The Stacks project

Proof. We will use the results of Lemma 46.4.3 without further mention. Denote $h : \textit{Alg}_ A \to \textit{Sets}$ the functor given by $h(C) = \mathop{\mathrm{Mor}}\nolimits _ A(B[N], C)$. Similarly for $h_1$ and $h_2$. The map $B[N] \to B[N_2]$ corresponding to the surjection $N \to N_2$ is surjective. It corresponds to a map $h_2 \to h$ such that $h_2(C) \to h(C)$ is injective for all $A$-algebras $C$. On the other hand, there are two maps $p, q : h \to h_1$, corresponding to the zero map $N_1 \to N$ and the injection $N_1 \to N$. Note that

is an equalizer diagram. Denote $\mathcal{O}_ h$ the module-valued functor $C \mapsto \bigoplus _{h(C)} C$. Similarly for $\mathcal{O}_{h_1}$ and $\mathcal{O}_{h_2}$. Note that

where $\mathcal{P}$ is the category of module-valued functors on $\textit{Alg}_ A$. We claim there is an equalizer diagram

in $\mathcal{P}$. Namely, suppose that $C \in \mathop{\mathrm{Ob}}\nolimits (\textit{Alg}_ A)$ and $\xi = \sum _{i = 1, \ldots , n} c_ i \cdot f_ i$ where $c_ i \in C$ and $f_ i : B[N] \to C$ is an element of $\mathcal{O}_ h(C)$. If $p(\xi ) = q(\xi )$, then we see that

where $z, y : B[N_1] \to B[N]$ are the maps $z : (b, m_1) \mapsto (b, 0)$ and $y : (b, m_1) \mapsto (b, m_1)$. This means that for every $i$ there exists a $j$ such that $f_ j \circ z = f_ i \circ y$. Clearly, this implies that $f_ i(N_1) = 0$, i.e., $f_ i$ factors through a unique map $\overline{f}_ i : B[N_2] \to C$. Hence $\xi $ is the image of $\overline{\xi } = \sum c_ i \cdot \overline{f}_ i$. Since $I$ is injective, it transforms this equalizer diagram into a coequalizer diagram

This diagram is compatible with the direct sum decompositions $I(B[N]) = I(B) \oplus TI(B, N)$ and $I(B[N_ i]) = I(B) \oplus TI(B, N_ i)$. The zero map $N \to N_1$ induces the zero map $TI(B, N) \to TI(B, N_1)$. Thus we see that the coequalizer property above means we have an exact sequence $TI(B, N_1) \to TI(B, N) \to TI(B, N_2) \to 0$ as desired. $\square$