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Stream: practice: terminology & notation

Topic: Morphisms that are "as mono as each other"

Amar Hadzihasanovic (Sep 16 2024 at 20:19):

Consider the following equivalence relation on morphisms with the same domain $c$ in a category: $f \sim g$ if and only if, for all parallel pairs $h, k$ with codomain $c$, we have $fh = fk$ if and only if $gh = gk$.

Then $f \sim \mathrm{id}_c$ if and only if $f$ is mono.

Is there a name for this equivalence relation, or for the property of belonging to the same equivalence class?

John Baez (Sep 16 2024 at 20:52):

I don't know a name for that equivalence relation, but I think it would be useful to consider the preorder on morphisms with domain $c$ where $g \le f$ iff for all parallel pairs $h,k$ with codomain $c$, we have $f h = fk \implies gh = gk$.

Roughly this says "$f$ is at least as monic as $g$ is".

John Baez (Sep 16 2024 at 20:53):

Then $f \sim g$ iff $f \le g$ and $g \le f$.

John Baez (Sep 16 2024 at 20:58):

Note that if $g$ factors through $f$ then $g \le f$. But I don't see why $g \le f$ should imply that $g$ factors through $f$. Am I missing a trick? If not, there seem to be two different preorders here: the "factoring through" preorder and the one described above.

Kevin Carlson (Sep 16 2024 at 21:13):

Isn't the name for this equivalence relation "$f$ and $g$ have the same kernel pair" in the case kernel pairs exist?

Kevin Carlson (Sep 16 2024 at 21:18):

And yes, these are different preorders. John's preorder $g\le f$ is called "$\mathrm{ker} f \le \mathrm{ker} g$", especially in abelian-land. And it's easy for two things to even have the same kernel while not factoring through each other. You can also see the orders are different from Amar's observation that all monos are $\le$ each other though of course they don't all factor through each other.

Kevin Carlson (Sep 16 2024 at 21:23):

Put yet another way, two morphisms that have the same kernel have isomorphic images, but not necessarily isomorphic as subobjects of the codomain, which is what morphisms that factor through each other get.

Amar Hadzihasanovic (Sep 16 2024 at 21:40):

Yes, you are right about "having the same kernel pair" when kernel pairs exist. Now I'm wondering if there is a good way to extend the terminology in a compatible way when that's not the case.

Kevin Carlson (Sep 16 2024 at 21:54):

Freyd uses the dual relation in Concreteness. He calls an equivalence class a generalized regular subobject, so you could call an equivalence class for this relation a generalized regular quotient. That doesn't immediately suggest an obviously better name for this relation than "representing the same regular quotient", though...

fosco (Sep 16 2024 at 21:55):

See Remark 2.4 here https://arxiv.org/pdf/1704.00303

This (and the dual) condition was first introduced by John Isbell in 1964 and used by PJ Freyd in his concreteness papers

fosco (Sep 16 2024 at 21:55):

Kevin Carlson said:

Freyd uses the dual relation in Concreteness. He calls an equivalence class a generalized regular subobject, so you could call an equivalence class for this relation a generalized regular quotient. That doesn't immediately suggest an obviously better name for this relation than "representing the same regular quotient", though...

ouch, for a split second!

fosco (Sep 16 2024 at 21:56):

(we use the co-Freyd condition in our "homotopical algebra is not concrete" paper)

Kevin Carlson (Sep 16 2024 at 21:58):

Maybe you could get away with "equifibered".

Mike Shulman (Sep 17 2024 at 01:51):

As usual, even if kernel pairs don't exist in $C$, they exist in $[C^{\rm op},\rm Set]$, and this preorder can equivalently be detected by those non-representable kernel pairs. So I think "having the same kernel" can be used even when the category doesn't have kernels.