Cartesian closed functors #
Define the exponential comparison morphisms for a functor which preserves binary products, and use them to define a cartesian closed functor: one which (naturally) preserves exponentials.
Define the Frobenius morphism, and show it is an isomorphism iff the exponential comparison is an isomorphism.
TODO #
Some of the results here are true more generally for closed objects and for closed monoidal categories, and these could be generalised.
References #
https://ncatlab.org/nlab/show/cartesian+closed+functor https://ncatlab.org/nlab/show/Frobenius+reciprocity
Tags #
Frobenius reciprocity, cartesian closed functor
The Frobenius morphism for an adjunction L ⊣ F
at A
is given by the morphism
L(FA ⨯ B) ⟶ LFA ⨯ LB ⟶ A ⨯ LB
natural in B
, where the first morphism is the product comparison and the latter uses the counit
of the adjunction.
We will show that if C
and D
are cartesian closed, then this morphism is an isomorphism for all
A
iff F
is a cartesian closed functor, i.e. it preserves exponentials.
Equations
- One or more equations did not get rendered due to their size.
Instances For
If F
is full and faithful and has a left adjoint L
which preserves binary products, then the
Frobenius morphism is an isomorphism.
Equations
- ⋯ = ⋯
The exponential comparison map.
F
is a cartesian closed functor if this is an iso for all A
.
Equations
- One or more equations did not get rendered due to their size.
Instances For
The exponential comparison map is natural in A
.
The functor F
is cartesian closed (ie preserves exponentials) if each natural transformation
exp_comparison F A
is an isomorphism
- comparison_iso : ∀ (A : C), CategoryTheory.IsIso (CategoryTheory.expComparison F A)
Instances
If the exponential comparison transformation (at A
) is an isomorphism, then the Frobenius morphism
at A
is an isomorphism.
If the Frobenius morphism at A
is an isomorphism, then the exponential comparison transformation
(at A
) is an isomorphism.
If F
is full and faithful, and has a left adjoint which preserves binary products, then it is
cartesian closed.
TODO: Show the converse, that if F
is cartesian closed and its left adjoint preserves binary
products, then it is full and faithful.