On Tue, Apr 10, 2012 at 6:49 PM, Jordan Tirrell <jordantirr...@gmail.com> wrote:
> I tried to answer a few of the remaining category theory questions as best
> I could. This thread is a week old, but I've been away and I saw some
> unanswered questions. I tried looking online for good resources to help
> explain precisely what a category is but most of what I found was either
> vague or assumed significant knowledge of pure math.
Thank you.
> On Tue, Apr 3, 2012 at 4:17 PM, Raul Miller <rauldmil...@gmail.com> wrote:
>> On Tue, Apr 3, 2012 at 3:55 PM, Marshall Lochbaum <mwlochb...@gmail.com>
>> wrote:
>> > "The category of sets" is defined to be the category of sets and the
>> > functions between them. It is an unambiguous term. If there is another
>> > category whose elements are sets (i.e. the category of sets containing a
>> > "base" element and the functions between them fixing that element), it
>> will
>> > be properly delineated.
>>
>> Then this category must contain all possible arrows?
>
> "The category of sets" is a common shorthand for "the category whose
> objects are ALL sets and whose morphisms are ALL functions between sets".
Yes, I had come to that conclusion also: that "the
category of ___" can be the complete category and
other "categories of ____" are possible, which do
not contain all possible arrows.
> I think this should answer your question. Yes, we've included all arrows
> that are functions between our objects (as sets).
>
> Arrows don't have to be functions in general though. So we can throw in
> additional arrows and create a larger category on the same objects (if we
> define composition appropriately), but in this new category we would no
> longer have arrows that represent actual functions.
I had also come to this conclusion. If Arrows can represent
relations then by definition there will be arrows that are
not functions.
>> > An arrow is only associated with one source and one destination. Thus
>> there
>> > is no arrow which leads from 0 to 0 and from 1 to 1. If we are in the
>> > category of sets, this means each function has precisely one domain and
>> > codomain.
>>
>> I think you are telling me that an arrow cannot represent
>> an arbitrary function.
>>
>> Consider the function F(x) = x+1
>>
>> F(1) = 2
>> F(2) = 3
>>
>> There can be no arrow that leads from 1 to 2 which also
>> leads from 2 to 3.
>>
>> We can have an arrow which leads from the set of all
>> integers to the set of all integers, but that arrow
>> cannot distinguish between the above function and any
>> other function on integers.
>>
>> But this does not match what I read when I read about arrows/morphisms.
>
> Let me give two different ways of formalizing natural numbers with
> categories that I think help clarify precisely what categories are.
>
> Let N1 be a category with only one object (ie a monoid), which is the set
> N={0,1,2,3,...}, and arrows which are given by all functions from this set
> to itself (composition is normal function composition). Your function f(x)
> = x + 1 is a valid arrow in this category, we would write f: N -> N to
> specify its domain and codomain. It composes nicely with itself and with
> any other arrows (composition of arrows in our category is given to us by
> composition of functions in our set-theory interpretation).
>
> The key observation here is that the categorical structure is the abstract
> structure of objects, arrows, and composition (N, functions on N, and their
> composition). It doesn't capture "our objects are sets" or "our arrows are
> functions". In the strictly categorical structure of N1, "0" and "1" are
> nothing; they are neither objects nor arrows, we only use them to define it.
I have several difficulties with your presentation, here.
First, any exposition which involves infinity is ambiguous.
But we can resolve this difficulty by considering a non-infinite
set of natural numbers. I shall arbitrarily pick: natural
numbers modulo 5. N= {0,1,2,3,4}
But if we consider our function to be a transformation on N,
there is no distinction between f(x)= x+1 and g(x)= x-1. In
both cases the transformation on N yields N.
I have an additional difficulty, here, since I do not see
how an arrow can represent an arbitrary relation.
Finally, technically speaking, if we go back to dealing with
the infinite set of natural numbers, the codomain of f is not
the set of natural numbers. It's a different set. It's the
set of integers which are greater than 0.
> Let N2 be a category whose objects are the numbers 0,1,2,3,... themselves,
> so we now have infinitely many objects (I'm not specifying what arrows are
> in this category yet). It seems like the earlier confusion arose from the
> desire to have a single arrow in a category like this which represents f(x)
> = x + 1. But we also must define its domain and codomain, so some f0: 0 ->
> 1 would be distinct from some f1: 1 -> 2. The function f(x) = x + 1 can't
> be represented by a single arrow in this category because its domain and
> codomain are not individual numbers, which are our objects. We might be
> able to get a functor out of it, but this depends on our choice of arrows.
Here, my issue is that an arrow cannot represent a
function. Only a collection of arrows can represent a
function. But that collection cannot be a category.
Because if two arrows in a category can be composed the
category automatically includes the composition of those
arrows.
So, here, we cannot, for example, distinguish between f x = x + 1
and h x = x + 2
> A function, like any mathematical thing, can be represented in terms of
> category theory in many different ways and you have to be careful when
> doing so.
It seems to me that, if your statements here are correct,
then category theory cannot fully represent functions nor
relations -- instead we have to first choose which aspect
of the function (or whatever) we are representing and then
build our categories to represent just that aspect (and
not the complete definition of the function).
Put differently: the presentation here gave two different
representations of aspects of a single function, and neither
presentation fully represented that function.
--
Raul
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