Bourbaki

Exercise 1, § 1

Commutative Algebra, I, 39

In the commutative diagram (10) suppose that the ordered pair (u′,u′)(u', u') is an exact sequence and that v∘u=0v \circ u = 0. Show that

Im⁡(b)∩Im⁡(u′)=b(Ker⁡(c∘u))andKer⁡(b)+Ker⁡(v)=b−1(Im⁡(u′∘a)). \operatorname{Im}(b) \cap \operatorname{Im}(u') = b(\operatorname{Ker}(c \circ u)) \quad \text{and} \quad \operatorname{Ker}(b) + \operatorname{Ker}(v) = b^{-1}(\operatorname{Im}(u' \circ a)).
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Transcribed from the printed volume. Not translated.