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: practice: terminology & notation

Topic: Functors sending morphisms to isomorphisms

Josselin Poiret (Jun 30 2023 at 15:52):

Is there a specific adjective for functors that send any morphism to an isomorphism? I want to be able to talk about it in the following way: the functor $(x,y) \mapsto F(x,y)$ is ? in $x$.

Morgan Rogers (he/him) (Jun 30 2023 at 15:57):

As in $F$ is a functor with two variables, and it's morphisms in the first variable which are sent to isomorphisms?

Josselin Poiret (Jun 30 2023 at 16:04):

yes, as in $F(f, id_y)$ is always an isomorphism

Josselin Poiret (Jun 30 2023 at 16:05):

it's in between being simply functorial and "parametric" (which would be that $F(x,y) \simeq F(x', y)$ for all $x,x',y$)

John Baez (Jun 30 2023 at 22:58):

I've never heard of a name of functors that map all morphisms to isomorphisms (much less do so in one argument). They are extremely rare in my experience except when the target is a groupoid. So, I'm not surprised that I don't know of a name for such functors. I guess you can make one up. "Isomorphizing?" As in "the functor $(x,y) \to F(x,y)$ is iromorphizing in $x$". But whatever name you use, you have to explain it.

Mike Shulman (Jun 30 2023 at 23:07):

Or "inverting" or "localizing", perhaps.

John Baez (Jun 30 2023 at 23:46):

By the way, I wouldn't make up a name unless I needed to say this > 3 times. It's pretty quick to say "$F$ maps morphisms in the first variable to isomorphisms".

Josselin Poiret (Jul 01 2023 at 06:42):

I'm still at an exploratory stage but I do think I need a small name for it, because I'll need some notation to go with it as well! Thanks for the ideas, I think localizing might be the more understandable word.

Martti Karvonen (Jul 01 2023 at 12:54):

If you had a single-variable functor with this property, you could also say that the functor actually lands in the core of the codomain.

Josselin Poiret (Jul 01 2023 at 14:51):

Martti Karvonen said:

If you had a single-variable functor with this property, you could also say that the functor actually lands in the core of the codomain.

right, but here it's for multivariable functors, I really want an adjective describing some action, and not constrain the codomain

Heiko Braun (Jul 02 2023 at 00:25):

Let $\mathit{F} \colon \mathit{X} \times \!\mathit{Y} \to \mathit{Z}$. You can say that $\mathit{x} \mapsto (\mathit{y} \mapsto F(\mathit{x}, \mathit{y}))$ lands in the core of $\mathsf{Fun}(\!\mathit{Y}, \mathit{Z}\,)$. Otherwise you can also define the subcategory $\mathit{W} \subseteq \mathit{X} \times \!\mathit{Y}$ consisting of morphisms of the type $(\,\mathit{f}, 1_ \mathit{y}\,)$ and say that the functor is $\mathit{W}$-localizing as suggested above. This also be somewhat standard.

Jonathan Beardsley (Jul 10 2023 at 22:03):

I am looking at Brian Day's thesis, and he is describing taking coends of functors, but it seems like his variance is not the standard one:
image.png
does anyone know about this?

Jonathan Beardsley (Jul 10 2023 at 22:04):

he seems to be looking at a tensor product of two different functors A→V, rather than A→V and A°→V

Kevin Arlin (Jul 10 2023 at 22:57):

I think the notation is that $F(AA-)$ is a functor that's contravariant in one $A,$ covariant in another, and also maybe there are more variables in the $-$. And $T=S\otimes R$ further down needs $S$ to be contravariant in $A$ and $R$ covariant, or vice versa.

Jonathan Beardsley (Jul 10 2023 at 23:00):

Yeah, right, okay. Where he says "different variances in A" or whatever. Hrmph. This doesn't seem to line up with what he does elsewhere.

Kevin Arlin (Jul 11 2023 at 00:13):

I think it should always be interpretable as the thing it's supposed to be but if you see somewhere else that looks wrong I'd be happy to take a look. Day wasn't always the easiest to read.

Jonathan Beardsley (Jul 11 2023 at 06:39):

Thanks. Yeah I'll find it tomorrow. It's the conditions he writes down for a functor to be promonoidal. They involve various tensor products like this and I can't parse them.