[proofplan] Choose a projective resolution of $M$ and compute $\operatorname{Ext}_R^1(M,N)$ as the first cohomology of the complex $\operatorname{Hom}_R(P_\bullet,N)$. An extension determines a cocycle by lifting the map $P_0\to M$ through the extension and measuring the failure of that lift to vanish on the first syzygies. Conversely, a cocycle $c:P_1\to N$ builds an extension by quotienting $N\oplus P_0$ by the relations $d_1x=-c(x)$. These two constructions are inverse to each other, and addition of cocycles matches Baer sum. [/proofplan]

[step:Fix a projective resolution and identify the cocycles computing $\operatorname{Ext}_R^1(M,N)$] Choose a projective resolution of $M$ by left $R$-modules: \begin{align*} \cdots \xrightarrow{d_3} P_2 \xrightarrow{d_2} P_1 \xrightarrow{d_1} P_0 \xrightarrow{\varepsilon} M \to 0. \end{align*} Thus each $P_k$ is projective, $\varepsilon$ is surjective, $\ker \varepsilon=\operatorname{im}d_1$, and $\ker d_1=\operatorname{im}d_2$. Apply the contravariant functor $\operatorname{Hom}_R(-,N)$ to the resolution. Define \begin{align*} C^0&:=\operatorname{Hom}_R(P_0,N),& C^1&:=\operatorname{Hom}_R(P_1,N),& C^2&:=\operatorname{Hom}_R(P_2,N), \end{align*} and define the coboundary maps \begin{align*} \delta^0:C^0&\to C^1,& a&\mapsto a\circ d_1,\\ \delta^1:C^1&\to C^2,& c&\mapsto c\circ d_2. \end{align*} Since $d_1\circ d_2=0$, we have $\delta^1\circ\delta^0=0$. Hence \begin{align*} \operatorname{Ext}_R^1(M,N) = \ker(\delta^1)/\operatorname{im}(\delta^0). \end{align*} Thus an element of $\operatorname{Ext}_R^1(M,N)$ is represented by an $R$-[linear map](/page/Linear%20Map) $c:P_1\to N$ satisfying $c\circ d_2=0$, with two representatives $c,c':P_1\to N$ identified when there exists an $R$-linear map $a:P_0\to N$ such that \begin{align*} c'-c=a\circ d_1. \end{align*} [/step]

[step:Send an extension to the cocycle measuring a projective lift] Let \begin{align*} \xi:\quad 0 \to N \xrightarrow{i} E \xrightarrow{\pi} M \to 0 \end{align*} be a short exact sequence of left $R$-modules. Since $P_0$ is projective and $\pi:E\to M$ is surjective, there exists an $R$-linear map $s:P_0\to E$ such that \begin{align*} \pi\circ s=\varepsilon. \end{align*} Because $\varepsilon\circ d_1=0$, we have \begin{align*} \pi\circ s\circ d_1=\varepsilon\circ d_1=0. \end{align*} Thus $s\circ d_1$ has image in $\ker\pi=\operatorname{im}i$. Since $i:N\to E$ is injective, there is a unique $R$-linear map $c_\xi:P_1\to N$ satisfying \begin{align*} i\circ c_\xi=s\circ d_1. \end{align*} Now \begin{align*} i\circ c_\xi\circ d_2=s\circ d_1\circ d_2=0. \end{align*} Since $i$ is injective, $c_\xi\circ d_2=0$. Therefore $c_\xi\in\ker\delta^1$. If $s':P_0\to E$ is another lift of $\varepsilon$, then $\pi\circ(s'-s)=0$, so $s'-s$ has image in $\operatorname{im}i$. Hence there is a unique $R$-linear map $a:P_0\to N$ such that \begin{align*} s'-s=i\circ a. \end{align*} The corresponding cocycle $c_\xi'$ satisfies \begin{align*} i\circ c_\xi' =s'\circ d_1 =s\circ d_1+i\circ a\circ d_1 =i\circ(c_\xi+a\circ d_1). \end{align*} Injectivity of $i$ gives $c_\xi'=c_\xi+a\circ d_1$, so $c_\xi$ and $c_\xi'$ define the same class in $\operatorname{Ext}_R^1(M,N)$. If two extensions are equivalent through an isomorphism $\varphi:E\to E'$ commuting with $N$ and $M$, then $\varphi\circ s$ is a lift for the second extension and produces the same cocycle. Hence the rule \begin{align*} [\xi]\mapsto [c_\xi] \end{align*} is a well-defined map $\operatorname{YExt}_R^1(M,N)\to\operatorname{Ext}_R^1(M,N)$. [/step]

[step:Construct an extension from a cocycle] Let $c:P_1\to N$ be an $R$-linear map satisfying $c\circ d_2=0$. Define a submodule $S_c\subset N\oplus P_0$ by \begin{align*} S_c:=\{(-c(x),d_1(x)):x\in P_1\}. \end{align*} Define the left $R$-module \begin{align*} E_c:=(N\oplus P_0)/S_c, \end{align*} and let $q_c:N\oplus P_0\to E_c$ be the quotient map. Define $R$-linear maps \begin{align*} i_c:N&\to E_c,& n&\mapsto q_c(n,0),\\ \pi_c:E_c&\to M,& q_c(n,p)&\mapsto \varepsilon(p). \end{align*} The map $\pi_c$ is well-defined because if $(n,p)$ is changed by $(-c(x),d_1(x))$, then $\varepsilon(p+d_1(x))=\varepsilon(p)$. We prove exactness. The equality $\pi_c\circ i_c=0$ follows from $\varepsilon(0)=0$. The map $\pi_c$ is surjective because $\varepsilon:P_0\to M$ is surjective. If $q_c(n,p)\in\ker\pi_c$, then $\varepsilon(p)=0$, so $p=d_1(x)$ for some $x\in P_1$. Therefore \begin{align*} q_c(n,p) =q_c(n,d_1(x)) =q_c(n+c(x),0) =i_c(n+c(x)), \end{align*} so $\ker\pi_c\subset\operatorname{im}i_c$. The reverse inclusion follows from $\pi_c\circ i_c=0$. It remains to show that $i_c$ is injective. Suppose $i_c(n)=0$. Then $(n,0)\in S_c$, so there exists $x\in P_1$ such that \begin{align*} (n,0)=(-c(x),d_1(x)). \end{align*} Thus $d_1(x)=0$. Since $\ker d_1=\operatorname{im}d_2$, there exists $y\in P_2$ with $d_2(y)=x$. Hence \begin{align*} n=-c(x)=-c(d_2(y))=0, \end{align*} because $c\circ d_2=0$. Therefore \begin{align*} 0\to N\xrightarrow{i_c}E_c\xrightarrow{\pi_c}M\to0 \end{align*} is a short exact sequence. [/step]

[step:Show cohomologous cocycles give equivalent extensions] Suppose $c,c':P_1\to N$ are cocycles and that there exists an $R$-linear map $a:P_0\to N$ such that \begin{align*} c'=c+a\circ d_1. \end{align*} Define an $R$-linear map \begin{align*} T:N\oplus P_0&\to N\oplus P_0,& (n,p)&\mapsto (n-a(p),p). \end{align*} For $x\in P_1$, \begin{align*} T(-c(x),d_1(x)) =(-c(x)-a(d_1(x)),d_1(x)) =(-c'(x),d_1(x)). \end{align*} Thus $T(S_c)\subset S_{c'}$, and $T$ descends to an $R$-linear map $\overline{T}:E_c\to E_{c'}$. The inverse is induced by \begin{align*} (n,p)\mapsto (n+a(p),p), \end{align*} so $\overline{T}$ is an isomorphism. Moreover \begin{align*} \overline{T}\circ i_c=i_{c'} \quad\text{and}\quad \pi_{c'}\circ\overline{T}=\pi_c. \end{align*} Therefore cohomologous cocycles determine equivalent extensions, and the construction gives a well-defined map \begin{align*} \operatorname{Ext}_R^1(M,N)\to\operatorname{YExt}_R^1(M,N). \end{align*} [/step]

[step:Verify that the two constructions are inverse] Start with an extension \begin{align*} 0\to N\xrightarrow{i}E\xrightarrow{\pi}M\to0 \end{align*} and choose a lift $s:P_0\to E$ of $\varepsilon:P_0\to M$. Let $c_\xi:P_1\to N$ be the associated cocycle, so $i\circ c_\xi=s\circ d_1$. Construct $E_{c_\xi}$. Define \begin{align*} \Theta:E_{c_\xi}&\to E,& q_{c_\xi}(n,p)&\mapsto i(n)+s(p). \end{align*} This is well-defined because for $x\in P_1$, \begin{align*} i(-c_\xi(x))+s(d_1(x))=-i(c_\xi(x))+i(c_\xi(x))=0. \end{align*} The map $\Theta$ commutes with the maps from $N$ and to $M$. It is an isomorphism by exactness: surjectivity follows because every $e\in E$ has $\pi(e)=\varepsilon(p)$ for some $p\in P_0$, so $e-s(p)\in\ker\pi=\operatorname{im}i$; injectivity follows from the same computation used to prove exactness of $E_c$. Hence the reconstructed extension is equivalent to the original one. Conversely, start with a cocycle $c:P_1\to N$. In the extension \begin{align*} 0\to N\xrightarrow{i_c}E_c\xrightarrow{\pi_c}M\to0, \end{align*} use the lift \begin{align*} s_c:P_0&\to E_c,& p&\mapsto q_c(0,p). \end{align*} Then for $x\in P_1$, \begin{align*} s_c(d_1(x)) =q_c(0,d_1(x)) =q_c(c(x),0) =i_c(c(x)). \end{align*} Thus the cocycle recovered from $E_c$ is exactly $c$. The two maps are inverse bijections. [/step]

[step:Identify addition with Baer sum] Let $c,b:P_1\to N$ be cocycles. The extension associated to $c+b$ is \begin{align*} 0\to N\xrightarrow{i_{c+b}}E_{c+b}\xrightarrow{\pi_{c+b}}M\to0. \end{align*} The Baer sum of the extensions $E_c$ and $E_b$ is obtained by first taking the pullback over the diagonal map $M\to M\oplus M$ and then pushing out along the addition map $N\oplus N\to N$. Its middle term is represented by pairs $(e,e')\in E_c\oplus E_b$ satisfying $\pi_c(e)=\pi_b(e')$, modulo the relation \begin{align*} (i_c(n),-i_b(n))\sim 0 \quad\text{for all } n\in N. \end{align*} Define \begin{align*} \Lambda:E_{c+b}&\to E_c+_{\mathrm{Baer}}E_b,& q_{c+b}(n,p)&\mapsto [(q_c(n,p),q_b(0,p))]. \end{align*} For $x\in P_1$, the defining relation of $E_{c+b}$ maps to \begin{align*} (q_c(-(c+b)(x),d_1(x)),q_b(0,d_1(x))) =(i_c(-b(x)),i_b(b(x))), \end{align*} which is zero in the pushout quotient. Hence $\Lambda$ is well-defined. It commutes with the inclusion of $N$ and the projection to $M$, and the standard representatives in the pullback-pushout construction show that it is an isomorphism of extensions. Therefore addition of cocycles corresponds exactly to Baer sum. The zero cocycle gives the split extension $0\to N\to N\oplus M\to M\to0$, and replacing $c$ by $-c$ gives the Baer inverse. Hence the bijection is an isomorphism of abelian groups. The construction uses only pullbacks, pushouts, projective lifts, and composition with module homomorphisms, so it is natural in both $M$ and $N$. This proves the theorem. [/step]