Definition 12.4.1. Let $\mathcal{C}$ be a preadditive category. We say $\mathcal{C}$ is Karoubian if every idempotent endomorphism of an object of $\mathcal{C}$ has a kernel.
12.4 Karoubian categories
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The dual notion would be that every idempotent endomorphism of an object has a cokernel. However, in view of the (dual of the) following lemma that would be an equivalent notion.
Lemma 12.4.2. Let $\mathcal{C}$ be a preadditive category. The following are equivalent
$\mathcal{C}$ is Karoubian,
every idempotent endomorphism of an object of $\mathcal{C}$ has a cokernel, and
given an idempotent endomorphism $p : z \to z$ of $\mathcal{C}$ there exists a direct sum decomposition $z = x \oplus y$ such that $p$ corresponds to the projection onto $y$.
Proof. Assume (1) and let $p : z \to z$ be as in (3). Let $x = \mathop{\mathrm{Ker}}(p)$ and $y = \mathop{\mathrm{Ker}}(1 - p)$. There are maps $x \to z$ and $y \to z$. Since $(1 - p)p = 0$ we see that $p : z \to z$ factors through $y$, hence we obtain a morphism $z \to y$. Similarly we obtain a morphism $z \to x$. We omit the verification that these four morphisms induce an isomorphism $x = y \oplus z$ as in Remark 12.3.6. Thus (1) $\Rightarrow $ (3). The implication (2) $\Rightarrow $ (3) is dual. Finally, condition (3) implies (1) and (2) by Lemma 12.3.10. $\square$
Lemma 12.4.3. Let $\mathcal{D}$ be a preadditive category.
If $\mathcal{D}$ has countable products and kernels of maps which have a right inverse, then $\mathcal{D}$ is Karoubian.
If $\mathcal{D}$ has countable coproducts and cokernels of maps which have a left inverse, then $\mathcal{D}$ is Karoubian.
Proof. Let $X$ be an object of $\mathcal{D}$ and let $e : X \to X$ be an idempotent. The functor
is representable if and only if $e$ has a kernel. Note that for any abelian group $A$ and idempotent endomorphism $e : A \to A$ we have
where
Moreover, $\Phi $ has the right inverse
Hence (1) holds. The proof of (2) is dual (using the dual definition of a Karoubian category, namely condition (2) of Lemma 12.4.2). $\square$
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