Lemma 2

If in the commutative diagram of A-modules and homomorphisms

one has $\alpha \circ \alpha' = 0, \mathrm{Ker}\ \beta = \mathrm{Im}\ \beta'$ and if $M'$ is projective, there exists an A-homomorphism $u': M' \to N'$ such that $\beta' \circ u' = u \circ \alpha'$.

It is enough to put $k'' = u'', k = -u, f = 0, f' = 0$ and $u' = k'$ in Lemma 1.

Lemma 1 bis. — Consider a diagram of A-modules and homomorphisms

such that $f \circ \alpha' = \beta \circ f'$, $\mathrm{Ker}\, \alpha = \mathrm{Im}\, \alpha'$, $\beta \circ \beta' = 0$, and $f' = k \circ \alpha' + \beta' \circ k'$ and where $N''$ is injective. There exists an $A$-homomorphism $k'' : M'' \to N''$ such that $f = k'' \circ \alpha + \beta \circ k$.

In fact, put $g = f - \beta \circ k$, we have

This implies $\mathrm{Ker}\, g \supset \mathrm{Im}\, \alpha' = \mathrm{Ker}\, \alpha$. Since $N''$ is injective, there therefore exists (X, p. 16, remark) an $A$-homomorphism $k' : M'' \to N''$ such that $g = k'' \circ \alpha$, whence the lemma.

Lemma 2 bis. — If, in the commutative diagram of $A$-modules and homomorphisms

one has $\mathrm{Ker}\, \alpha = \mathrm{Im}\, \alpha', \beta \circ \beta' = 0$ and if $N''$ is injective, there exists an $A$-homomorphism $u'' : M'' \to N''$ such that $u'' \circ \alpha = \beta \circ u$. It is enough to put $u' = k', u = -k, f = 0, f' = 0$ and $k'' = u''$ in Lemma 1 bis.