Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

The context of this question is given below but I don't think it is of essence here, so I will try to formulate the question for maps between sets.

Given two sets $A$ and $B$, denote the set of all maps $f\colon A\to B$ by $B^A$. Then, given a set $C$, a map $h\colon B\to C$ "lifts" to a map $h^A\colon B^A\to C^A$ defined by $h^A(f)(a)=h(f(a))$ for all $f\in B^A$ and $a\in A$.

This construct allows to "change the final codomain" in a "nested" map $F\colon X\to Z^Y$ with a simple composition operation: given a map $H\colon Z\to W$, the composition $H^Y\circ F$ is well defined and creates a map $X\to W^Y$ where the "final codomain" $Z$ has been changed to $W$. The point here is that $H^Y\circ F$ still only takes a single argument from $X$ and after evaluation then a single argument from $Y$.

In case you are curious, I want to use something like this to calculate with Hessian operators for vector-valued maps on Riemannian manifolds and their duals without using explicit basis notation. The above construct allows to do that quite conveniently, including certain types of "chain rules".

The questions are: does this construct have a name? Maybe something from category theory? Is this idea of "lifting" $h$ to $h^A$ standard? Does it have a name?

share|improve this question
    
Probably you intended to write $h^A(f)(a)=h(f(a))$ in the second paragraph. –  Joel David Hamkins Jun 12 '12 at 12:46
2  
This is exactly the statement that $-^A$ is a functor: en.wikipedia.org/wiki/Functor ; ncatlab.org/nlab/show/functor –  Finn Lawler Jun 12 '12 at 14:52
1  
(Well, no, not exactly that statement, but part of it anyway: that the mapping acts on morphisms in a way that respects its action on objects.) –  Finn Lawler Jun 12 '12 at 17:42
    
@Joel David Hamkins: yes, thank you. I have edited the question accordingly. –  Jochen Trumpf Jun 13 '12 at 1:26
1  
'Exponential' or 'mapping functor' are reasonable. –  David Roberts Jun 13 '12 at 1:54

1 Answer 1

up vote 0 down vote accepted

I think you want to say something like the following:

Let $J,K:{\cal Sets}\times{\cal Sets}\times{\cal Sets}\rightarrow{\cal Sets}$ be defined by $$J(X,Y,Z)=Hom(X,Hom(Y,Z))$$ $$K(X,Y,Z)=Nat(Hom(Z,-),Hom(X,Hom(Y,-))$$ where "Nat" means the set of natural transformations between two functors.

(These will be contravariant in the first two variables and covariant in the third.)

Then the map $F\mapsto H^Y\circ F$ defines a natural transformation from $J$ to $K$.

This contains a bit more content because it says that your construction commutes with compositions with maps into and out of $X$, $Y$ and $Z$.

share|improve this answer
    
Ahem, this is heavy stuff ... I would love to accept your answer if only I understood it ... My understanding is that for $F\mapsto H^Y\circ F$ to define a natural transformation it needs to (for given $(X,Y,Z)$) constitute a morphism from $Hom(X,Hom(Y,Z))$ to $Nat(Hom(Z,-),Hom(X,Hom(Y,-)))$. It's this last bit that I don't get. In what sense is $H^Y\circ F$ a natural transformation between $Hom(Z,-)$ and $Hom(X,Hom(Y,-))$? –  Jochen Trumpf Jun 14 '12 at 11:35
    
Ah, I think I get it now. For given $W$, $H$ is an element of the image of $W$ under the functor $Hom(Z,-)$, and it gets mapped to $H^Y\circ F$, the image of $W$ under the functor $Hom(X,Hom(Y,-))$. Need to think of $F\mapsto H^Y\circ F$ as a "nested" map with additional argument $H$ then it makes sense. Great. –  Jochen Trumpf Jun 15 '12 at 4:44

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.