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Stream: learning: questions

Topic: Do only monoidal functors compose to give monoidal functors?

Asad Saeeduddin (Apr 06 2020 at 12:26):

Imagine any two functors $F : \mathcal{B} \to \mathcal{C}$ and $G : \mathcal{A} \to \mathcal{B}$ that are monoidal (i.e. each coheres appropriately the "transported" monoidal structure of its domain with the monoidal structure of its codomain, for some choice of monoidal structures on $\mathcal{A}$, $\mathcal{B}$ and $\mathcal{C}$).

Their composite $F \circ G : \mathcal{A} \to \mathcal{C}$ is also a monoidal functor that coheres the transported monoidal structure from $\mathcal{A}$ with the structure in $\mathcal{C}$.

But given only the fact that some functor $F \circ G : \mathcal{A} \to \mathcal{C}$ is monoidal, can we assume a) that the "traced out" category that is the codomain of $G$ and the domain of $F$ is monoidal, and b) that $F$ and $G$ are themselves monoidal functors?

If not in general, is it true for some specialization of the situation at hand? For example when $F \dashv G$, or when talking about strong monoidal functors instead of lax monoidal ones?

Sam Tenka (Apr 06 2020 at 19:36):

@Asad Saeeduddin There are non-monoidal functors that compose to monoidal functors:

0. For example, take A=B=C=(complex vector spaces with tensor), and let F=(coproduct with a 1D space) and G=(send everything to 0). F is not monoidal, but GF is.
1. Or, A=B=C=(topological spaces with product), and F=(product with the circle) and G=(product with countably infinitely many copies of the circle). F is not monoidal, but GF is.

In both cases above, one of the factors is not monoidal and the other factor seems to be monoidal. Can you come up with example where both are not monoidal?

2. (edit: if we think of a ring as a category with one object (so the morphisms are ring elements that compose by addition, and the monoid operation is ring multiplication), then any additive involution that is not a ring map will provide an example.)

This doesn't answer your next few questions, but hopefully these concrete examples are a start! :-)

Sam Tenka (Apr 06 2020 at 19:50):

In general, it would be cool to identify "prime" properties: properties that, if they hold of a composition, must hold of a factor.

John Baez (Apr 06 2020 at 19:56):

What are some examples?

Sam Tenka (Apr 06 2020 at 20:02):

John Baez said:

What are some examples?

For examples of "prime" properties:

The only ones that come to mind come from contrapositive-afying compositional laws. For example, if two functors compose into a non-equivalence, then at least one must have been a non-equivalence. Then again, the notion of "prime" in commutative algebra has this nature, too, when we think about prime ideals as the complements of multiplicative systems. Perhaps the asker @Asad Saeeduddin has more examples in mind?

Asad Saeeduddin (Apr 06 2020 at 21:17):

@Sam Tenka (naive student) Thank you, those are exactly the kinds of examples I was looking for!

Asad Saeeduddin (Apr 06 2020 at 21:28):

I should have thought more about constant functors. It seems they essentially act as absorbing elements: any constant functor to a monoid object is monoidal and can be "decomposed" into the same constant functor composed onto an arbitrary functor to an arbitrary intermediate category (neither of which need be monoidal)

Sam Tenka (Apr 06 2020 at 21:32):

Yep. There seems to be more than just constant functors, though. examples 1 and 2 don't involve constant functors, and example 2 doesn't involve absorbtion (in the sense of idempotents)