# For every exact sequence $0 \to N \xrightarrow{f} M \xrightarrow{g} L \to 0$ the following are equivalent

For every exact sequence $$0 \to N \xrightarrow{f} M \xrightarrow{g} L \to 0$$ are equivalent:

a) $$f$$ is a split monomorphism

b) $$g$$ is a split epimorphism

c) $$\ker g$$ is a direct summand of $$M$$

d) M $$\cong N \times L$$

b) $$\implies$$ c)

Its because how $$f$$ is a split homomorphism then exist $$f':M \to N$$ such that $$f'f=1_{N}$$

Then $$f'$$ is epi, if $$x=f(y) \in\ker f' \cap \operatorname{im}f$$ then

$$0=f'(x)=ff(y)=y\text{ and }x=f(y)=0$$

if $$x\in N$$ then

$$f'(x-ff'(x))=f(x)-f(x)=0\text{ and }x=(x-ff'(x)+ff'(x)) \in \ker f' \cap \operatorname{im}f$$

And the same for b) implies c)

But im stuck in a) implies b) and c) implies d). Any hint?

• Commented Dec 4, 2022 at 7:21
• @MarianoSuárez-Álvarez Isn't the given equivalence just the splitting lemma (at least (a),(b) and (d))? What makes the usual argument not work with these assumptions? Commented Dec 4, 2022 at 19:45
• @MarianoSuárez-Álvarez Ah, that make sense and goes well with the argument I had in mind. This is an interesting, though in hindsight really unsurprising, caveat. Commented Dec 5, 2022 at 6:26

As mentioned in the commets by Mariano Suárez-Álvarez, you have to be careful how to state this theorem. If stated correctly, this is (or at least part of the) Splitting Lemma. The part missing here is that the isomorphism $$M\cong N\times L$$ has to fit into the right diagrams, i.e. is not abstract but compatible with the initial short exact sequence

$$\require{AMSc} 0 \to N \xrightarrow{f} M \xrightarrow{g} L \to 0$$

meaning that composed with $$f$$ and $$g$$ corresponds precisely the the inclusion and projections of $$N\times L$$ (you can find the precise wording on Wikipedia).

Let's talk about the the points (a),(b) and (d). In fact, (a) and (b) behave very similar (being dual statements of each other). I will give you the idea of how to prove the equivalence of (a) and (d). You should stare at the following diagram:

$$\require{AMScd} \begin{CD} 0 @>>> N @>{f}>> M @>{g}>> L @>>>0 \\ @. @V{1_N}VV @. @V{1_L}VV \\ 0 @>>> N @>{\iota}>> N\times L @>{\pi}>> L @>>>0 \end{CD}$$

If given either a splitting of $$f$$ (or a splitting of $$g$$) or a compatible isomorphism $$h\colon M\to N\times L$$, you can complete this diagram with a vertical arrow $$M\to N\times L$$ or $$N\times L\to M$$. Then some diagram chasing will give you the desired conclusion.

The equivalence of (b) and (c) can be approached in a similar manner by looking at $$M\cong\ker g\oplus K$$ (note that $$\ker g\cong N$$ and $$K\cong L$$ by exactness). In encourage you to find the correct diagrams to consider in these cases.