How do I know the derived category is NOT abelian? I have heard the claim that the derived category of an abelian category is in general additive but not abelian.  If this is true there should be some toy example of a (co)kernel that should be there but isn't, or something to that effect (for that matter, I could ask the same question just about the homotopy category).  
Unless I'm mistaken, the derived category of a semisimple category is just a ℤ-graded version of the original category, which should still be abelian.  So even though I have no reason to doubt that this is a really special case, it would still be nice to have an illustrative counterexample for, say, abelian groups.
 A: The following nicely does the trick I think...
Lemma Every monomorphism in a triangulated category splits.
Proof: Let $T$ be a triangulated category and suppose that $f\colon x\to y$ is a monomorphism. Complete this to a triangle
$x \stackrel{f}{\to} y \stackrel{g}{\to} z \stackrel{h}{\to} \Sigma x$
then $f\circ \Sigma^{-1}h = 0$ as we can rotate backward and maps in triangles compose to zero. Since $f$ is a monomorphism we deduce that $\Sigma^{-1}h$ and hence $h$ are zero. But this implies that $y\cong x\oplus z$ (a proof of this can be found in the first part of my answer here so that $f$ is a split monomorphism. █
Since every kernel is a monomorphism we get the following counterexample. The map
$\mathbb{Z}/p^2\mathbb{Z} \to \mathbb{Z}/p\mathbb{Z}$
does not have a kernel in $D(Ab)$ by virtue of the fact that $\mathbb{Z}/p^2\mathbb{Z}$ is indecomposable. Of course the same thing works in the homotopy category.
A: Even the homotopy category K(A) is not abelian. given any f in C(A) the 1: cone(f) ---> cone(f) is homotopic to 0 morphism. but homotopy for their kernels doesn't make sense.
Well, this is kind of wague, but can be made precise.
