Tensor-hom adjunction

In mathematics, the tensor-hom adjunction is that the tensor product $$- \otimes X$$ and hom-functor $$\operatorname{Hom}(X,-)$$ form an adjoint pair:
 * $$\operatorname{Hom}(Y \otimes X, Z) \cong \operatorname{Hom}(Y,\operatorname{Hom}(X,Z)).$$

This is made more precise below. The order of terms in the phrase "tensor-hom adjunction" reflects their relationship: tensor is the left adjoint, while hom is the right adjoint.

General statement
Say R and S are (possibly noncommutative) rings, and consider the right module categories (an analogous statement holds for left modules):


 * $$\mathcal{C} = \mathrm{Mod}_S\quad \text{and} \quad \mathcal{D} = \mathrm{Mod}_R .$$

Fix an $$(R,S)$$-bimodule $$X$$ and define functors $$F \colon \mathcal D \rightarrow \mathcal C$$ and $$G \colon \mathcal C \rightarrow \mathcal D$$ as follows:


 * $$F(Y) = Y \otimes_R X \quad \text{for } Y \in \mathcal{D}$$


 * $$G(Z) = \operatorname{Hom}_S (X, Z) \quad \text{for } Z \in \mathcal{C}$$

Then $$F$$ is left adjoint to $$G$$. This means there is a natural isomorphism


 * $$\operatorname{Hom}_S (Y \otimes_R X, Z) \cong \operatorname{Hom}_R (Y, \operatorname{Hom}_S (X, Z)).$$

This is actually an isomorphism of abelian groups. More precisely, if $$Y$$ is an $$(A,R)$$-bimodule and $$Z$$ is a $$(B,S)$$-bimodule, then this is an isomorphism of $$(B,A)$$-bimodules. This is one of the motivating examples of the structure in a closed bicategory.

Counit and unit
Like all adjunctions, the tensor-hom adjunction can be described by its counit and unit natural transformations. Using the notation from the previous section, the counit


 * $$\varepsilon : FG \to 1_{\mathcal{C}}$$

has components


 * $$\varepsilon_Z : \operatorname{Hom}_S (X, Z) \otimes_R X \to Z$$

given by evaluation: For


 * $$\phi \in \operatorname{Hom}_S (X, Z) \quad \text{and} \quad x \in X,$$


 * $$\varepsilon(\phi \otimes x) = \phi(x).$$

The components of the unit


 * $$\eta : 1_{\mathcal{D}} \to GF$$


 * $$\eta_Y : Y \to \operatorname{Hom}_S (X, Y \otimes_R X)$$

are defined as follows: For $$y$$ in $$Y$$,


 * $$\eta_Y(y) \in \operatorname{Hom}_S (X, Y \otimes_R X)$$

is a right $$S$$-module homomorphism given by


 * $$\eta_Y(y)(t) = y \otimes t \quad \text{for } t \in X.$$

The counit and unit equations can now be explicitly verified. For $$Y$$ in $$\mathcal{D}$$,



\varepsilon_{FY}\circ F(\eta_Y) : Y \otimes_R X \to \operatorname{Hom}_S (X, Y \otimes_R X) \otimes_R X \to Y \otimes_R X $$

is given on simple tensors of $$Y \otimes X$$ by


 * $$\varepsilon_{FY}\circ F(\eta_Y)(y \otimes x) = \eta_Y(y)(x) = y \otimes x.$$

Likewise,


 * $$G(\varepsilon_Z)\circ\eta_{GZ} :

\operatorname{Hom}_S (X, Z) \to \operatorname{Hom}_S (X, \operatorname{Hom}_S (X, Z) \otimes_R X) \to \operatorname{Hom}_S (X, Z). $$

For $$\phi$$ in $$ \operatorname{Hom}_S (X, Z)$$,


 * $$G(\varepsilon_Z)\circ\eta_{GZ}(\phi)$$

is a right $$S$$-module homomorphism defined by


 * $$G(\varepsilon_Z)\circ\eta_{GZ}(\phi)(x) = \varepsilon_{Z}(\phi \otimes x) = \phi(x)$$

and therefore


 * $$G(\varepsilon_Z)\circ\eta_{GZ}(\phi) = \phi.$$

The Ext and Tor functors
The Hom functor $$\hom(X,-)$$ commutes with arbitrary limits, while the tensor product $$-\otimes X$$ functor commutes with arbitrary colimits that exist in their domain category. However, in general, $$\hom(X,-)$$ fails to commute with colimits, and $$-\otimes X$$ fails to commute with limits; this failure occurs even among finite limits or colimits. This failure to preserve short exact sequences motivates the definition of the Ext functor and the Tor functor.