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
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
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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