The splitting lemma for short exact sequences of modules

Suppose $s$ is a section and define $\mathrm{\Phi }:A\oplus C\to B$ by $\mathrm{\Phi }(a,c)=i(a)+s(c)$; [L2] makes this a homomorphism.

Suppose assertion 3 holds. Define $r:B\to A$ as the first coordinate of ${\mathrm{\Phi }}^{-1}$; then $\mathrm{\Phi }(a,0)=i(a)$ gives $r(i(a))=a$, so $r$ is a retraction.

Suppose instead that $r$ is a retraction. For each $c\in C$, choose any $b$ with $p(b)=c$ using surjectivity and put $k=b-i(r(b))$. Then $r(k)=0$ and $p(k)=c$.

For $b\in B$, the element $b-s(p(b))$ lies in $\ker p=\mathrm{im}i$, so by injectivity of $i$ there is a unique $a\in A$ with $i(a)=b-s(p(b))$; hence $b=\mathrm{\Phi }(a,p(b))$ and $\mathrm{\Phi }$ is surjective.

If $\mathrm{\Phi }(a,c)=0$, applying $p$ gives $c=0$, and then injectivity of $i$ gives $a=0$; thus $\mathrm{\Phi }$ is injective and satisfies the compatibility conditions in assertion 3.

The element $k$ of step 1.3 is unique in $\ker r$ with image $c$: if $k^{\mathrm{\prime }}\in \ker r$ and $p(k^{\mathrm{\prime }})=c$, then $k-k^{\mathrm{\prime }}\in \ker p=\mathrm{im}i$, say $k-k^{\mathrm{\prime }}=i(a)$, and applying $r$ gives $a=0$. Therefore the rule $s(c)=k$ is independent of the temporary lift $b$, is linear by uniqueness, and satisfies $p\circ s={\mathrm{id}}_C$.

Steps 1.1, 2.1, and 2.2 prove $1\Rightarrow 3$, step 1.2 proves $3\Rightarrow 2$, and steps 1.3 and 2.3 prove $2\Rightarrow 1$. The formula for $\mathrm{\Phi }$ also yields the internal direct sum $B=i(A)\oplus s(C)$.