Bourbaki

Example 2

Tag 0DZH, permanent. Cite it as [Bourbaki, Tag 0DZH].

Algebra, chapter X, § 7. Produit de composition

If u∈HomA(N,P)u \in \mathrm{Hom}_A(N, P), v∈ExtA(M,N)v \in \mathrm{Ext}_A(M, N), then

u∘v=ExtA(lM,u)(v)∈ExtA(M,P): u \circ v = \mathrm{Ext}_A(l_M, u)(v) \in \mathrm{Ext}_A(M, P):

analogously, if u∈ExtA(N,P)u \in \mathrm{Ext}_A(N, P), v∈HomA(M,N)v \in \mathrm{Hom}_A(M, N), then

u∘v=ExtA(v,lP)(u)∈ExtA(M,P). u \circ v = \mathrm{Ext}_A(v, l_P)(u) \in \mathrm{Ext}_A(M, P).

This follows from the definitions and the remarks of X. p. 88.

If Q,M,N,PQ, M, N, P are four left AA-modules, and if

u∈ExtA(N,P),v∈ExtA(M,N),w∈ExtA(Q,M), u \in \mathrm{Ext}_A(N, P), \quad v \in \mathrm{Ext}_A(M, N), \quad w \in \mathrm{Ext}_A(Q, M),

then (u∘v)∘w=u∘(v∘w)(u \circ v) \circ w = u \circ (v \circ w): the composition product is associative; we will therefore denote the composition of several elements without parentheses. In particular, by Example 2:

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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