Lemma 10.131.6. In diagram (10.131.4.1), suppose that $S \to S'$ is surjective with kernel $I \subset S$. Then $\Omega _{S/R} \to \Omega _{S'/R'}$ is surjective with kernel generated as an $S$-module by the elements $\text{d}a$, where $a \in S$ is such that $\varphi (a) \in \beta (R')$. (This includes in particular the elements $\text{d}(i)$, $i \in I$.)

**Proof.**
Consider the map of presentations above. Clearly the right vertical map of free modules is surjective. Thus the map is surjective. A diagram chase shows that the following elements generate the kernel as an $S$-module for sure: $i\text{d}a, i\in I, a \in S$, and $\text{d}a$, with $a \in S$ such that $\varphi (a) = \beta (r')$ for some $r' \in R'$. Note that $\varphi (i) = \varphi (ia) = 0 = \beta (0)$, and that $\text{d}(ia) = i\text{d}a + a \text{d}i$. Hence $i\text{d}a = \text{d}(ia) - a \text{d}i$ is an $S$-linear combination of elements of the second kind.
$\square$

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