Remark 91.2.6. Let $A \to B \to C$, $M$, $N$, $c : M \to N$, $0 \to M \to B' \to B \to 0$, $\xi \in \mathop{\mathrm{Ext}}\nolimits ^1_ B(\mathop{N\! L}\nolimits _{B/A}, M)$, $0 \to N \to C' \to C \to 0$, and $\zeta \in \mathop{\mathrm{Ext}}\nolimits ^1_ C(\mathop{N\! L}\nolimits _{C/A}, N)$ be as in Lemma 91.2.5. Using pushout along $c : M \to N$ we can construct an extension

$\xymatrix{ 0 \ar[r] & N \ar[r] & B'_1 \ar[r] & B \ar[r] & 0 \\ 0 \ar[r] & M \ar[u]^ c \ar[r] & B' \ar[u] \ar[r] & B \ar[u] \ar[r] & 0 }$

by setting $B'_1 = (N \times B')/M$ where $M$ is antidiagonally embedded. Using pullback along $B \to C$ we can construct an extension

$\xymatrix{ 0 \ar[r] & N \ar[r] & C' \ar[r] & C \ar[r] & 0 \\ 0 \ar[r] & N \ar[u] \ar[r] & B'_2 \ar[u] \ar[r] & B \ar[u] \ar[r] & 0 }$

by setting $B'_2 = C' \times _ C B$ (fibre product of rings). A simple diagram chase tells us that there exists an $A$-algebra map $B' \to C'$ compatible with $B \to C$ and $c$ if and only if $B'_1$ is isomorphic to $B'_2$ as $A$-algebra extensions of $B$ by $N$. Thus to see Lemma 91.2.5 is true, it suffices to show that $B'_1$ corresponds via the bijection of Lemma 91.2.3 to the image of $\xi$ by the map $\mathop{\mathrm{Ext}}\nolimits ^1_ B(\mathop{N\! L}\nolimits _{B/A}, M) \to \mathop{\mathrm{Ext}}\nolimits ^1_ B(\mathop{N\! L}\nolimits _{B/A}, N)$ and that $B'_2$ correspond to the image of $\zeta$ by the map $\mathop{\mathrm{Ext}}\nolimits ^1_ C(\mathop{N\! L}\nolimits _{C/A}, N) \to \mathop{\mathrm{Ext}}\nolimits ^1_ B(\mathop{N\! L}\nolimits _{B/A}, N)$. The first of these two statements is immediate from the construction of the class in Remark 91.2.4. For the second, choose a commutative diagram

$\xymatrix{ Q \ar[r]_\beta & C \\ P \ar[u]^\varphi \ar[r]^\alpha & B \ar[u] }$

of $A$-algebras, such that $\alpha$ is a presentation of $B$ over $A$ and $\beta$ is a presentation of $C$ over $A$. See Remark 91.2.4 and references therein. Set $J = \mathop{\mathrm{Ker}}(\alpha )$ and $K = \mathop{\mathrm{Ker}}(\beta )$. The map $\varphi$ induces a map of complexes $\mathop{N\! L}\nolimits (\alpha ) \to \mathop{N\! L}\nolimits (\beta )$ and in particular $\bar\varphi : J/J^2 \to K/K^2$. Choose $A$-algebra homomorphism $\beta ' : Q \to C'$ which is a lift of $\beta$. Then $\alpha ' = (\beta ' \circ \varphi , \alpha ) : P \to B'_2 = C' \times _ C B$ is a lift of $\alpha$. With these choices the composition of the map $K/K^2 \to N$ induced by $\beta '$ and the map $\bar\varphi : J/J^2 \to K/K^2$ is the restriction of $\alpha '$ to $J/J^2$. Unwinding the constructions of our classes in Remark 91.2.4 this indeed shows that $B'_2$ correspond to the image of $\zeta$ by the map $\mathop{\mathrm{Ext}}\nolimits ^1_ C(\mathop{N\! L}\nolimits _{C/A}, N) \to \mathop{\mathrm{Ext}}\nolimits ^1_ B(\mathop{N\! L}\nolimits _{B/A}, N)$.

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