Is a functor which has a left adjoint which is also its right adjoint an equivalence ? - MathOverflow

Is a functor which has a left adjoint which is also its right adjoint an equivalence ?

2009-12-05T22:27:50Z

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)

Answer by Ben Webster for Is a functor which has a left adjoint which is also its right adjoint an equivalence ?

2009-12-05T22:42:46Z

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.

Answer by YBL for Is a functor which has a left adjoint which is also its right adjoint an equivalence ?

2009-12-05T22:50:17Z

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).

Answer by nicojo for Is a functor which has a left adjoint which is also its right adjoint an equivalence ?

2009-12-05T23:36:25Z

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.

Answer by Tom Leinster for Is a functor which has a left adjoint which is also its right adjoint an equivalence ?

2009-12-05T23:36:28Z

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.

Answer by Gigel Militaru for Is a functor which has a left adjoint which is also its right adjoint an equivalence ?

2010-04-28T13:12:39Z

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