Category Theory
Zulip Server
Archive

You're reading the public-facing archive of the Category Theory Zulip server.
To join the server you need an invite. Anybody can get an invite by contacting Matteo Capucci at name dot surname at gmail dot com.
For all things related to this archive refer to the same person.

Stream: learning: questions

Topic: Karoubian completion

Davit (May 05 2025 at 02:47):

Hi, I am trying to learn about Karoubian completions, and I don't manage to show that an idempotent (X,p) -q-> (X,p) always has a kernel.
Sure, (X,p) nor (X,q) work because qp and qq=q are nonzero, so I ran out of candidates for the kernel.
Any help please ?

Kevin Carlson (May 05 2025 at 04:25):

Hi Davit, this notation is really pretty ambiguous. If you could define what $(X,p)$ means, and especially what $\stackrel{q}{\to}$ means, it would be easier to see where you're stuck. And, in what category are you asking for this map to have a kernel?

Davit (May 05 2025 at 04:32):

Sorry for the uncompleteness of my question. So I start with a pre-additive category C, and take its Karoubian completion, which is a category whose objects are pairs (X,p) with X object of C and p:X->X an idempotent in C, and whose morphisms f : (X,p) -> (Y,q) are those morphisms f : X -> Y in C such that f = qf = fp. Composition is just the one in C.
So every textbooks and online articles say that in the Karoubian completion of a pre-additive category, every idempotent (so a morphism q : (X,p) -> (X,p) such that q^2=q) has a kernel (in the Karoubian completion).
So this is the kernel I am looking for.
Thank you.

Morgan Rogers (he/him) (May 05 2025 at 07:27):

Out of interest, where did you encounter the terminology "Karoubian"? I've seen "Karoubi completion" before, but this variant is new to me.

Davit (May 05 2025 at 07:35):

I think the term « Karoubian » mostly occurs in French litterature, like in the fourse on derived algebraic geometry of Sophie Morel.

Morgan Rogers (he/him) (May 05 2025 at 07:39):

Do they say "complétion Karoubienne"? Curious!

To answer the question, it might be useful to observe that:

• $q$ must be idempotent in the original category
• Assuming pre-additive means enriched in abelian groups (bad historical name), you can check that $1-q$ is also idempotent (for $1$ the identity on $X$).

Morgan Rogers (he/him) (May 05 2025 at 07:44):

(Those hints might be enough ;) )

Davit (May 05 2025 at 07:44):

How didn’t I think of it !! Yes (X,1-q) totally does the work !
Thank you so much !