Lemma 12.4.2. Let $\mathcal{C}$ be a preadditive category. The following are equivalent

1. $\mathcal{C}$ is Karoubian,

2. every idempotent endomorphism of an object of $\mathcal{C}$ has a cokernel, and

3. 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. $\square$

Comment #539 by Nuno on

Minor typo: "induce an isomorphsm"

Comment #9223 by on

One could add another equivalent property:

(4) Every idempotent endomorphism $e$ is of the form $e=i\circ p$, where $p:z\to x$ and $i:x\to z$ is a section of $p$.

Proof. (3)$\Rightarrow$(4). Clear.

(4)$\Rightarrow$(3). Suppose that $p:z\to x$ is a retraction of $i:x\to z$. Then $i p$ is an idempotent endomorphism and thus so is $1_z-i p$. Hence, $1_z-i p=j q$, where $q:z\to y$ and $j:y\to z$ is a section of $q$. We already know that $1_z=ip+jq$, $pi=1_x$, $qj=1_y$, so by Remark 12.3.6 it suffices to see $pj=0$, $qi=0$. Note $ipj=(1_z-jq)j=j-jqj=j-j=0$, thus $pj=0$ by monicity of $i$. Analogously, $qi=0$.

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