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Stream: learning: questions

Topic: Commutative diagram

Nick Smith (Jul 22 2021 at 07:09):

If someone tells me this diagram commutes, what do they actually mean? There are no parallel paths unless you consider the == bars as merging the objects into one, in which case there are infinitely-many parallel paths.

Screen-Shot-2021-07-22-at-5.07.05-pm.png

James Wood (Jul 22 2021 at 07:25):

I think the ==es point outwards, so $\delta; \epsilon \triangleleft c = 1$ and symmetrically on the other side.

Nick Smith (Jul 22 2021 at 07:37):

That’s what I’d pieced together. So it’s fair to say the diagram is malformed?

James Wood (Jul 22 2021 at 07:40):

Maybe it would be clearer to note the direction on the ==es, but I also think it's somewhat standard to draw a diagram like this and make the reader infer the direction (particularly where there's only one direction that makes sense, like in this case).

Nick Smith (Jul 22 2021 at 07:44):

Oh it’s a standard thing? I guess I’m going to continue to be baffled then, when I stumble onto diagrams like this. 😮

Nick Smith (Jul 22 2021 at 07:45):

The use of c instead of a notation for the identity map confuses me too tbh

Chad Nester (Jul 22 2021 at 07:58):

I imagine the idea here is to emphasize that $c$, $y \triangleleft c$, and $c \triangleleft y$ are equal as objects.

Chad Nester (Jul 22 2021 at 08:03):

If we just write $\delta ; c \triangleleft \epsilon = 1_c = \delta ; \epsilon \triangleleft c$ then that's hidden!

Javier Prieto (Jul 22 2021 at 09:05):

I believe this is standard notation, see e.g. the unit law for monads on Wikipedia. As in your example, the source, target, and direction of the arrows must be inferred.

Nick Smith (Jul 22 2021 at 12:32):

Hm, that's a good example, thank you! I guess I'll have to get used to the notation.

Nathanael Arkor (Jul 22 2021 at 12:37):

Nick Smith said:

The use of c instead of a notation for the identity map confuses me too tbh

If our category has finite products, and we have a morphism $f : A \to B$, then we can denote the canonical morphism $A \times A \to A \times B$ either as $A \times f$ or $\mathrm{id}_A \times f$. In the first case, we're using the functor $A \times ({-}) : \mathbf C \to \mathbf C$ and in the second case, we're using the functor $({-}) \times ({-}) : \mathbf C^2 \to \mathbf C^2$. Of course, they have the same effect, so it doesn't matter which we use, and it becomes a matter of notation. I think it's helpful to note that both notations are entirely reasonable from this perspective.

Nick Smith (Jul 22 2021 at 12:41):

Oh, so that's the logic behind it. I guess that makes sense, though $A \times f$ still feels like an abuse of notation to me. It looks misleadingly like a binary operation.

Nathanael Arkor (Jul 22 2021 at 12:48):

It can be thought of as a binary operator. Remember that functors have actions on objects and morphisms. A bifunctor (i.e. a functor from a product category) therefore has actions on objects and morphisms in each argument. $A \times f$ is the mixed action of a bifunctor, on the left side on an object, and on the right side on a morphism.

Nick Smith (Jul 22 2021 at 12:52):

I thought a bifunctor would be accepting pairs of objects and pairs of morphisms... how does it make sense that it accepts one of each? The thing it spits out should be undefined.

Nick Smith (Jul 22 2021 at 12:54):

Unless you're operating under the philosophy that objects and identity morphisms are the same thing.

Joe Moeller (Jul 22 2021 at 12:56):

It's a convenient abuse of notation, or it can be justified by what you just said. You could think of A as a shorthand for id_A. It really starts paying off when you're whiskering a 2-cell with a 1-cell, imo.

James Wood (Jul 22 2021 at 12:58):

If you Curry the bifunctor, then you can mix objects and morphisms in applications.

Kenji Maillard (Jul 22 2021 at 12:58):

You also want your definition of bifunctor to be compatible with a notion of currying: bifunctors $F : \mathbf{C} \times \mathbf{D} \to \mathbf{E}$ should be equivalent to functors $F' : \mathbf{C} \to [\mathbf{D}, \mathbf{E}]$ to the functor category, and in that case $F'$ does associate to each object $A \in \mathbf{C}$ a functor $F'(A) : \mathbf{D} \to \mathbf{E}$