The Stacks project

Proof. Let $\delta _1, \delta _2 \in \Gamma _0$ correspond to $\theta _1, \theta _2$ as above and the composition $\delta = \delta _1 \circ \delta _2$ in $\Gamma _0$ correspond to $\theta $. We have to show that $\theta = \theta _1 + \theta _2$. Recall (More on Morphisms of Spaces, Lemma 76.17.2) that $\theta _1, \theta _2, \theta $ correspond to derivations $D_1, D_2, D : e_0^{-1}\mathcal{O}_{R_0} \to \mathcal{C}_{U_0/U}$ given by $D_1 = \theta _1 \circ \text{d}_{R_0/U_0}$ and so on. It suffices to check that $D = D_1 + D_2$.

We may check equality on stalks. Let $\overline{u}$ be a geometric point of $U$ and let us use the local rings $A, B, C$ introduced in Section 79.4. The morphisms $\delta _ i$ correspond to ring maps $\delta _ i : B \to A$. Let $K \subset A$ be the ideal of square zero such that $A/K = \mathcal{O}_{U_0, \overline{u}}$. In other words, $K$ is the stalk of $\mathcal{C}_{U_0/U}$ at $\overline{u}$. The fact that $\delta _ i \in \Gamma _0$ means exactly that $\delta _ i(I) \subset K$. The derivation $D_ i$ is just the map $\delta _ i - e : B \to A$. Since $B = s(A) \oplus I$ we see that $D_ i$ is determined by its restriction to $I$ and that this is just given by $\delta _ i|_ I$. Moreover $D_ i$ and hence $\delta _ i$ annihilates $I^2$ because $I = \mathop{\mathrm{Ker}}(I)$.

To finish the proof we observe that $\delta $ corresponds to the composition

where the first arrow is $c$ and the second arrow is determined by the rule $b_1 \otimes b_2 \mapsto \delta _2(t(\delta _1(b_1))) \delta _2(b_2)$ as follows from (79.5.0.1). By Lemma 79.4.1 we see that an element $\zeta $ of $I$ maps to $\zeta \otimes 1 + 1 \otimes \zeta $ plus higher order terms. Hence we conclude that

However, by Lemma 79.5.1 the action of $\delta _2 \circ t$ on $K = \mathcal{C}_{U_0/U, \overline{u}}$ is the identity and we win. $\square$