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Stream: learning: questions

Topic: hom is lax monoidal

Jules Hedges (Jan 11 2022 at 12:03):

More of a comment than a question: If $\mathcal C$ is a monoidal category, then $\hom : \mathcal C^\mathrm{op} \times \mathcal C \to \mathbf{Set}$ is lax monoidal (for the cartesian product on $\mathbf{Set}$), but it is NOT strong monoidal. That is, there is a non-invertible natural transformation $\hom (x, y) \times \hom (x', y') \to \hom (x \otimes x', y \otimes y')$. It takes about 5 seconds thinking to realise this, but I made exactly this error in one of my papers, and peer review didn't catch it either

Jules Hedges (Jan 11 2022 at 12:05):

See also: https://mathoverflow.net/questions/137359/what-are-the-conditions-for-the-hom-functor-to-be-strong-monoidal

Paolo Perrone (Jan 11 2022 at 13:18):

The failure of this functor from being strong monoidal could be a good description for the "complexity" of things.

David Egolf (Jan 11 2022 at 15:46):

Paolo Perrone said:

The failure of this functor from being strong monoidal could a good description for the "complexity" of things.

It would be really interesting if someone could illustrate this with a small example.

Jules Hedges (Jan 11 2022 at 16:06):

For $\mathbf{Set}$ with cartesian product, the elements of $\hom (X \times X', Y \times Y')$ in the image of the laxator are exactly those functions that can be factored as $f (x, x') = (f_1 (x), f_2 (x'))$. So "complexity" in this example is the ability of every output of a function to depend on every input

David Egolf (Jan 11 2022 at 16:19):

Jules Hedges said:

For $\mathbf{Set}$ with cartesian product, the elements of $\hom (X \times X', Y \times Y')$ in the image of the laxator are exactly those functions that can be factored as $f (x, x') = (f_1 (x), f_2 (x'))$. So "complexity" in this example is the ability of every output of a function to depend on every input

Interesting! This is similar to a situation in engineering where we say "there is no cross-talk". If we view $f$ as a message sent across a channel from two sources $x$ and $x'$, then the condition $f (x, x') = (f_1 (x), f_2 (x'))$ says that the message from the two sources decomposes into messages generated by each source individually.
So, I suppose the more "complex" case would correspond to when there is cross-talk.

John Baez (Jan 12 2022 at 17:05):

This business about the laxator also shows up in the theory of decorated cospans, where we use it to describe, e.g. how there are more chemical reactions involving molecules A,B,C,D,E than those the consist of a reaction involving A,B,C and a reaction involving D,E.