"Abstract nonsense" proof of the splitting lemma

Solution 1:

Notations: The short exact sequence splits on the left

$$ 0 \longrightarrow A \underset{\underset{\phi}{\longleftarrow}}{\overset{f}{\longrightarrow}} B \overset{g}{\longrightarrow} C \longrightarrow 0 \quad, \quad \phi\circ f =\mathrm{id}_A$$

Let's show that $B \cong A\oplus C$

$\underline{1)}\ $● $\enspace \phi\circ f = \mathrm{id}_A\ \Rightarrow\ P:= f\circ \phi $ is an idempotent ($P\circ P=P$).

• Any arrow in an abelian category has an image so that $P$ splits into $ B \overset{j_P}{\longrightarrow} \mathrm{Im}(P) \overset{i_P}{\longrightarrow} B $ with $j_P$ epi and $i_P$ mono (cf. the equivalent def. of image and the property that $j_P$ is epi in categories with finite limits and colimits)

• $j_P$ is actually a left inverse of $i_P$: $$i_P\, j_P\,i_P\, j_P= P^2 = P = i_P\, j_P\quad\underset{i_p\ \text{mono}}{\overset{j_P\ \text{epi}}{\Longrightarrow}}\quad j_P\,i_P= \mathrm{id}_{\mathrm{Im}(P)}$$

• $j_P\circ f$ is actually an isomorphism $A \cong \mathrm{Im}(P)\ $: $f\circ \phi$ is a factorization of $P$ so that by the universal property of the image (in the sense of "initial factorization") there exists a unique morphism $v: \mathrm{Im}(P)\to A$ such that $i_P =f\, v$ and $\phi =v\, j_P$ whence

  $$(j_P\, f)\, v = j_P\, i_P = \mathrm{id}_{\mathrm{Im}(P)}\quad ,\quad v\,(j_P\, f)= \phi \, f = \mathrm{id}_A$$

(This is Proposition 10.2 p.12 from "Theory of Categories", Barry Mitchell: In a balanced category, if a morphism has an image and an (Epi-Mono) factorization then the mono is the image)

$\underline{2)} $ Hom sets of an abelian category are abelian groups so that it makes sense to define $$Q := \mathrm{id}_B -P\quad \in\big(\mathrm{Hom}(B,B),+\big)$$

As previously, $Q$ factorizes as $Q=i_Q \circ j_Q$ and with $i_P, i_Q$ mono, $j_P, j_Q$ epi $$ QP = QP = 0\quad \Longrightarrow\quad j_P\circ i_Q = 0 \enspace ,\enspace j_Q \circ i_P = 0 \qquad (Eq)$$

$\underline{3)} $ The two maps $\ i_P: \mathrm{Im}(P)\rightarrow B\ ,\ i_Q: \mathrm{Im}(Q)\rightarrow B $ define by the universal property of the coproduct a map $\Phi: \mathrm{Im}(P)\coprod \mathrm{Im}(Q) \rightarrow B$ (this could also be written $i_P+i_Q$, + symbolically if we were not in an abelian category )

and the two maps $j_P : B \rightarrow \mathrm{Im}(P)\ ,\ j_Q : B \rightarrow \mathrm{Im}(Q) $ a map $\Psi: B\rightarrow \mathrm{Im}(P)\prod \mathrm{Im}(Q)$

Using $j_P\circ i_{P} = \mathrm{id}_{\mathrm{Im}(P)},\ j_Q\circ i_Q = \mathrm{id}_{\mathrm{Im}(Q)},\ (Eq)$ and the definition of biproduct, one checks the isomorphism $B\cong \mathrm{Im}(P)\oplus \mathrm{Im}(Q)$, i.e. $$ \Phi\circ \Psi =P+Q = \mathrm{id}_B\quad \text{and} \quad \Psi\circ \Phi =\mathrm{id}_{\mathrm{Im}(P)\oplus\mathrm{Im}(Q)}$$

Solution 2:

Just saw this: there's a far simpler and more intuitive proof using the short five lemma.

Letting $i:A\to A\oplus C$, $j:C\to A\oplus C$, $p:A\oplus C\to A$, $q:A\oplus C\to C$ be the $4$ morphisms characterizing $A\oplus C$, consider the following diagram

Clearly, the upper short exact sequence splits on the left. $\phi$ is defined as the unique morphism such that $t=p\circ\phi$, $g=q\circ\phi$. It's easy to show that this diagram commutes. Thus, by the short five lemma, $\phi$ is an isomorphism.

A pretty similar proof works for splitting on the right as well.