I've read about split idempotents, and I know they are morphisms $e:X\to X$ such that there exists $Y$ and morphisms $i:Y\to X$, $p:X\to Y$ with $pi=\mathrm{id}_Y$, $ip=e$.
In the setting of triangulated categories (which are additive but not abelian), many sources give this notion (or at least a notion with the same name) saying that $e$ provides a splitting of $X=\mathrm{Im}(e)\oplus\mathrm{Ker}(e)$. That would means $\mathrm{Ker}(e)$ and $\mathrm{Im}(e)$ exist, and give a decomposition of $X$.
I don't know how to prove that a splitting idempotent $e$ has a kernel and an image, and neither that these give such a decomposition of $X$.
My intuition says that $Y$ should be the image of $e$, but I'm stuck trying to compute the kernel and the cokernel. It seems $X$ with morphism $\mathrm{id}_X-e$ works as cocone for both the pullback a the pushout diagrams defining kernel and cokernel of $e$, but is it limiting? My geometric intuition of an idempotent as a projection endomorphism of, say, a vector space, says not, and I don't know how to proceed.
Thanks in advance for any help
Ok, it has to be said slighly better. In a category where any idempotent split, for any idempotent $e:X\to X$ there is a canonincal decomposition $X=W\oplus Y$ where Y is the splitting of $e$ (image of $e$), and $W$ is the splitting of $\mathrm{id}_X-e$ (the kernel of $e$).
Canonical monomorphisms of the two splits provide the inclusion morphisms for the coproduct, and for morphisms $h:Y\to Z$ and $h':W\to Z$ the unique morphism $X\to Z$ proving $X$ to be actually the coproduct is the sum $h'p'+hp$, where $p$ and $p'$ are the epimorphisms of the two splits.