Are adjunctions unique?

Question

Let us fix two categories $\mathcal{C}$ and $\mathcal{D}$, as well as two functors $F:\mathcal{C}\rightarrow \mathcal{D}$ and $G:\mathcal{D}\rightarrow \mathcal{C}$. By an adjunction between $F$ and $G$ I mean a natural isomorphism of bifunctors $\Phi _{X,Y}:\mathrm{Mor}_{\mathcal{D}}(F(X),Y)\rightarrow \mathrm{Mor}_{\mathcal{C}}(X,G(Y))$. If there is some adjunction between $F$ and $G$, we say that $(F,G)$ is an adjointable pair.

Are there examples of adjointable pairs $(F,G)$, but for which the adjunction between them is not unique? If so, is there a sense in which they are all unique (analogous to, for example, how if a functor admits a left-adjoint, that functor is not literally unique, but it is unique up to natural isomorphism)?

My first thought is "Yes, there do exist such pairs.", and that you can find other adjunctions of the same adjointable pair by finding automorphisms of the identity functors on $\mathcal{C}$ and $\mathcal{D}$ (though I have not checked this in detail). If this is so, are adjunctions unique up to automorphisms of the identity functors, or we can find more adjunctions still?

I object to this terminology; "adjoint pair" should include both the data of $F$ and $G$ and the data of the adjunction between them (rather than just the claim that there exists an adjunction). — Qiaochu Yuan, Sep 10 '15 at 23:21
What do you mean? I agree that the thing that matters is the the functors together with the adjunction, but I specifically needed a term that 'forgot' about any specific adjunction so as to be able to actually ask the question (at least in the first way that came to mind). — Jonathan Gleason, Sep 10 '15 at 23:36
Okay, I like this. I have updated the question accordingly. You might then rephrase the question as "Is information lost in passing from the adjoint pair (i.e. with the data of the particular adjunction included) to the adjointable pair?" If the adjunction itself were unique, then you could recover it from just $F$ and $G$ alone. — Jonathan Gleason, Sep 10 '15 at 23:42

score 11 · Accepted Answer · answered Sep 10 '15 at 23:24

11

Adjoint functors are not only unique up to isomorphism, they are unique up to unique isomorphism; that is, if $F$ is left adjoint to $G$ and also to $G'$, then there is a unique isomorphism $G \cong G'$ compatible with the data of the two adjunctions. In particular, this is true if $G = G'$. So it's indeed the case that adjunction data is unique up to twisting by an automorphism of $G$, or dually up to twisting by an automorphism of $F$ (in particular, $F$ and $G$ have the same automorphism group, but there's no reason that this should be trivial).

answered Sep 10 '15 at 23:24

Qiaochu Yuan

468,795

For an explicit example to mull over, take $C = D = \text{Vect}$ and take $F = V \otimes (-), G = \text{Hom}(V, -)$. The automorphism group of both of these functors is $\text{Aut}(V)$. – Qiaochu Yuan Sep 10 '15 at 23:32
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What? I don't mean to ask whether the functors are unique, I mean to ask whether the adjunction that relates them is unique. Does uniqueness of the functors imply uniqueness of the natural isomorphism $\Phi _{X,Y}$? If so, I do not see it . . . – Jonathan Gleason Sep 10 '15 at 23:41
@Jonathan: I'm claiming that it's not unique, but that it is unique up to automorphisms of $F$ or $G$. The reason is because of the uniqueness of adjoints: if you have some adjunction data relating $F$ and $G$, and some adjunction data relating $F$ and $G'$, then you get a unique isomorphism $G \cong G'$ intertwining these. This is in particular the case if $G'$ happens to be equal to $G$. Of course this is all just exactly what I wrote above. – Qiaochu Yuan Sep 10 '15 at 23:44
Ahhh, okay. I had to write more details down to see what you meant, but now I see how it follows. Thanks! – Jonathan Gleason Sep 11 '15 at 00:14
@QiaochuYuan Could you please explain why adjoint functors have the same automorphism groups? – Saal Hardali Nov 05 '17 at 19:11
@Saal: taking adjoints is contravariantly functorial, with inverse given by taking adjoints in the other direction. – Qiaochu Yuan Nov 06 '17 at 03:22
@QiaochuYuan I don't understand this. Take the projection from a set with 2 elements to a set with one element. Taking sheaves of vector spaces we get $Vect$ and $Vect^{\times 2}$ and an adjoint pair: $V \mapsto (V,V)$ and $(V,W) \mapsto V \oplus W$. The second of these has an automorphism corresponding to a permutation of $V$ and $W$ but the second one doesn't appear to have any automorphisms at all. Perhaps I don't understand the meaning of automorphism here. – Saal Hardali Nov 06 '17 at 09:19
@Saal: an automorphism of a functor is a natural transformation from that functor to itself which is invertible. Permuting $V$ and $W$ is a natural transformation to a different (but equivalent) functor, namely $(V, W) \mapsto W \oplus V$. In fact I believe the automorphism group of both of these functors is $(k^{\times})^2$ acting on both factors by scalar multiplication, where $k$ is the underlying field. Feel free to ask a separate question about this and I can give more detail and examples. – Qiaochu Yuan Nov 06 '17 at 19:38
@QiaochuYuan Of course you're right, thanks for the patience. Do you happen to know if your answer to this question is valid in the infinity categorical world? I know that given a fixed functor an adjoint (to gether with the adjunction data) if it exists is unique (up to contractible choice). However what happens if we fix two functors and ask whether they are adjoint? In other words what is the homotopy type of the groupoid of adjunction data between two given functors? Is this just the classifying space of the automorphism group of one of them? Or can it be more complicated? – Saal Hardali Nov 06 '17 at 22:45
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@Saal: yes, the same argument applies, it's the classifying space of the automorphism group of either of them. – Qiaochu Yuan Nov 06 '17 at 22:48
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Good answer, however it would be improved if either some proofs (or links to them) were supplied. – goblin GONE Feb 05 '18 at 00:32

Are adjunctions unique?

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