - f\circ \varepsilon_P \\ &= 0. \end{align*} Therefore $a_\bullet$ is a chain map lifting the zero map $M\to N$. It remains to construct maps $s_n:P_n\to Q_{n+1}$ satisfying \begin{align*} a_n=d_{n+1}^Q\circ s_n+s_{n-1}\circ d_n^P \end{align*} for all $n\geq 0$, where $s_{-1}:=0$. [/step] [step:Lift the degree-zero map through $d_1^Q$] Because $\varepsilon_Q\circ a_0=0$, the image of $a_0$ is contained in $\ker \varepsilon_Q$. 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*} Let \begin{align*} \iota_0:\ker \varepsilon_Q \to Q_0 \end{align*} denote the inclusion map. There is a unique $R$-module homomorphism \begin{align*} \bar a_0:P_0 \to \ker \varepsilon_Q \end{align*} such that $\iota_0\circ \bar a_0=a_0$. The map \begin{align*} d_1^Q:Q_1\to \ker \varepsilon_Q \end{align*} is surjective when its codomain is restricted to $\ker \varepsilon_Q$, because $\operatorname{im}d_1^Q=\ker\varepsilon_Q$. Since $P_0$ is projective, the map $\bar a_0:P_0\to\ker\varepsilon_Q$ lifts through this surjection. Hence there exists an $R$-module homomorphism \begin{align*} s_0:P_0\to Q_1 \end{align*} such that \begin{align*} d_1^Q\circ s_0=\bar a_0. \end{align*} Composing with the inclusion $\iota_0$ gives \begin{align*} d_1^Q\circ s_0=a_0. \end{align*} Since $s_{-1}:=0$, this is exactly the homotopy identity in degree $0$: \begin{align*} a_0=d_1^Q\circ s_0+s_{-1}\circ \varepsilon_P. \end{align*} [guided] The degree-zero case is where the assumption that $g_\bullet$ and $h_\bullet$ lift the same map is used. Since \begin{align*} \varepsilon_Q\circ g_0=f\circ\varepsilon_P \quad\text{and}\quad \varepsilon_Q\circ h_0=f\circ\varepsilon_P, \end{align*} their difference satisfies \begin{align*} \varepsilon_Q\circ a_0=0. \end{align*} This means that $a_0:P_0\to Q_0$ lands in the submodule $\ker\varepsilon_Q\subset Q_0$. Exactness of the projective resolution $Q_\bullet\to N$ at $Q_0$ says precisely that \begin{align*} \ker\varepsilon_Q=\operatorname{im}d_1^Q. \end{align*} Thus the map $a_0$ factors uniquely through the inclusion $\iota_0:\ker\varepsilon_Q\to Q_0$ as a map \begin{align*} \bar a_0:P_0\to \ker\varepsilon_Q. \end{align*} Now regard \begin{align*} d_1^Q:Q_1\to \ker\varepsilon_Q \end{align*} as a surjective map onto its image, which is $\ker\varepsilon_Q$. Since $P_0$ is projective, every map from $P_0$ into the target of a surjection lifts through that surjection. Applying projectivity to $\bar a_0:P_0\to\ker\varepsilon_Q$ gives an $R$-module homomorphism \begin{align*} s_0:P_0\to Q_1 \end{align*} with \begin{align*} d_1^Q\circ s_0=\bar a_0. \end{align*} As a map into $Q_0$, this equality says \begin{align*} d_1^Q\circ s_0=a_0. \end{align*} This is the required homotopy formula in degree $0$, since the term involving $s_{-1}$ is declared to be zero. [/guided] [/step] [step:Inductively lift each higher defect through exactness of $Q_\bullet$] Assume that for some $n\geq 1$ the maps \begin{align*} s_0:P_0\to Q_1,\quad \dots,\quad s_{n-1}:P_{n-1}\to Q_n \end{align*} have been constructed and satisfy \begin{align*} a_k=d_{k+1}^Q\circ s_k+s_{k-1}\circ d_k^P \end{align*} for every $0\leq k\leq n-1$. Define the $n$th defect map \begin{align*} b_n:P_n &\to Q_n \\ x &\mapsto a_n(x)-s_{n-1}(d_n^P(x)). \end{align*} We prove that $b_n$ lands in $\ker d_n^Q$. Using the chain-map identity for $a_\bullet$, the induction hypothesis in degree $n-1$, and the complex identity $d_{n-1}^P\circ d_n^P=0$, we obtain \begin{align*} d_n^Q\circ b_n &=d_n^Q\circ a_n-d_n^Q\circ s_{n-1}\circ d_n^P\\ &=a_{n-1}\circ d_n^P-d_n^Q\circ s_{n-1}\circ d_n^P\\ &=\left(a_{n-1}-d_n^Q\circ s_{n-1}\right)\circ d_n^P\\ &=s_{n-2}\circ d_{n-1}^P\circ d_n^P\\ &=0. \end{align*} Here, when $n=1$, the expression $s_{n-2}$ means $s_{-1}=0$. Let \begin{align*} \iota_n:\ker d_n^Q\to Q_n \end{align*} denote the inclusion map. Since $d_n^Q\circ b_n=0$, there is a unique $R$-module homomorphism \begin{align*} \bar b_n:P_n\to \ker d_n^Q \end{align*} such that $\iota_n\circ \bar b_n=b_n$. Exactness of $Q_\bullet$ at $Q_n$ gives \begin{align*} \ker d_n^Q=\operatorname{im}d_{n+1}^Q. \end{align*} Thus \begin{align*} d_{n+1}^Q:Q_{n+1}\to \ker d_n^Q \end{align*} is surjective after restricting its codomain to $\ker d_n^Q$. Since $P_n$ is projective, the map $\bar b_n:P_n\to\ker d_n^Q$ lifts through this surjection. Therefore there exists an $R$-module homomorphism \begin{align*} s_n:P_n\to Q_{n+1} \end{align*} such that \begin{align*} d_{n+1}^Q\circ s_n=\bar b_n. \end{align*} As a map into $Q_n$, this gives \begin{align*} d_{n+1}^Q\circ s_n=b_n=a_n-s_{n-1}\circ d_n^P. \end{align*} Rearranging yields the homotopy identity in degree $n$: \begin{align*} a_n=d_{n+1}^Q\circ s_n+s_{n-1}\circ d_n^P. \end{align*} [guided] Suppose the homotopy maps have already been built through degree $n-1$. The next goal is to find a map \begin{align*} s_n:P_n\to Q_{n+1} \end{align*} such that \begin{align*} a_n=d_{n+1}^Q\circ s_n+s_{n-1}\circ d_n^P. \end{align*} The term $s_{n-1}\circ d_n^P$ is already known, so the only part still needing construction is the difference \begin{align*} b_n:=a_n-s_{n-1}\circ d_n^P. \end{align*} This defines an $R$-module homomorphism $b_n:P_n\to Q_n$. To lift $b_n$ through $d_{n+1}^Q$, we must first prove that $b_n$ lands in $\ker d_n^Q$, because exactness identifies $\ker d_n^Q$ with the image of $d_{n+1}^Q$. We compute: \begin{align*} d_n^Q\circ b_n &=d_n^Q\circ a_n-d_n^Q\circ s_{n-1}\circ d_n^P. \end{align*} Since $a_\bullet$ is a chain map, $d_n^Q\circ a_n=a_{n-1}\circ d_n^P$. Hence \begin{align*} d_n^Q\circ b_n &=a_{n-1}\circ d_n^P-d_n^Q\circ s_{n-1}\circ d_n^P\\ &=\left(a_{n-1}-d_n^Q\circ s_{n-1}\right)\circ d_n^P. \end{align*} The induction hypothesis in degree $n-1$ says \begin{align*} a_{n-1}=d_n^Q\circ s_{n-1}+s_{n-2}\circ d_{n-1}^P. \end{align*} Substituting this identity gives \begin{align*} d_n^Q\circ b_n &=s_{n-2}\circ d_{n-1}^P\circ d_n^P. \end{align*} Because $P_\bullet$ is a chain complex, $d_{n-1}^P\circ d_n^P=0$, so \begin{align*} d_n^Q\circ b_n=0. \end{align*} When $n=1$, the same formula is valid because $s_{-1}$ was defined to be the zero map. Thus $b_n$ factors through the inclusion $\iota_n:\ker d_n^Q\to Q_n$ as a map \begin{align*} \bar b_n:P_n\to\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*} Therefore \begin{align*} d_{n+1}^Q:Q_{n+1}\to\ker d_n^Q \end{align*} is surjective when regarded as a map onto $\ker d_n^Q$. Since $P_n$ is projective, the map $\bar b_n$ lifts through this surjection. Hence there exists an $R$-module homomorphism \begin{align*} s_n:P_n\to Q_{n+1} \end{align*} with \begin{align*} d_{n+1}^Q\circ s_n=\bar b_n. \end{align*} Viewing both sides as maps into $Q_n$, this says \begin{align*} d_{n+1}^Q\circ s_n=b_n=a_n-s_{n-1}\circ d_n^P. \end{align*} Rearranging gives exactly \begin{align*} a_n=d_{n+1}^Q\circ s_n+s_{n-1}\circ d_n^P. \end{align*} [/guided] [/step] [step:Assemble the homotopy between the original chain maps] By the degree-zero construction and the induction step, there exist $R$-module homomorphisms \begin{align*} s_n:P_n\to Q_{n+1} \end{align*} for every $n\geq 0$ such that \begin{align*} a_n=d_{n+1}^Q\circ s_n+s_{n-1}\circ d_n^P \end{align*} for every $n\geq 0$, with $s_{-1}:=0$. Since $a_n=g_n-h_n$, this becomes \begin{align*} g_n-h_n=d_{n+1}^Q\circ s_n+s_{n-1}\circ d_n^P \end{align*} for every $n\geq 0$. Hence $(s_n)_{n\geq 0}$ is a chain homotopy from $g_\bullet$ to $h_\bullet$. Therefore the two chain maps lifting $f:M\to N$ are chain homotopic. [/step]