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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?
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You should settle on one convention for the order of composition of functions. In your statement of Proposition 1, $qt$ means to first apply $q$ and then $t$, but $ru$ means to first apply $u$ and then $r$. –  Andreas Blass May 8 at 23:08
    
@AndreasBlass: That's an error, thanks for spotting it. My intention was to use the usual "compose right to left" convention. Should be fixed now. –  Mike F May 8 at 23:12

1 Answer 1

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.

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Thanks for the reply! Viewed this way, though, one still seems to miss out on a slightly stronger statement. Given an exact sequence $0 \to A \overset{q}{\to} B \overset{r}{\to} C \to 0$ and a map $t : B \to A$ such that $tq = \mathrm{id}_A$, there is precisely one isomorphism of $B$ with $A \oplus C$ such that $q$ is identified with the inclusion $A \to A \oplus C$, $r$ is identified with the projection $A \oplus C \to C$ and $t$ is identified with the projection $A \oplus C \to A$. –  Mike F May 8 at 23:22
    
Yes, exactly. And then your finding can be seen as a consequence of this. –  Berci May 8 at 23:22

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