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?
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 $\varphi:e\to e$.