TITLE: Why is the Category of Correspondences not pseudo abelian?
QUESTION [5 upvotes]: i've just gotten into the theory of motives.I understand the construction of the Karoubian envelope (pseudo-abelian completion) to ensure that morphisms have kernels and images in order to get certain decompositions of the diagonal of Motives in $\mathcal{M}_k$.But i don't see why the morphisms in $Corr_k$ which are elements of $CH(X \times Y)$ for the objects $X,Y$ don't have kernels and/or images in general.
A second one: In the category of effective Chow-Motives, do you know an example of non isomorphic varieties,that have the same motive,except for a totally split quadric and projective spaces?

REPLY [2 votes]: Answer 1. Take the projective line $\mathbb{P}^{1}_{k}$, and let $p$ be the correspondence projecting to a point. Then this has an image, namely $\mathrm{Spec}(k)$. However, it does not have a kernel. One can see this by looking at cohomological realisations. If the kernel existed, it would only have an $\mathrm{H}^{2}$; since the $\mathrm{H}^{0}$ is accounted for by $\textrm{Spec}(k)$.
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I realise I am made this more difficult then necessary. Suppose that all idempotents have kernels. Then the category of correspondences is equivalent to its Karoubian envelope. In the Karoubian envelope we have a decomposition $\mathbb{P}^{1}_{k} = \textrm{ker}(p) \oplus \textrm{ker}(1 - p)$. The direct sum in this category is given by the disjoint union of the underlying schemes and correspondences. Under our assumption of equivalence with the category of correspondences, this exhibits the connected scheme $\mathbb{P}^{1}_{k}$ as the disjoint union of $\textrm{Spec}(k)$ and some other scheme. Contradiction.
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Answer 2. Since the Hom-sets have $\mathbb{Q}$-coefficients, every isogeny becomes an isomorphism. But maybe you find this answer a bit cheating.