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Stream: learning: questions

Topic: transporting structure across an adjunction

David Egolf (Dec 03 2023 at 18:48):

Let us assume we have an adjunction $F \dashv G$ where $F: C \to D$ and $G: D \to C$. Then, for each object $c \in C$ and object $d \in D$ we have an isomorphism in $\mathsf{Set}$: $D(Fc, d) \cong C(c, Gd)$.

Our adjunction has already told us about some structure in $C$ in terms of structure in $D$. That is, we have learned that the set $C(c,Gd)$ of morphisms from $c$ to $Gd$ in $C$ is isomorphic to the set $D(Fc, d)$.

David Egolf (Dec 03 2023 at 18:49):

I'm wondering if in some cases we can get more structure on the data of $C$ from structure in $D$. For example, consider the case where $d = F(c)$. Then we have an isomorphism in $\mathsf{Set}$: $D(Fc, Fc) \cong C(c, GFc)$. I notice in this case that $D(Fc,Fc)$ supports more structure than that of a set. That is, $D(Fc,Fc)$ can be turned into a monoid by using composition of the endomorphisms of $Fc$.

David Egolf (Dec 03 2023 at 18:50):

I am curious if this monoid structure can be transported across the adjunction. To compose $g: c \to GFc$ after $f: c \to GFc$ we could maybe use this process:

• start with $f: c \to GFc$ and $g: c \to GFc$
• move each of them across the adjunction bijection to get $f^\sharp: Fc \to Fc$ and and $g^\sharp: Fc \to Fc$
• compose $g^\sharp$ after $f^\sharp$, getting $g^\sharp \circ f^\sharp: Fc \to Fc$
• move this back across the adjunction, getting $(g^\sharp \circ f^\sharp)^\flat: c \to GFc$

(Here $\sharp: C(c, GFc) \cong D(Fc, Fc)$ and $\flat: D(Fc, Fc) \cong C(c, GFc)$ are inverse functions in $\mathsf{Set}$).

Do we actually get a monoid on $C(c, GFc)$ in this way using the monoid structure on $D(Fc,Fc)$? I think we have an identity element, given by $1_{Fc}^\flat: c \to GFc$.

David Egolf (Dec 03 2023 at 18:50):

Note that for any $f: c \to GFc$, $f \circ 1_{Fc}^\flat =( f^\sharp \circ (1_{Fc}^\flat)^\sharp)^\flat = (f^\sharp \circ 1_{Fc})^\flat = (f^\sharp)^\flat=f$. We use the fact that $\sharp$ and $\flat$ are inverse functions in the last step.

We still need to check composition is associative. That is, do we have $(g \circ f) \circ h = g \circ (f \circ h)$ for arbitrary $g,f,h:c \to GFc$?

• $(g \circ f) \circ h = (((g^\sharp \circ f^\sharp)^\flat)^\sharp \circ h^\sharp)^\flat$ . Noting that $\sharp$ and $\flat$ are inverse functions, this simplifies to: $(g^\sharp \circ f^\sharp \circ h^\sharp)^\flat$
• $g \circ (f \circ h) = (g^\sharp \circ ((f^\sharp \circ h^\sharp)^\flat)^\sharp)^\flat$ . Noting that $\sharp$ and $\flat$ are inverse functions, this simplifies to: $(g^\sharp \circ f^\sharp \circ h^\sharp)^\flat$ .
• We conclude that composition is associative.

So, I think that the monoid structure on $D(Fc, Fc)$ induces a monoid structure on $C(c, GFc)$, thanks to our adjunction.

David Egolf (Dec 03 2023 at 18:50):

I'm curious if this can be generalized. Consider again a bijection provided by our adjunction: $D(Fc, d) \cong C(c, Gd)$. Let us imagine the case where $D = \mathsf{Ab}$, the category of commutative groups, and where $C = \mathsf{Set}$. We can take $F: \mathsf{Set} \to \mathsf{Ab}$ to be the functor that sends a set to the free commutative group on that set, and $G: \mathsf{Ab} \to \mathsf{Set}$ to be the forgetful functor the sends a commutative group to its underlying set.

In this setting, our adjunction tells us $\mathsf{Ab}(Fc, d) \cong \mathsf{Set}(c, Gd)$. Now, the category $\mathsf{Ab}$ is special in that every hom-set is also a commutative group. Can we use a similar procedure as described above to induce a commutative group structure on $\mathsf{Set}(c, Gd)$?

More generally, for an adjunction $F \dashv G$ with $F: C \to D$ and $G: D \to C$, if the hom-sets in $D$ have some additional structure, when can that additional structure be transferred to the corresponding hom-sets in $C$?

Todd Trimble (Dec 03 2023 at 19:27):

I have only a few seconds to write this down, but in this regard, you might enjoy reading up on the Kleisli category construction, specifically composition of Kleisli morphisms, which is closely connected to what you're bringing up.

Mike Shulman (Dec 03 2023 at 19:29):

Another quick comment is that all mathematical structure can be transported across any isomorphism. So if you have an isomorphism $X\cong Y$ and $X$ has some structure, whether it's a monoid or an abelian group or a topological space or what-have you, then you can transport it across that isomorphism to give $Y$ the same structure. In particular, therefore, you can do this when $X$ and $Y$ are hom-sets and the isomorphism comes from an adjunction.