For the sake of concreteness, let's restrict discussion to the category of abelian groups. Throughout, $$ 0 \to A \overset{q}{\to} B \overset{r}{\to}C \to 0$$ is a short exact sequence. One part of the splitting lemma states:
Proposition 1: If there exists a map $t : B \to A$ such that $tq = \mathrm{id}_A$, then there also exists a map $u: C \to B$ such that $r u = \mathrm{id}_C$.
However, as the above proposition is stated, the map $u$ is not unique! For example, if \begin{align*} A = C = \mathbb{Z} && B = \mathbb{Z} \times \mathbb{Z} &&q(a) = (a,0) && r(a,c) = c && t(a,c) = a \end{align*}
then $u(b) = (b,b)$ is just as legitmate a choice as the more obvious $u(b) = (0,b)$. But, the proof of Proposition 1 on wikipedia is constructive and produces a particular $u$. This particular $u$ is unique if you impose one more condition. To be explicit, we have the following:
Proposition 2: If $t : B \to A$ is a map such that $tq = \mathrm{id}_A$, then there is a unique map $u: C \to B$ such that $r u = \mathrm{id}_C$ and $$ 0 \leftarrow A \overset{t}{\leftarrow} B \overset{u}{\leftarrow}C \leftarrow 0$$ is an exact sequence.
Proof: The conclusions above just say that $u$ should be an isomorphism of $C$ onto $\ker(t) \subset B$ with inverse $r \big\vert_{\ker(t)}$, so $u$ is uniquely determined. To see $u$ exists, we just need to check the restriction of $r$ is an isomorphism $\ker(t) \to C$ is an isomorphism. Injectivity: if $x \in \ker(t)$ also belongs to $\ker(r)$, then $x = q(a)$ for $a \in A$ and $a = tq(a) = t(x) = 0$ so that $x = 0$. Surjectivity: if $c \in C$ is arbitrary, then choose any $b \in B$ with $r(b) = c$ and then note $b' = b - qt(b)$ has $r(b') = r(b) - rqt(b) = c$ but also $t(b') = t(b) - tqt(b) = t(b) - t(b) = 0$.
My question is, why does the splitting lemma not bother with this uniqueness statement? It seems to come at essentially no extra cost... A few possiblities:
- Is Proposition 2 simply not very useful?
- Does Proposition 2 fail in more general categories?
Good finding!
It does hold in more general categories as well.
The reason might be that if $0\to A\to B\to C\to 0$ splits (in either side), then we get $B\cong A\oplus C$, and the original sequence is equivalent to $$0\to A\to A\oplus C\to C\to 0$$ which is easier to work with, generally.