Bourbaki

Exercise 1, § 1

Algebra, X, p. 168

Soit

M⟶N→tP⟶Q↓u↓v↓w↓sM′⟶N′→t′P′⟶Q′ \begin{array}{ccccccc} M &\longrightarrow& N &\xrightarrow{t}& P &\longrightarrow& Q\\ \downarrow u && \downarrow v && \downarrow w && \downarrow s\\ M'&\longrightarrow&N'&\xrightarrow{t'}&P'&\longrightarrow&Q' \end{array}

a commutative diagram of AA-modules in which the rows are exact; suppose that uu is surjective and ss is injective. Now prove that:

Ker⁡(w)=t(Ker⁡(v))Im⁡(v)=t′−1(Im⁡(w)). \operatorname{Ker}(w)=t(\operatorname{Ker}(v))\qquad \operatorname{Im}(v)={t'}^{-1}(\operatorname{Im}(w)).
Solution

No solution has been written yet.

Machine translation of the English, by gpt-5-mini. Not checked by a person. That model is a cut down version of the one the rest of this language was written by.