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Stream: learning: questions

Topic: Uniqueness of Natural Isomorphism in Adjunction?

Suraaj K S (Jun 07 2024 at 14:33):

If I understand correctly, an adjunction between two functors F and G would be isomorphisms between Hom(FX,Y) and Hom(X,GY) for all X,Y (which is also natural).

I was wondering if these isomorphisms are unique? Playing around, the isomorphisms seem to be unique. I was wondering if this is a theorem?

John Baez (Jun 07 2024 at 14:49):

I doubt it.

John Baez (Jun 07 2024 at 15:34):

It should be fun to find the smallest counterexample! Can we do it when F and G are the identity functor on a category with one object?

Peva Blanchard (Jun 07 2024 at 16:51):

Indeed, if we take the category with one object $x$, and two morphisms $hom(x, x) = \{id, f\}$ with $f \neq id$ and $f^2 = id$, then I think we have two distinct bijections $hom(x, x) \cong hom(x, x)$ natural in $x$ in both positions (covariant, and contravariant).

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John Baez (Jun 07 2024 at 17:24):

@Peva Blanchard - if we take an arbitrary groupoid with one object $x$, so that $\mathrm{hom}(x,x)$ is an arbitrary group, is it true that we get one bijection $\mathrm{hom}(x,x) \cong \mathrm{hom}(x,x)$, natural in $x$ in both positions, from each element of the center of this group?

Mike Shulman (Jun 07 2024 at 18:21):

I think in fact the center of the group is isomorphic to the group of such doubly-natural bijections. We can see this with end calculus: for any category $C$, the Yoneda lemma says

$$\begin{align*} \mathrm{Set}^{C^{\mathrm{op}}\times C}(\hom,\hom) &= \int_{x,y\in C} \mathrm{Set}(\hom(x,y),\hom(x,y)) \\ &\cong \int_{x\in C} \hom(x,x)\\ & = C^C (\mathrm{Id}_C,\mathrm{Id}_C) \end{align*}$$

and the latter is the [[center of a category]], which is so-named because the center of a one-object groupoid is the center of the corresponding group.

Rémy Tuyéras (Jun 07 2024 at 19:27):

Mike Shulman said:

We can see this with end calculus: for any category $C$, the Yoneda lemma says

$$\begin{align*} \mathrm{Set}^{C^{\mathrm{op}}\times C}(\hom,\hom) &= \int_{x,y\in C} \mathrm{Set}(\hom(x,y),\hom(x,y)) \\ &\cong \int_{x\in C} \hom(x,x)\\ & = C^C (\mathrm{Id}_C,\mathrm{Id}_C) \end{align*}$$

Nice! The calculation could also be used to further answer @Suraaj K S's question by using the following generalizations:

1. choosing one direction of the isomorphism gives us a unit:
   $\mathbf{Set}^{C^{\mathsf{op}}\times D}(\mathsf{hom}(F(-),-),\mathsf{hom}(-,G(-))) \cong C^C(\mathsf{id}_C,GF)$

2. choosing the other direction of the isomorphism gives us a counit:
   $\mathbf{Set}^{C^{\mathsf{op}}\times D}(\mathsf{hom}(-,G(-)),\mathsf{hom}(F(-),-)) \cong D^D(FG,\mathsf{id}_D)$

Using the previous isomorphisms, we can say that given $(F,G)$ and a pair of units/counits, there is a unique way to choose the natural bijections $\mathsf{hom}(F(-),-)\leftrightarrow \mathsf{hom}(-,G(-))$.

I have not checked the details, but is true that the two isomorphisms in the case $\mathsf{hom}(x,x) = \{\mathsf{id},f\}$ gives us the pairs $(f,f)$ and $(\mathsf{id},\mathsf{id})$ for the different possible unit/counit pairs?

Peva Blanchard (Jun 07 2024 at 19:56):

This is interesting. But, I think we cannot choose options 1 and 2 independently: the unit and counit must satisfy the zig-zag identities.

It's not a complete argument, but in the example of $hom(x, x) = \{id, f\}$, I think this rules out the choices $(id, f)$ and $(f, id)$.

Rémy Tuyéras (Jun 07 2024 at 20:01):

Peva Blanchard said:

This is interesting. But, I think we cannot choose options 1 and 2 independently: the unit and counit must satisfy the zig-zag identities.

Oh yeah, I definitely agree

Rémy Tuyéras (Jun 07 2024 at 20:02):

And as you said, it is interesting because now we can explore more general situations like weak retractions (e.g. only $\mathsf{id} \Rightarrow GF$ is given), etc.

Rémy Tuyéras (Jun 07 2024 at 20:24):

Peva Blanchard said:

It's not a complete argument, but in the example of $hom(x, x) = \{id, f\}$, I think this rules out the choices $(id, f)$ and $(f, id)$.

Hmm, I see. So, I think we have the following natural inclusions:

$\mathbf{Set}^{C^{\mathsf{op}} \times D}(\mathsf{hom}(F(-),-),\mathsf{hom}(-,G(-)) \hookrightarrow \mathbf{Rel}^{C^{\mathsf{op}} \times D}(\mathsf{hom}(F(-),-),\mathsf{hom}(-,G(-))$

$\mathbf{Set}^{C^{\mathsf{op}} \times D}(\mathsf{hom}(-,G(-),\mathsf{hom}(F(-),-)) \hookrightarrow \mathbf{Rel}^{C^{\mathsf{op}} \times D}(\mathsf{hom}(-,G(-),\mathsf{hom}(F(-),-))$

and a natural isomorphism:

$\mathbf{Rel}^{C^{\mathsf{op}} \times D}(\mathsf{hom}(F(-),-),\mathsf{hom}(-,G(-)) \cong \mathbf{Rel}^{C^{\mathsf{op}} \times D}(\mathsf{hom}(-,G(-),\mathsf{hom}(F(-),-))$

My guess is that the pullback of the previous two inclusions (up to the isomorphism) gives us the homset of natural bijections we are after:

$\mathbf{Bij}^{C^{\mathsf{op}} \times D}(\mathsf{hom}(F(-),-),\mathsf{hom}(-,G(-))$

The relations involved in the construction of this pullback probably determines (to some large extent) the relationship between the unit and counit.

Rémy Tuyéras (Jun 07 2024 at 20:28):

However, not sure yet how we can use that pullback to formalize that relationship