Lemma 40.5.1. The map $I/I^2 \to J/J^2$ induced by $c$ is the composition

where the second arrow comes from the equality $J = (I \otimes B + B \otimes I)C$. The map $i : B \to B$ induces the map $-1 : I/I^2 \to I/I^2$.

Lemma 40.5.1. The map $I/I^2 \to J/J^2$ induced by $c$ is the composition

\[ I/I^2 \xrightarrow {(1, 1)} I/I^2 \oplus I/I^2 \to J/J^2 \]

where the second arrow comes from the equality $J = (I \otimes B + B \otimes I)C$. The map $i : B \to B$ induces the map $-1 : I/I^2 \to I/I^2$.

**Proof.**
To describe a local homomorphism from $C$ to another local ring it is enough to say what happens to elements of the form $b_1 \otimes b_2$. Keeping this in mind we have the two canonical maps

\[ e_2 : C \to B,\ b_1 \otimes b_2 \mapsto b_1s(e(b_2)),\quad e_1 : C \to B,\ b_1 \otimes b_2 \mapsto t(e(b_1))b_2 \]

corresponding to the embeddings $R \to R \times _{s, U, t} R$ given by $r \mapsto (r, e(s(r)))$ and $r \mapsto (e(t(r)), r)$. These maps define maps $J/J^2 \to I/I^2$ which jointly give an inverse to the map $I/I^2 \oplus I/I^2 \to J/J^2$ of the lemma. Thus to prove statement we only have to show that $e_1 \circ c : B \to B$ and $e_2 \circ c : B \to B$ are the identity maps. This follows from the fact that both compositions $R \to R \times _{s, U, t} R \to R$ are identities.

The statement on $i$ follows from the statement on $c$ and the fact that $c \circ (1, i) = e \circ t$. Some details omitted. $\square$

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