Consider these two functions
f:Z x Z -> Z f(x,y) = x+y
g:Z x Z -> Z f(x,y) = x
There's a fundamental difference between them. In some sense f "makes use" of the fact that we're applying functions to integers, but even though g is acting on integers, it's not using the fact that we're acting on integers. But this is a vague idea. How can we make it precise?
Consider the same in the context of computer programs. This applies in many programming languages, but I'll use Haskell here
> f (x,y) = x+y
> g (x,y) = x
When we ask ghci for the types of these it tells us that f is of type (Num t) => (t, t) -> t but that g is of type (t, t1) -> t. ghci has spotted that f is making use of an interface exposed by integer types (the type class Num) but that g makes no assumptions at all about its arguments. So in a programming language (with static types) we can make the notion above precise by having a type signature that explicitly says something like "I'm not going to make any assumptions about my arguments". But how can we say something like this in the language of set theory, say? Set theory doesn't come with a mechanism for making such promises. (Well, ZF out of the box doesn't, anyway.)
So consider a set-theoretical function like this
h:Z x Z -> Z x Z x Z
h(a,b) = (b,a,a)
What does it mean to say that h doesn't use specific facts about its arguments? It clearly doesn't mean that h doesn't depend on its arguments because the right hand side clearly depends on a and b. We need to say that the right hand side depends on the arguments, but that it somehow doesn't make specific use of them.
Consider an example:
h(1,2) = (2,1,1).
Part of what we mean by not using specific knowledge is that if we were to replace one argument by another value we expect to perform the same substitution on the right. For example, replacing 1 by 3 should give:
h(3,2) = (2,3,3)
So for any 'substitution' function k:Z -> Z we want
h(k(1),k(2)) = (k(2),k(1),k(1))
More generally, we want:
h(k(a),k(b)) = (k(b),k(a),k(a))
Note that we've made use of the functions (a,b) -> (k(a),k(b)) and (a,b,c) -> (k(a),k(b),k(c)) to implement the idea of applying the same substitution to each of the elements. We might want to work with a wide variety of mathematical 'containers' like tuples and sequences as well as more abstract objects. Can we generalise to all of these in one go?
The function that coverts a set X to X x X, or X x X x X forms part of a functor. The other part maps functions between sets to functions between containers. So we can work with the usual ordered pair functor L with LX = X x X and Lf(x,y) = (f(x),f(y)). Similarly we can use MX = X x X x X and Mf(x,y,z) = (f(x),f(y),f(z)). We can stop talking about 'containers' now and talk about functors instead. We can rewrite our rule above as
h(Lk(a,b)) = Mk(h(a,b)).
I've been working with h acting on containers of integers, but if h truly doesn't rely on specific properties of its arguments then there is a version of h for any set X, ie. for any X we have a function
hX : X x X -> X x X x X
hX(a,b) = (b,a,a)
and when we use a function k to substitute values in the arguments there's no reason to substitute values from the same set. For example, if we let A be the set of the 26 letters of the alphabet (using your favourite way to encode letters as sets), then we could let k be the function that maps n to the nth letter (modulo 26) and still expect
hA(Lk(1,2)) = Mk(hZ(1,2))
to hold. For example
hA('a','b') = ('b','a','a')
More generally, h is a family of functions such that for any k:X -> Y we have:
hY o Lk = Mk o hX
Note that there's nothing set-specific about this, it's a statement that could hold in any category.
We can take this as a definition. A natural transformation between functors L,M:C -> D is a family of morphisms hX, one for each object X in C, such that for any morphism k:X -> Y in C we have:
hY o Lk = Mk o hX
And there we have it. Starting with the idea of a function that doesn't 'peek inside' its arguments, we're led inexorably to the idea of substitutability of arguments and from there to a categorical definition of a natural transformation.
As the definition has been generalised from Set to any category you can have fun unpacking the definition again in other categories. For Set-like categories with functions as morphisms you'll often find that natural transformations are still functions that don't make too many assumptions about their arguments, but in each case the rule about what exactly constitutes 'too many assumptions' will be different. The category of vector spaces is a good place to start.
One last point: Category theory was more or less invented for natural transformations. And it's curious how it fits so nicely with the idea of abstraction in computer science, ie. the idea of a function that is limited to using a particular interface on an object.
One interesting result relating to natural transformations is the Yoneda lemma. Note also that the Theorems for Free are also largely about similar kinds of substitutions on function arguments.
Indeed, historically natural transformations were defined before functors or categories - 1942 versus 1945. Eilenberg and Mac Lane started with natural transformations in the category of groups, and then took three years to find the appropriate language to describe them elsewhere.
ReplyDelete"But how can we say something like this in the language of set theory, say? Set theory doesn't come with a mechanism for making such promises."
ReplyDeleteYou can define such "functions" in ZF, although they are not called functions. The powerset (set of subsets) or an ordered pair are basic examples. In Mizar they are called "functors".
In that first set of functions, is that supposed to be g(x,y) = x?
ReplyDeleteSean,
ReplyDeleteYes, well spotted.
Thanks for the help in my struggles with category theory . :P .Much appreciated .
ReplyDeleteSo, what is the definition of a natural transformation unfolded in the category Vect of vectorspaces over a fixed field k?
ReplyDeleteI suspect for A to be a natural transformation in Vect it has to be defined in a polymorphic way only using scalar multiplication and vector addition, i.e. everything of the Haskell type (Vec a) => a -> a, with Vec being a suitable typeclass containing zero, add and mult.
Am I correct?
This is a really great blog btw :-) And that's coming on the heels of many great websites I've stumbled upon these past 2 weeks after discovering Haskell.
ReplyDeleteThis post really hit me right where I was coming to terms with some thoughts on functors. My brain always tries to map to the functor concept anytime I have a "function" defined on proper classes (groups to groups, say) and/or anytime I have something that seems to be defined naturally (like defining a group using only general group operations).
But I couldn't always map this to a functor! This was very discouraging because those two things I actually care about, whereas the functor definition is just abstract stuff.
A common problem I had for example is something like the center of a group. It's a map of type group -> group sending a group to it's center. The set builder definition is as "purely" or "naturally" defined as you could hope, but it is *not* a functor: given a morphism of groups there is really only 1 candidate for the induced morphism of centers, but it isn't a map from center to center (it fails to always lie inside the range).
The trick I think is that the center is a functor if you only consider surjections to be morphisms. You can see that this is exactly what you need to deal with the "for all" quantifier in the definition of the center. Likewise, I imagine a quotient group construction instead of a subgroup construction would necessitate throwing away the surjections and keeping the inclusions. I still haven't completely wrapped my head around it, but I'm closer.
As per the last comment I'm not sure if the notion of functor always maps nicely to "constructive operations" which feels like a stickier notion involving logic and set theory.