I have seen quoted here in a few places that having a sequence:
$$0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0$$ with maps $f: A \rightarrow B$ and $g: B \rightarrow C$ is split exact if and only if there exist maps $h: C \rightarrow B$ and $i: B \rightarrow C$ such that
$$\begin{align*} g \circ f = 0\\
g \circ h = Id_C\\ i \circ f = Id_A\\ h \circ g + f \circ i = Id_B. \end{align*}$$
The only thing I'm having trouble seeing is how the last equation is implied by the sequence being split exact, any help would be greatly appreciated.
Best Answer
If the sequence is split exact, that means that (up to isomorphism) we have $B=A\oplus C$ with $f$ the inclusion and $g$ the projection. Taking $h$ and $i$ to be the other inclusion and other projection, then $$(h\circ g+f\circ i)(a,c)=h(g(a,c))+f(i(a,c))=h(c)+f(a)=(0,c)+(a,0)=(a,c)$$ for any $(a,c)\in B$. Thus $h\circ g+f\circ i=Id_B$,