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Stream: learning: questions

Topic: The left orthogonal of monos is not always made of epis


view this post on Zulip fosco (Dec 30 2025 at 10:33):

The usual definition of the class of "strong epimorphisms" in a completely general category is Epis:=Epi(Mono)Epi^s := Epi\cap {}^\perp(Mono): a "strong epi" in CC is an epi p:EBp : E \to B which, in addition, is also left orthogonal to all monos.

The conditions under which asking pp to be an epi is redundant are, however, extremely weak: it is enough for CC to have equalizers, and in fact it seems to be enough that the diagonal BB×BB\to B\times B exists, and that pp is orthogonal to it, to ensure pp is an epimorphism.

How to build a counterexample, then? Can one find a (possibly finite) category CC such that there exists a pp, which is not an epimorphism, and yet p(Mono)p\in {}^\perp(Mono)?

Does this work? Take the free category on the graph [image], subject to the relations fp=gpf\circ p=g\circ p, pap=pbpp\circ a_p=p\circ b_p, faf=fbff\circ a_f = f\circ b_f, gag=gbgg\circ a_g=g\circ b_g, chosen in order to make pp not an epi, and p,f,gp,f,g not monos; in particular, gafgbfg\circ a_f\ne g\circ b_f, fagfbgf\circ a_g\neq f\circ b_g, etc.: so, pp is not an epi, and there are no nontrivial lifting problems to solve against the monos (for example, the arrows ax,bx:Zx...a_x,b_x : Z_x \to ... are vacuously mono, but hom(E,Zx)\hom(E,Z_x) is empty).

Is there a simpler counterexample?

image.png

view this post on Zulip Ivan Di Liberti (Dec 30 2025 at 10:34):

Man if you dont have products, are you even a category?

view this post on Zulip fosco (Dec 30 2025 at 10:38):

Products, in this economy!

view this post on Zulip Morgan Rogers (he/him) (Dec 31 2025 at 09:24):

I don't think you need the objects ZxZ_x to be distinct, so you could have fewer arrows, but conceptually this is pretty much the simplest example: you need enough arrows around to stop the relevant ones from being epi/mono, after all.