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Stream: theory: category theory

Topic: constant + cocontinuous

Tom Hirschowitz (Jun 02 2021 at 10:22):

(Sorry, I can't resist mimicking @Reid Barton's last topic, but + doesn't mean the same thing.)

What would you call a functor which is a coproduct of a constant functor and a cocontinuous one?

Calling them affine appears to conflict with existing terminology, but is the latter sufficiently global for this to be a problem?

Morgan Rogers (he/him) (Jun 02 2021 at 10:26):

What's the intuition behind thinking of the functors you're describing as affine?

Fawzi Hreiki (Jun 02 2021 at 10:32):

I guess because an affine transformation is a linear transformation + a constant

Fawzi Hreiki (Jun 02 2021 at 10:32):

But I don't really see how co-continuous functors are like linear transformations.

Tom Hirschowitz (Jun 02 2021 at 10:38):

Tough question, @Morgan Rogers (he/him) :sweat:

They pop up in a work in progress with @Ambroise, but the intuition is in progress too, I'm afraid.

The rough idea is that you start with a free monad $S = \Sigma^\star$, and you want to define an operation by induction on $S(0)$, i.e., a morphism $F(S(0)) \to S(0)$ for some endofunctor $F$. We take $F(X)$ to have the shape $G(X,X)$ for some $G$ which is cocontinuous in its first argument, and (cheating a bit) describe the inductive definition as a natural transformation $G(\Sigma(Y),X) \to S ( G(Y,X) + Y + X )$, where the $\ldots + Y + X$ is used for describing base cases. And the functor $Y \mapsto G(Y,X) + Y + X$ is affine, which is used a lot in the development (for now).

Tom Hirschowitz (Jun 02 2021 at 10:39):

@Fawzi Hreiki I'm guessing @Morgan Rogers (he/him) means the intuition for such functors, not for their candidate name.

Tom Hirschowitz (Jun 02 2021 at 10:42):

And yes, such cocontinuous functors are like linear maps because they preserve $+$ and $0$. Well, ok, they also preserve coequalisers... I think I've seen this analogy mentioned somewhere, maybe on the ncafé?

Morgan Rogers (he/him) (Jun 02 2021 at 10:43):

Yes; I personally much prefer names that give intuition for what a functor does rather than what the definition looks like :wink:
but it sounds like this does both! A good choice of name in my book.

Zhen Lin Low (Jun 02 2021 at 10:47):

Well, in a situation where you have the Eilenberg–Watts theorem a cocontinuous functor really will be "multiplication" (tensoring) by a constant.