One calls a category $\mathcal{C}$ karoubian if it is additive and for $X$ every idempotent map $e: X \to X$ (idempotent means $e = e^2$) splits, therefore there exist $Y$, $p:X\to Y$ and $q:Y \to X$ with properties \begin{equation} \tag{$\ast$} p \circ q = \mathrm{id}_Y \quad\text{and}\quad q \circ p = e. \end{equation}

Consider to an (from now) arbitrary $\mathcal{C}$ category the category $P(\mathcal{C})$ with

  • objects: idempotent morphisms in $\mathcal{C}$,

  • morphisms: for $e:X \to X$, $f: Y \to Y \in P(\mathcal{C})$ the maps $\psi: X \to Y$ with $f \circ \psi = \psi \circ e$.

My question is how to see that $P(\mathcal{C})$ is a karoubian category?

My attempts:

Let call the property \begin{equation} \tag{$\ast\ast$} f \circ \psi = \psi \circ e. \end{equation}

Fix a idempotent $e: X \to X$ and let $\psi: X \to X$ be idempotent in $P(\mathcal{C})$. Especially $\psi$ is idempotent in $\mathcal{C}$ and fulfills $(**)$.

Therefore we are looking for idempotent $f:Y \to Y$ and $p: X \to Y$ and $q: Y \to X$ satisfying $(*)$ and $(**)$ simultanteously.

In order to fulfill $(*)$ and $(**)$ I tried it with $Y = X$ and $f = \mathrm{id}_X$ since it seems to me beeing “canonic” in some way. But here occurs the problem for the choice of $p$ and $q$. By $(*)$ they must be different.

Can anybody help me in finding $p$ and $q$? Do I need here the property of additive categories?

By the way: Is the choice for $Y$ and $f$ as above correct?

  • $\begingroup$ I don't understand, is $\cal C$ Karoubian? If yes, just split the idempotent $\psi$... $\endgroup$ – fosco Sep 7 '18 at 8:09
  • $\begingroup$ No, from 5. line $\mathcal{C}$ is arbitrary. Otherwise the statement would be indeed trivial :) Sorry for ambiguous formulation, I fixed it. $\endgroup$ – KarlPeter Sep 7 '18 at 11:51

First, $(**)$ for $\psi$ being $e\to f$ in $P(\mathcal C) $ implies $f\psi=f\psi e=\psi e$, because $f\psi e=\psi ee=\psi e$.
Second, we also need the condition that $f\psi=\psi=\psi e$, so that the new objects will have identities (for an idempotent $e$, its identity will be itself, so we need $\psi e=\psi$ to hold for every arrow $\psi:e\to f$).

Now let $\varphi:e\to e$ be an idempotent in $P(\mathcal C)$ where $e:A\to A$ is idempotent in $\mathcal C$, and $e\varphi e=\varphi$.
Then it is also an idempotent morphism $A\to A$ in $\mathcal C$, hence it is an object of $P(\mathcal C)$, with two natural morphisms $\varphi:e\to \varphi$ and $\varphi:\varphi\to e$ which will actually split $\varphi:e\to e$.

  • $\begingroup$ Hi. Thank you for your answer. The point concerning the identity of a $e \in P(C)$ seems unclear to me: If $e: X \to X \in P(C)$ why it's identity map $id_e \in Hom_{P(C)}(e,e)$ is $e$ itself and not the identity map $id_X \in Hom_C(X,X)$? Indeed, $id_X$ would trivialy fulfill $\psi id_X = \psi$ for all $\psi$. But according your argument $\psi e = \psi$ only holds if we already know that $e$ is the identity of $e:X \to X \in P(C)$, right? Or is there an error in my reasonings? $\endgroup$ – KarlPeter Sep 7 '18 at 12:31
  • $\begingroup$ Well, the way you defined $P(C)$ indeed $1_A$ would be the identity on $e$, but then splitting an idempotent in $P(C)$ will be literally the same, hence equivalent to splitting it in $C$, so this result wouldn't hold. $\endgroup$ – Berci Sep 7 '18 at 14:50
  • $\begingroup$ Yes, that's my thinking barrier in this example. In your first comment you showed that $e$ as morphism in $P(C)$ is the identity of $e: A \to A$ (here $e$ as element) (so $e =id_e$). On the other hand it seems that $id_A = 1_A$ fulfills obviously also the property of a identity. By uniqueness of identity we must have $e = id_A$ BUT only as morphisms in $Hom_{P(C)}(e,e)$. But then - as you correctly pointed it out in your last comment - splittings in $C$ and $P(C)$ should be the same. But that seems absurd since $C$ is an arbitrary category, $P(C)$ is (or what we intend to show) a karoubian. $\endgroup$ – KarlPeter Sep 7 '18 at 15:19
  • $\begingroup$ But this is strange since in ncatlab.org/nlab/show/Karoubian+category the caroubian envelope of $C$ is constructed in this way. I think that here something is running wrong. Do you see what? $\endgroup$ – KarlPeter Sep 7 '18 at 15:22
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    $\begingroup$ It writes $f=q\circ f\circ p$ for the condition, which is just my first point.. $\endgroup$ – Berci Sep 7 '18 at 16:55

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