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Stream: learning: questions

Topic: endomorphisms of a presheaf

Joshua Meyers (Jan 04 2022 at 21:07):

Is there a presheaf $X:C^{\textrm op}\to {\sf Set}$ which has non-monic endomorphisms but no non-epic endomorphisms? (Cross-posted in a special case to Math.SE https://math.stackexchange.com/q/4348894/90658)

Morgan Rogers (he/him) (Jan 04 2022 at 22:43):

For any C whatsoever?

Morgan Rogers (he/him) (Jan 04 2022 at 22:50):

Assuning the answer to the above is yes: Consider the monoid M with generators a,b,c subject to the equation ab = ac, so that a is not monic. It's tedious but straightforward to verify that all elements of this monoid are epic. The right action of M on itself is a presheaf on M, and the endomorphisms monoid of this presheaf is again M.

Morgan Rogers (he/him) (Jan 04 2022 at 22:51):

I'll copy that to MO tomorrow, assuming it answers your question.

Zhen Lin Low (Jan 04 2022 at 22:54):

Does that really work? The Yoneda embedding doesn't preserve general epimorphisms, only split epimorphisms.

Morgan Rogers (he/him) (Jan 04 2022 at 22:55):

I've used this same trick with @Jens Hemelaer to answer some problems regarding endomorphism monoids before, it's pretty neat ;)

Morgan Rogers (he/him) (Jan 04 2022 at 22:56):

Ah I suppose they might no longer be epic in the presheaf topos, but i had assumed the epicness was only required in the endomorphism monoid

Morgan Rogers (he/him) (Jan 04 2022 at 23:06):

What you're suggesting makes more sense @Zhen Lin Low as what Joshua was probably after; I'll have to think harder about that

Joshua Meyers (Jan 04 2022 at 23:32):

Morgan Rogers (he/him) said:

For any C whatsoever?

Ideally I'm interested in the case where ${\sf Ob}(C)=A\sqcup B$ and all non-identity morphisms are from an object in $A$ to and object in $B$ (presheafs of these categories model instances of relational signatures). But I'll take an example for any $C$

Joshua Meyers (Jan 05 2022 at 00:44):

BTW, here is the version I posted on Math.SE, in case it seems more tangible to anyone:

Let $G$ be an infinite graph (directed, loops okay, no multi-edges, so essentially a set with a binary relation). $G$ is called core if every one of its endomorphisms is surjective. Does this ever happen in a non-trivial way, i.e. where $G$ is core and not all its endomorphisms are isomorphisms? I am aware that the answer is no for finite graphs, hence I am just asking about the infinite case.

Joshua Meyers (Jan 05 2022 at 02:58):

I found an example! Take the category with objects $X,R,A,B$ and morphisms $r_1,r_2:X\to R$, $a:X\to A$, and $b:X\to B$. Take the presheaf $I$ defined by $IX=\{\ldots,-3,-2,-1,0,1,1',2,2',\ldots\}$, $IR=\{(n,n-1),(n',(n-1)'),(1',0)\mid n\in\mathbb{Z}\}$, $IA=\{\ldots,-3,-2,-1,0,1,2,3,\ldots\}$, $IB=\{\ldots,-3,-2,-1,0,1',2',3',\ldots\}$, and by letting morphisms map to the canonical projections and inclusions.

Joshua Meyers (Jan 05 2022 at 03:02):

Still not sure how to do an example that satisfies the narrower Math.SE question

Morgan Rogers (he/him) (Jan 05 2022 at 10:45):

Consider the naturals with an edge $n \to n+1$, an edge $0 \to 0$ and no ther edges. Endomorphisms can only move things down.

Morgan Rogers (he/him) (Jan 05 2022 at 12:20):

Hmm but then we have an endomorphism sending everything to $0$....

Joshua Meyers (Jan 05 2022 at 16:05):

Yeah there can't be any loops for this reason

Morgan Rogers (he/him) (Jan 06 2022 at 17:28):

I found an example (post as an answer on the m.SE question)

Joshua Meyers (Jan 06 2022 at 17:53):

Great!