# A question on short exact sequences

Let $0\to L\stackrel{\alpha}\to M\stackrel{\beta}\to N\to 0$ be a splitting short exact sequence so it exist a morphism $r: M \to L$ such that $r \circ \alpha = Id_L$ and a morphism $s: N \to M$ such that $\beta \circ s = Id_N$. Is it true that then $0\to N\stackrel{s}\to M\stackrel{r}\to L\to 0$ is also a short exact sequence?

I have proved that it is at the endpoints, that is, $r$ is surjective and $s$ injective. But I cant prove that $\text{Im}(s)=\ker(r)$. Can anyone help me?

Facts that may be useful and that I have proved: $M= \alpha(L) \oplus \ker(r) = \alpha(L) \oplus s(N)$, also keep in mind that $\alpha$ is injective and $\beta$ is surjective.

It so many functions and relations to keep in mind, I cant make it to prove that the sequence is short exact, help please.

The result you are trying to prove is not true. This is because the section is not unique. Suppose $M$ is $\mathbb{Z}\times\mathbb{Z}$ and the other two groups are $\mathbb{Z}$. If the kernel of the projection is generated by $(1,0)$, meaning $\beta((1,0))=0$, then $1\mapsto (0,1)$ and $1\mapsto (1,1)$ are both valid sections. Those elements are mapped to the same element by $\beta$.
• Can you explain what you mean with your last sentence, what does "if the kernel of the projection is generated by $(1,0)$" mean? – user117449 Nov 12 '14 at 19:48
• 1 can be mapped to multiple possible elements, yielding multiple possible right inverses of $\beta$. If one of the sections makes the sequence exact, then none of the other choices do because their images are different. – Matt Samuel Nov 12 '14 at 20:32