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I am trying to understand the solution to the following exercise.

Let $\mathcal{C}$ be a category with idempotents $e:A\to A$ and $d:B\to B$, and a morphism $f: A \to B$.

The Karoubi envelope $\bar{\mathcal{C}}$ is a category which has as objects the idempotents of $\mathcal{C}$ and as morphism $e\to d$ those morphisms $f: \text{dom}\,e \to \text{dom}\, d$ in $\mathcal{C}$ for which $f\cdot e = f = d\cdot f$. The Karoubi envelope $\bar{\mathcal{C}}$ splits the idempotents in $\mathcal{C}$

We let the functor $E:\mathcal{C}\to \bar{\mathcal{C}}$ be the functor that maps $A\mapsto 1_A$, and the identity on morphisms. It embeds $\mathcal{C}$ in $\bar{\mathcal{C}}$.

Now if $e:A\to A$ is an idempotent in $\mathcal{C}$, in $\bar{\mathcal{C}}$ there is a pair of maps $\check{e}:1_A\to e$ and $\hat{e}: e\to 1_A$ which split the idempotent.

The claim is: any arbitrary functor $F:\mathcal{C}\to \mathcal{D}$ can be factored as $F=\tilde{F}E$ $\iff$ it sends idempotents of $\mathcal{C}$ to split idempotents of $\mathcal{D}$.


  • Forward direction: should be fairly trivial. All idempotents split in $\bar{\mathcal{C}}$, and functors preserve split idempotents.
  • Reverse direction: here I have some problems. For objects, I let $\tilde{F}A = \text{dom}(FA)$. This way $\tilde{F}(E(A)) = \tilde{F}(1_A) =\text{dom}(1_{FA})$ as required. For morphisms, since for each $f:A\to B$ in $\mathcal{C}$ there is a $f:e\to d$ in $\bar{\mathcal{C}}$, I would just set $\tilde{F}(f) = F(f)$. So $\tilde{F}(f:e\to d) = F(f:A\to B)$.

However, my solutions give the following.

Suppose $F$ splits idempotents $e$ as $\hat{e}\check{e}$. Then $\tilde{F}(f) = \check{d}\, F(f)\,\hat{e}$, where $f$ is defined as above. The solutions claim that it is the only way to chose $\tilde{F}$ such that it preserves splits. I do not understand why this is necessary. What's wrong with setting $\tilde{F}f = Ff$?

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Your definition of the morphisms in $\bar{\mathcal{C}}$ is incorrect. $\DeclareMathOperator{\dom}{dom}$ It consists of morphisms $f : \dom e \to \dom d$ in $\mathcal{C}$ such that $d f e = f$. This means $e : \dom e \to \dom e$ is the identity morphism in $\bar{\mathcal{C}}$. – Zhen Lin Oct 24 '12 at 6:50

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