Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I am looking for a counter-example of two functors F : C -> D and G : D->C such that

1) F is left adjoint to G

2) F is right adjoint to G

3) F is not an equivalence (ie F is not a quasi-inverse of G)

share|improve this question
add comment

5 Answers

Yes, there are many such functors. They are usually called "biadjoint." A good example is tensor product with a vector space $V$ in the category of finite dimensional vector spaces. This is actually adjoint to itself.

This is a little funny since to find this adjunction you have to pick an isomorphism $V\cong V^*$, but that's OK; adjunction of functors only makes sense up to isomorphism anyways.

Another good example is induction and restriction for an inclusion of finite groups.

share|improve this answer
3  
More generally, tensoring with any object of a symmetric monoidal category which is dualizable but not invertible. –  Reid Barton Dec 5 '09 at 23:02
add comment

There are lots of examples. Here's what I think is in some sense the minimal one.

Let $C$ be the terminal category $\mathbf{1}$ (one object, and only the identity arrow). Then for any category $D$, a left adjoint to the unique functor $G: D \to \mathbf{1}$ is an initial object of $D$, and a right adjoint is a terminal object. So, we're looking for a category $D$ that has a zero object (one that is both initial and terminal), but is not equivalent to the terminal category.

There are plenty of such categories $D$, e.g. $\mathbf{Vect}$. But I guess the minimal one is the category $D$ generated by a split epimorphism. In other words, it consists of two objects, $0$ and $d$, and non-identity arrows $$ p: d \to 0, \ \ \ i: 0 \to d, \ \ \ ip: d \to d, $$ satisfying $pi = 1_0$. Then $0$ is a zero object but $D$ is not equivalent to the terminal category.

share|improve this answer
add comment
up vote 9 down vote accepted

The answer of Ben Webster, can be made easier. Consider the functor F : (A-mod) -> (A-mod) which maps any A-module on (0). Then, F is a left adjoint to F ; and so, is a also a right adjoint to F. This is clear because for all A-modules N, M, one has Hom_A(0,N)=Hom_A(M,0). But, F is not an equivalence.

share|improve this answer
add comment

A pair of functors with this property where called Frobenius functors in S. Caenepeel, G. Militaru and S. Zhu, Doi-Hopf modules, Yetter-Drinfel'd modules and Frobenius type properties, {\sl Trans. Amer. Math. Soc.} {\bf 349} (1997), 4311--4342.

And main examples are given for cateogory o generalized Hopf modules, Yetter-Drinfel'd modules.

A detalied study of the you can find in :

S. Caenepeel, G. Militaru and Shenglin Zhu, {Frobenius Separable Functors for Generalized Module Categories and Nonlinear Equations}, {\sl Lect. Notes Math.} {\bf 1787} Springer Verlag, Berlin, 2002.

Cheers! Gigel Militaru

share|improve this answer
add comment

Edit: Misread the question

Take $j:U\to X$ an immersion of topological spaces. Then the restriction of sheaves of $A$-modules $j^* : Sh(X,A)\to Sh(U,A)$ has a right adjoint $j_*$ and a left adjoint $j_!$ (extension by 0).

share|improve this answer
    
Are there any examples where those are the same? Seem unlikely. You don't seem to have read the question very carefully. –  Ben Webster Dec 5 '09 at 22:56
    
I misread the question in the same way due to the title, which I now fixed. –  Reid Barton Dec 5 '09 at 23:05
1  
These coincide when the embedding is also closed, for example the embedding of a connected component. –  Jonathan Wise Dec 5 '09 at 23:31
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.