Splitting lemma for many (at least 3) components I am interested in such version of splitting lemma:
So given short exact sequence
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we have three equivalent statements:

*

*short exact sequence is right split, i.e there is map $t: B\to A$ such that $tq$ is the identity on $A$.

*short exact sequence is left split, i.e there is map $u: C\to B$ such that $ru$ is the identity on $C$.

$\hskip2.5in$

*

*There exists an isomorphism from $B$ to $A\oplus C$ such that $q$ and $r$ correspond to natural inclusion and projection respectively. (We may assume here that we are in some fixed abelian category and now the last statement means that $B$ is isomorphic to the biproduct of $A$ and $C$).


Can you see any nice looking modification of splitting lemma so that the statement "$B$ is isomorphic to direct sum of $B_1,\dots,B_n$" would be one of the equivalent conditions?

The point is that I like to do the following things:
Start with surjective mapping $r:B\to C,$ set $A=\ker r$ and $q:A\hookrightarrow B$ to be an inclusion. Then for obtained exact sequence
$$0\to A\hookrightarrow B \stackrel{r}\to C\to 0$$
I try to find left splitting $u:C\to B.$ If I succeed I have second projection $t:B\to A$ given by $$t=I_B-ru.$$
The advantage of this approach is that most of the things you just check algebraically.
So it would be nice to know if I can show that $B=B_1\oplus\dots\oplus B_n$ finding just splittings. Even more, obtaining in return projections on components with explicit formulas.
Btw. I know that I can do one splitting and then repeat the procedure for components, but maybe there is some sneaky way to grasp it globally.
 A: Here is the closest analogue I can see.  Suppose you have a finite set of objects $C_1,\dots C_n$, another object $B$, and maps $u_i:C_i\to B$ and $r_i:B\to C_i$ such that $r_iu_i=1_{C_i}$ for all $i$ and $r_ju_i=0$ for all $i\neq j$.  Let $q:A\to B$ be a kernel of the map $\sum u_ir_i:B\to B$.  Then there is a map $t:B\to A$ such that $tq=1_A$, and there is an isomorphism $B\cong A\oplus \bigoplus C_i$ that turns all of these maps into the inclusions and projections.  Furthermore, $t$ is obtained by simply factoring the map $1_B-\sum u_ir_i$ through $q$.
(This corresponds to the "left split" case of the splitting lemma; for the "right split" case, take a cokernel of $\sum u_ir_i$ instead of a kernel.)
Of course, if you let $C=\bigoplus C_i$, this is just a rehashing of the usual splitting lemma.   The hypotheses on the $r_i$ and $u_i$ say exactly that they combine to maps $r:B\to C$ and $u:C\to B$ such that $ru=1_C$, and the definition of $A$ says that we have an exact sequence $0\to A\to B\to C$.
