Pseudo-Abelian Completion in the constrution of Motifs (by Y. Manin)

Question

I reading Yuri Manin's famous paper on "CORRESPONDENCES, MOTIFS AND MONOIDAL TRANSFORMATIONS" and struggle with his definition for so called pseudo-abelian completion given on page 453 by a reason I would like to explain below. first of all the setting:

Definition 1: An additive category $\mathscr{D}$ is called pseudo-abelian if the following condition holds in it:

(P)For any projector $p \in Hom(X,X)$ (recall projector = idempotent map satisfying $p=p^2$) there exists a kernel $Ker p$, and the canonical isomorphism $Ker(p) \oplus Ker(id_X-p) \cong X$.

Definition 2: Let $\mathscr{D}$ be an additive category. Its pseudo-abelian completion is the category $\tilde{\mathscr{D}}$ defined by the following data:

-Objects $Ob(\tilde{\mathscr{D}})$ are pairs $(X,p)$ with $X \in Ob(\mathscr{D})$ and $p$ projector

-Morphisms are defined by

$$Hom_{\tilde{\mathscr{D}}}((X,p),(Y,q)):= \\ \Big\{\text{group of } f \in Hom_{\mathscr{D}}(X,Y) \text{ with } f \circ p=q \circ f \Big\} \Big / \Big\{\text{ the subgroup of } f \text{ for } \\ \text{ which } f \circ p=q \circ f=0 \Big\}$$

Then it is claimed that the "category" $\tilde{\mathscr{D}}$ is pseudo-abelian.

Problem 1: I'm not satisfied with the definition of morphisms. Essentially what is here done one associate to a category $\mathscr{D}$ a Karoubian hull. See here: https://ncatlab.org/nlab/show/Karoubian+category#properties or here: https://en.wikipedia.org/wiki/Pseudo-abelian_category#Pseudo-abelian_completion for details.

And I think that it not suffice to demand that the morphisms in $\tilde{\mathscr{D}}$ are $f \in Hom_{\mathscr{D}}(X,Y) \text{ with } f \circ p=q \circ f$ modulo certain relation.

I think it's neccessary to require that these morphism have to satisfy $f \circ p=q \circ f= f$ as well.

Why I think so: Let $(X, p)$ an object in this category and the question is what is the identity $id_{(X,p)}$ of it? If we only demand that the morphisms have to satisfy $f \circ p=q \circ f$, then we have a uniqueness problem of the identity:

$id_X$ and $p$ behave both well as identities of $(X,p)$ as one can easily check. And if $p \neq id_X$ that is a problem. I don't also see how the moduled out relations can repair this leck of uniqueness.

But if we require instead the extended relation $f \circ p=q \circ f= f$, we see that $p$ and not $id_X$ is unique identity $id_{(X,p)}$ on $(X,p)$.

Problem 2: What is the reason for taking quotient by $\{\text{ the subgroup of } f \text{ for which } f \circ p=q \circ f=0 \}$? The classical description of Karoubian completion also not require it.

Answer 1

This is an equivalent way of describing the same thing. To see that, notice that for any $f$ such that $q\circ f=f\circ p$, $$( f\circ p - q\circ f\circ p)\circ p =q\circ( f\circ p - q\circ f\circ p)=0$$ Therefore $[f\circ p]=[q\circ f\circ p]$ in our homomorphism group. Similarly, $[f]=[f\circ p]$ in our homomorphism group. Hence every homomorphism is represented by an element of the form $g=q\circ f\circ p$( and hence $q\circ g=g\circ p=g$).

On the other hand, if two maps $f,g:X\to Y$ are such that $q\circ f= f\circ p=f$, $q\circ g= g\circ p=g$, and represent the same homomorphism, then by definition $q\circ (f-g)=(f-g)\circ p=0$. Since composition is bilinear, this forces $f=g$.

So, the answer to your questions are 1.) this is a different way of describing the same thing, and 2.) to make the problems you described in 1. go away.