nLab cartesian closed enriched category

Cartesian closed enriched categories

Cartesian closed enriched categories


Let VV be a monoidal category with products and CC a VV-enriched category with products, in the enriched sense that we have a V V -natural isomorphisms of hom-objects in VV:

C(Z,X×Y)C(Z,X)×C(Z,Y). C(Z, X\times Y) \;\cong\; C(Z,X) \times C(Z,Y) \,.

We say CC is VV-cartesian-closed if each VV-functor (X×):CC(X\times -) \colon C \to C has a V V -enriched right adjoint.

Relation to ordinary cartesian closedness

If CC is VV-cartesian-closed, then its underlying ordinary category C 0C_0 is cartesian closed in the usual sense, since VV-enriched right adjoints have underlying ordinary right adjoints.

The converse is true in some cases, such as the following:

  • When V=SetV=Set, trivially.

  • More generally, whenever the underlying-set functor V(I,):VSetV(I,-) : V\to Set is conservative, since the morphism of hom-objects C(X,Y Z)C(X×Y,Z)C(X,Y^Z) \to C(X\times Y,Z) induced by the evaluation morphism Y Z×YZY^Z\times Y \to Z has invertible image in SetSet, hence is itself invertible if V(I,)V(I,-) is conservative.

  • When VV is cartesian monoidal and C=VC=V: a cartesian closed category is automatically enriched-cartesian-closed over itself. In other words, the defining isomorphisms V 0(X×Y,Z)V 0(X,Z Y)V_0(X\times Y,Z) \cong V_0(X, Z^Y) induce, by the Yoneda lemma, isomorphisms of exponential objects Z X×Y(Z Y) XZ^{X\times Y} \cong (Z^Y)^X.

However, the converse is false in general. Counterexamples can be found in this mathoverflow discussion.


Last revised on March 10, 2023 at 04:51:22. See the history of this page for a list of all contributions to it.