I have a couple of questions about a exercise I have:
Let $0\to L\stackrel{\alpha}\to M\stackrel{\beta}\to N \to 0$ be a splitting exact sequence and let $r$ be a retraction of $\alpha$ such that $r\circ\alpha=id_{L}$. I want to know more about the kernel of $r$. Now $\alpha$ is injective by definition and $r$ is surjective by construction. So the kernel must have the zero element of $M$, now I would like to know what $r$ is doing with the elements of $M$ outside $\alpha(L)$? Any help?
Take $\xi \in \alpha(L) \cap \ker(r)$. Then $\xi = \alpha(x)$ for some $x \in L$. Applying $r$ on both sides we get $0=r(\xi)=x$ and so $\xi=0$. Next, let $y \in M$. Then the element $y' = y - a(r(y))$ is in the kernel of $r$.
This shows that $M = \alpha(L) \oplus \ker(r)$.
Hence if $y \not\in \alpha(L)$, then $y = \alpha(x) + \xi$, where $x \in L, 0 \neq \xi \in \ker(r)$. So $r(y) = x$. Notice that $x$ is uniquely defined by $y$.