Adjoint functors are a general construct often used for describing universal properties (among other things).
Take two categories and .
Covariant functors and are said to be adjoint if we have isomorphisms
which are natural in A and B, when we regard both sides as functors .
We also say F is left adjoint to G and G is right adjoint to F.
Unwinding the definition, for any and , we have a bijection
such that the following diagram commutes for any and .
Suppose F and G are adjoint. For any , the covariant functor
is representable since it is isomorphic to . Similarly for each , the contravariant functor taking is representable.
There are various equivalent ways of defining adjoint functors but we will not delve into those here. Instead, let us contend ourselves with some examples before restricting to the case of modules.
“Adjoint functors arise everywhere.” – Saunders Mac Lane, Categories for the Working Mathematician.
Let take a set S to the free group F(S) on S and be the forgetful functor. By the universal property of free groups, we have a natural bijection
for any set S and group G.
Let take a set S to the (commutative) ring and be the forgetful functor. Again we get
for any set S and ring R.
In general, free objects are left adjoint to forgetful functors.
Let B be an A-algebra. Recall that for each A-module M, we have an induced B-module ; this gives a functor . On the other hand, we have the forgetful functor from the canonical ring homomorphism . By the universal property of induced B-modules
Thus the induced module construction is also left adjoint to a forgetful functor.
Let and . Take functors where and . In proposition 2 here we obtained a natural isomorphism . This translates to a natural isomorphism
In short is left adjoint to .
Let . Suppose the coproduct of any two objects in exist. Let take and take . By definition of coproduct in categories, we have
for any and . Hence the coproduct functor is left adjoint to the diagonal functor. Dually, the product is right adjoint to the diagonal functor.
More generally, let J be an index category and be the category of all diagrams in of type J. Assuming all colimits of type J exist in , let be this colimit functor. On the other hand, take the diagonal embedding which takes an object A to the diagram where all objects are A and all morphisms are . Then
for any diagram and object .
Let be the forgetful functor. Find a left adjoint and a right adjoint to U (these are two different functors of course).
Let be the inclusion functor, where is the category of abelian groups. Find a left adjoint functor to U.
Suppose is left adjoint to . If coproducts exist in both categories, then because we have natural isomorphisms
More generally, we have:
Let J be an index category; assume that colimits of diagrams of type J all exist and . Then for any diagram in of type J,
By the universal property of colimits and limits, we have, for any diagram in of type J, and any object ,
Now we apply this universal property twice to obtain:
and we are done. ♦
1. State and prove the dual of the above properties.
2. Prove that the forgetful functor has no right adjoint.
3. Prove that the inclusion functor has no right adjoint.
4. Prove that for any diagram of A-modules and A-module N, we have
As a special case, this implies tensor product is distributive over direct sums (already proved in proposition 1 here).
Exactness of Adjoint Functors
Now suppose instead of general categories we have additive functors
such that F is left adjoint to G. Further, we assume that the natural , for A-module M and B-module N, is an isomorphism of additive groups
In fact, one can show that if F and G are adjoint functors as above, then they must be additive and must preserve the additive structure.
The functors F and G are right-exact and left-exact respectively.
In summary, left adjoint functors are right-exact and vice versa.
Suppose is an exact sequence of A-modules. To prove is exact, by proposition 4 here it suffices to show:
is exact for every B-module N. But by adjointness, the above is naturally isomorphic to
which is exact since is a left-exact functor by proposition 3 here. The case for G is similar. ♦
1. By example 4, we see that for any A-module M, the functor is right-exact.
2. By example 5, for any A-algebra B, taking the induced module is right-exact.
could you please elaborate on the second isomorphism after the statement ” Now we apply this universal property twice”. in Proposition 1?
The second isomorphism follows from the previous paragraph with replaced by and replaced by . This is actually just the universal property of colimits once you unravel it. E.g. pick a simple diagram (such as that in example 6), then carefully unravel what both sides mean.