## Commutative Algebra 50

Adjoint functors are a general construct often used for describing universal properties (among other things).

Take two categories $\mathcal C$ and $\mathcal D$.

Definition.

Covariant functors $F:\mathcal D\to \mathcal C$ and $G: \mathcal C \to \mathcal D$ are said to be adjoint if we have isomorphisms

$A \in \mathcal C, B \in \mathcal D \implies \mathrm{hom}_{\mathcal C}(F(B), A) \cong \mathrm{hom}_{\mathcal D}(B, G(A))$

which are natural in A and B, when we regard both sides as functors $\mathcal D^{\text{op}} \times \mathcal C \to \mathbf{Set}$.

We also say F is left adjoint to G and G is right adjoint to F.

Unwinding the definition, for any $B\in \mathcal D$ and $A\in \mathcal C$, we have a bijection

$T(B, A) : \mathrm{hom}_{\mathcal C}(F(B), A) \stackrel\cong\longrightarrow \mathrm{hom}_{\mathcal D}(B, G(A))$

such that the following diagram commutes for any $g:B' \to B$ and $f:A\to A'$.

Note

Suppose F and G are adjoint. For any $B\in \mathcal D$, the covariant functor

$\mathcal C \longrightarrow \mathbf{Set}, \quad A \mapsto \mathrm{hom}_{\mathcal D}(B, G(A))$

is representable since it is isomorphic to $\mathrm{hom}_{\mathcal C}(F(B), -)$. Similarly for each $A\in\mathcal C$, the contravariant functor $\mathcal D \to \mathbf{Set}$ taking $B\mapsto \mathrm{hom}_{\mathcal C}(-, G(A))$ 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.

# Examples Galore

“Adjoint functors arise everywhere.” – Saunders Mac Lane, Categories for the Working Mathematician.

### Example 1

Let $F: \mathbf{Set} \to \mathbf{Gp}$ take a set S to the free group F(S) on S and $U: \mathbf{Gp} \to \mathbf{Set}$ be the forgetful functor. By the universal property of free groups, we have a natural bijection

$\mathrm{hom}_{\mathbf{Gp}}(F(S), G) \cong \mathrm{hom}_{\mathbf{Set}}(S, U(G))$

for any set S and group G.

### Example 2

Let $F : \mathbf{Set} \to \mathbf{Ring}$ take a set S to the (commutative) ring $\mathbb \mathbb Z[S] := Z[x_s : s\in S]$ and $U :\mathbf{Ring} \to \mathbf{Set}$ be the forgetful functor. Again we get

$\mathrm{hom}_{\mathbf{Ring}}(\mathbb Z[S], R) \cong \mathrm{hom}_{\mathbf{Set}}(S, U(R))$

for any set S and ring R.

In general, free objects are left adjoint to forgetful functors.

### Example 3

Let B be an A-algebra. Recall that for each A-module M, we have an induced B-module $M^B = B\otimes_A M$; this gives a functor $F: A\text{-}\mathbf{Mod} \to B\text{-}\mathbf{Mod}$. On the other hand, we have the forgetful functor $U : B\text{-}\mathbf{Mod} \to A\text{-}\mathbf{Mod}$ from the canonical ring homomorphism $A\to B$. By the universal property of induced B-modules

$\mathrm{hom}_{B\text{-}\mathbf{Mod}}(M^B, N) \cong \mathrm{hom}_{A\text{-}\mathbf{Mod}}(M, U(N)).$

Thus the induced module construction is also left adjoint to a forgetful functor.

### Example 4

Let $\mathcal C = A\text{-}\mathbf{Mod}$ and $N \in \mathcal C$. Take functors $F_N, G_N : \mathcal C \to \mathcal C$ where $F_N(M) = M\otimes_A N$ and $G_N(P) = \mathrm{Hom}_A(N, P)$. In proposition 2 here we obtained a natural isomorphism $\mathrm{Hom}_A(M\otimes N, P) \cong \mathrm{Hom}_A(M, \mathrm{Hom}_A(N, P))$. This translates to a natural isomorphism

$\mathrm{hom}_{\mathcal C}(F_N(M), P) \cong \mathrm{hom}_{\mathcal C}(M, G_N(P)).$

In short $- \otimes_A N$ is left adjoint to $\mathrm{Hom}_A(N, -)$.

### Example 5

Let $\mathcal D = \mathcal C \times \mathcal C$. Suppose the coproduct of any two objects in $\mathcal C$ exist. Let $\Sigma:\mathcal D \to \mathcal C$ take $(A, A')\mapsto A \amalg A'$ and $\Delta : \mathcal C \to \mathcal D$ take $A\mapsto (A, A)$. By definition of coproduct in categories, we have

\begin{aligned} \mathrm{hom}_{\mathcal D}((A, A'), \Delta(B)) &= \mathrm{hom}_{\mathcal C}(A, B) \times \mathrm{hom}_{\mathcal C}(A', B)\\ &\cong \mathrm{hom}_{\mathcal C}(A \amalg A', B)\\ &= \mathrm{hom}_{\mathcal C}(\Sigma(A, A'), B)\end{aligned}

for any $(A, A') \in \mathcal D$ and $B\in \mathcal C$. Hence the coproduct functor is left adjoint to the diagonal functor. Dually, the product is right adjoint to the diagonal functor.

### Example 6

More generally, let J be an index category and $\mathcal D = \mathcal C^J$ be the category of all diagrams in $\mathcal C$ of type J. Assuming all colimits of type J exist in $\mathcal C$, let $\mathrm{colim}_J : \mathcal D \to \mathcal C$ be this colimit functor. On the other hand, take the diagonal embedding $\mathcal C \to \mathcal D$ which takes an object A to the diagram where all objects are A and all morphisms are $1_A$. Then

$\mathrm{hom}_{\mathcal D}(D, \Delta(B)) \cong \mathrm{hom}_{\mathcal C}(\mathrm{colim} D, B)$

for any diagram $D\in \mathcal D$ and object $B \in \mathcal C$.

Exercise A

Let $U : \mathbf{Top} \to \mathbf{Set}$ be the forgetful functor. Find a left adjoint and a right adjoint to U (these are two different functors of course).

Let $U : \mathbf{AbGp} \to \mathbf{Gp}$ be the inclusion functor, where $\mathbf{AbGp}$ is the category of abelian groups. Find a left adjoint functor to U.

# Properties

Suppose $F:\mathcal D \to \mathcal C$ is left adjoint to $G : \mathcal C \to \mathcal D$. If coproducts exist in both categories, then $F(B) \amalg F(B') \cong F(B\amalg B')$ because we have natural isomorphisms

\begin{aligned} A \in\mathcal C \implies \mathrm{hom}_{\mathcal C}(F(B\amalg B'), A) &\cong \mathrm{hom}_{\mathcal D}(B \amalg B', G(A)) \\ &\cong \mathrm{hom} _{\mathcal D}(B, G(A)) \times \mathrm{hom}_{\mathcal D}(B', G(A)) \\ & \cong \mathrm{hom}_{\mathcal C}(F(B), A) \times \mathrm{hom}_{\mathcal C}(F(B'), A) \\ &\cong \mathrm{hom}_{\mathcal D}(F(B) \amalg F(B'), A).\end{aligned}

More generally, we have:

Proposition 1.

Let J be an index category; assume that colimits of diagrams of type J all exist $\mathcal C$ and $\mathcal D$. Then for any diagram $(B_i)_{i\in J}$ in $\mathcal D$ of type J,

$\mathrm{colim }_{i\in J} F(B_i) \cong F(\mathrm{colim}_{i\in J} B_i)$.

Proof

By the universal property of colimits and limits, we have, for any diagram $(A_i)_{i\in J}$ in $\mathcal C$ of type J, and any object $A\in \mathcal C$,

$\mathrm{hom}_{\mathcal C}(\mathrm{colim}_{i\in J} A_i, A) \cong \mathrm{lim}_{i\in J^{\text{op}}} (\mathrm{hom}_{\mathcal C}(A_i, A)).$

Now we apply this universal property twice to obtain:

\begin{aligned}A \in \mathcal C \implies \mathrm{hom}_{\mathcal C}(F(\mathrm{colim}_{i \in J} B_i), A) &\cong \mathrm{hom}_{\mathcal D}(\mathrm{colim}_{i \in J} B_i, G(A))\\ &\cong \mathrm{lim}_{i \in J^{\text{op}}} (\mathrm{hom}_{\mathcal D}(B_i, G(A))) \\ &\cong \mathrm{lim}_{i\in J^{\text{op}}} ( \mathrm{hom}_{\mathcal C} (F(B_i), A))\\ &\cong \mathrm{hom}_{\mathcal C} (\mathrm{colim}_{i\in J} F(B_i), A)\end{aligned}

and we are done. ♦

Exercise B

1. State and prove the dual of the above properties.

2. Prove that the forgetful functor $U:\mathbf{Gp} \to \mathbf{Set}$ has no right adjoint.

3. Prove that the inclusion functor $U:\mathbf{AbGp} \to \mathbf{Gp}$ has no right adjoint.

4. Prove that for any diagram of A-modules $(M_i)_{i \in J}$ and A-module N, we have

$\mathrm{colim}_{i \in J} (M_i \otimes_A N) \cong (\mathrm{colim}_{i\in J} M_i) \otimes_A N.$

As a special case, this implies tensor product is distributive over direct sums (already proved in proposition 1 here).

$F : B\text{-}\mathbf{Mod} \to A\text{-}\mathbf{Mod}, \quad G : A \text{-}\mathbf{Mod} \to B\text{-}\mathbf{Mod}$

such that F is left adjoint to G. Further, we assume that the natural $\mathrm{Hom}_B(F(N), M) \cong \mathrm{Hom}_A(N, G(M))$, for A-module M and B-module N, is an isomorphism of additive groups

Note

In fact, one can show that if F and G are adjoint functors as above, then they must be additive and $\mathrm{Hom}_B(F(N), M) \cong \mathrm{Hom}_A(N, G(M))$ must preserve the additive structure.

Proposition 2.

The functors F and G are right-exact and left-exact respectively.

In summary, left adjoint functors are right-exact and vice versa.

Proof

Suppose $M_1 \to M_2 \to M_3 \to 0$ is an exact sequence of A-modules. To prove $F(M_1) \to F(M_2) \to F(M_3) \to 0$ is exact, by proposition 4 here it suffices to show:

$0 \longrightarrow \mathrm{Hom}_B(F(M_3), N) \longrightarrow \mathrm{Hom}_B(F(M_2), N) \longrightarrow \mathrm{Hom}_B(F(M_1), N)$

is exact for every B-module N. But by adjointness, the above is naturally isomorphic to

$0 \longrightarrow \mathrm{Hom}_A(M_3, G(N)) \longrightarrow \mathrm{Hom}_B(M_2, G(N)) \longrightarrow \mathrm{Hom}_B(M_1, G(N))$

which is exact since $\mathrm{Hom}_A(-, G(N))$ is a left-exact functor by proposition 3 here. The case for G is similar. ♦

### Examples

1. By example 4, we see that for any A-module M, the functor $M\otimes -$ is right-exact.

2. By example 5, for any A-algebra B, taking the induced module $M\mapsto M^B$ is right-exact.

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### 2 Responses to Commutative Algebra 50

1. Vanya says:

could you please elaborate on the second isomorphism after the statement ” Now we apply this universal property twice”. in Proposition 1?

• limsup says:

The second isomorphism follows from the previous paragraph with $A_i$ replaced by $B_i$ and $A$ replaced by $G(A)$. 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.