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Direct limit question

  1. Aug 16, 2006 #1
    Hello,

    I'm a bit new to these direct limits (or colimits if you like) and have been using them for free abelian groups for about a month now and so am growing somewhat familiar with them. However I have a new question:

    If you have a case where an object M is the union of a nested family of subobjects U_lambda so your directed set is totally ordered, then it seems as though the direct limit of the U_lambda should be M, along with inclusion maps. That is to say, M and the inclusion maps fit into the definition's diagram quite nicely and it makes sense using an intuitive argument with cofinal sets (keep choosing smaller "tail end" cofinal subsets of your directed set and the direct limit is the same, but then you're using "larger" subobjects that "approach" M)

    However, I'm still a novice and am worried that I have overlooked something. As always, many thanks for your help.
     
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  3. Aug 16, 2006 #2

    Hurkyl

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    It depends, of course, on the category you're using.

    But in Set, you are right -- the nested union is a direct limit of a nested sequence of sets. This is true for any variety of universal algebras (such as groups, rings, or real vector spaces) and many other concrete categories.

    Have you tried giving a rigorous proof? It might be a good exercise. :smile: I think the easiest approach is {given a sequence of maps U_lambda -> N that commute with inclusions, construct the appropriate map M -> N and show it's unique}.
     
  4. Aug 16, 2006 #3

    mathwonk

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    thats right, a direct limit is just a generalized union.

    i.e. direct sum of sets is disjoint union, and a direct limt is the usual union with overlaps.

    this is very sharp of you to realize this!
     
  5. Aug 16, 2006 #4

    mathwonk

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    when trying to prove such thigns always use the mapping property. is that familiar or new to you? it is worth becoming friends with.
     
  6. Aug 16, 2006 #5

    mathwonk

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    i.e. a dirwct limit of sets is a bunch of sets, plus a big set, with maps from the (gee what is it?) little sets into the big sets, but lets see also maps among the little sets.

    oh yeah, thete is a prtial order among the little sets which is basically inclusion, so all the little maps are inclusions,

    well bnlah blah blah, the upshot is that a map out of the direct limit is a familt of amps out of each aset that sends elements that correspond to each other to the same element, i.e. it is a map out of the unino, so the direct limit must be the union.

    i apologize for this cruddy explanaton but it sems soe asy to me, i guess i had a better teacher than me.


    the point is the mapping property is everything. i.e. write down the proecise mapping property of a direct limit and then check that the union has it. done.
     
  7. Aug 17, 2006 #6

    mathwonk

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    heres a standard dirct limit: consider a given point p in the complex plane, and look at all complex analytic functions which are defiend and analytic on some open nset including p. define the ordering by saying a function is greater than another if it is a restriction of the other to a smaller set (which still contains p).

    then the direct limit is the set of germs of functions analytic at p, and coincides with the ring of power series centered at p.

    one can make the same definition for any other type of functions and get a ring of germs, but they do not have such simple descriptions since most other functions are not determined by their sequence of derivatives at p.


    in algebraic geometry this is the localization at p, of the ring of polynomial fiunctions on any affine neighborhood of p. these rings are called stalks then of the sheaf of regular functions on the variety.
     
  8. Aug 17, 2006 #7
    Thanks for your replies.

    Let X be the union of a nested collection of sets U_i

    I've done what I've thought was a formal proof. Here it is (not so formal): The natural set of maps from the items of the direct system U_i to the direct limit X, in this case, would be inclusion, I_i. Then take any other similiar object A along with its maps (I call them a compatible collection) say f_i and then its easy to define a map h:X->A that makes the appropriate diagram commute. Take an x in X and look at the image of x under one of the maps f_i(x) in A then define h(x) to be that value. It seems that X along with the I_i do the job.

    Now the only difficulty I could imagine having with this proof in other categories (mind you, I'm no expert on categories) is whether h also ends up being the appropriate "kind" of morphism.
     
  9. Aug 17, 2006 #8
    I was originally interested as to whether this held for topological spaces, i.e. the case where X was a union of a nested family of open sets. Of course, h is continous so it works. Now I'm interested in when it would not work. Could one of you clever folk provide me an instance where it fails?

    Many thanks,
     
  10. Aug 20, 2006 #9

    matt grime

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    Gah, I had just written a long reply when I inadvertently hit a keyboard shortcut for the back button. (The back space key in Safari.)

    Right, here's the gist:

    It depends on the category you're taking colimits in. One thing I notice is that you're sort of taking the object x first and then working out what it is the colimit of and asking if this fails. I see that as taking things in the wrong order.

    Consider R->R^2->R^3... the nested union of copies of the real numbers. Then this has no colimit in vect, but does have a colimit in Vect (lower case letters mean finite, upper case mean infinite in whatever appropriate sense - dimension here).

    Here is one possible example you might be interested in. Suppose that Comp means a category whose objects are compact metric spaces and where Mor(x,y) is the set of continuous maps. Let x_r = [1/r,1] with the obvious topologies and injections from x_r to x{r+1}, then I claim that the colimit of this exists in Comp and is [0,1], yet if we use the forgetful functor and take the colimit in the category of metric spaces it will be (0,1] (notice the left (I think) adjoint to the forgetful functor is the completion functor, and the completion of (0,1] is [0,1]).

    What is true is that in the category Set, or A-Mod (A a finite dimensional algebra) all objects can be written as direct limits of finite (dimensional) subobjects. Eg, in Set take X a set and start with any element x of X, now consider all possible ways to add one element to x, then consider all possible sets with three elements from X all ordered with inclusions where appropriate. The (huge, filtered) colimit is X.

    You can do the same with a module for A by picking a basis and considering modules generated by finite choices of basis elements. (Yes, I am assuming the axiom of choice at all times.) I do not know if the result holds true in Top, but I think it ought to since Top has quotients.

    One way to generate cases where colimits aren't what you might think is to start with a category C where they are, such as Vect (which is k-Mod for k the ground field) and then take quotient categories and show that the quotient functor does not preserve colimits.

    There are plenty of cases where interesting things happen with colimits, and there are questions such as 'do these things even exist'. These cases often involve categories where you have a choice in defining how one represents objects or morphisms, such as homotopy categoires. This colimit (you are learning, I hope, that colimits are functors from the category of direct graphs)

    * -> *
    |
    V
    *

    for instance does not exist in the homotopy category. The colimit, if it were to exist would be the object you get by joining the two objects at the ends of the arrows along the one at the tails. For instance you can imagine joining two discs along their boundary using it, so that is a sphere, but discs are contractible, so homotopically the colimit could be a point and points and spheres are not homotopic.
     
    Last edited: Aug 20, 2006
  11. Aug 20, 2006 #10
    Well Matt Grime, its clear you know a great deal more about categories than I, though I believe I follow the "gist" of your post. I am not, however, familiar with the notion that direct limits are functions from the category of direct graphs. I'm currently studiing Massey's book on Alg. Top. and using, as a reference, Osbourne's text on Homological Algebra. Where might I find a description of direct limits in terms of direct graphs? This sounds interesting.

    Thanks again,
     
  12. Aug 20, 2006 #11

    matt grime

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    Functors not functions. I don't know where you'd read about it (though this might be a start: http://www.maths.bris.ac.uk/~maxmg/maths/introductory/limits.html).

    It is, though, how one should tihnk about categories. A category 'is' a directed graph. Its objects are the vertices, and the morphisms are the arrows. That is what we have all come to accept is the 'correct' way of thinking about mathematics (if you ascribe to the category theoretic viewpoint), and we all think of functions these days like this: f:X-->Y, i.e. an arrow.

    Thus given a directed graph and a category we can assign to the graphs vertices some of the objects of the category and to each arrow a map. The colimit C of the diagram is then, if it exists, an object with a map from every object in the diagram to C such that all possible triangles commute.

    Thus the coproduct (direct sum) is really the functor associated to this diagram:

    * *

    (* means vertex in these diagrams) and the pull back squares come from the diagram

    *--->*<---*



    If you have that X is the nested union of its subobjects, then the diagram is just the diagram you get when you have one vertex for each subobject and an arrow between vertices if one corresponding subobject is contained in the other.
     
    Last edited: Aug 20, 2006
  13. Aug 21, 2006 #12

    mathwonk

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    It is loads of fun to define things in such abstraction that almost no one can understand them. Here is such a definition of a direct limit of any functor:

    Let I,C be any categories and F:I-->C a functor. We will define the direct limit of F. Given any object X in C, define a constant functor c(X):I-->C taking every object in I to X, and every map in I to the identity map of X. This defines a functor c from C to the category Fun(I,C), taking each object X to the corresponding constant functor c(X). Then a direct limit construction in this setting is an "adjoint functor" for c.


    this is presumably another way to state matt's definition.
     
    Last edited: Aug 21, 2006
  14. Aug 22, 2006 #13

    matt grime

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    What role does F play? It appears in the first line but not afterwards.
     
  15. Aug 22, 2006 #14

    mathwonk

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    sorry. i did not read my own post but copied it from somewhere in ym notes. obviously i did not understand it ymself, now ill have to think about it.

    the idea apparently, from the notation, is that F is analogous to an indexed system with index category I. ill be back shortly, by which time this hint will have probably enabled you to see it before me.
     
  16. Aug 22, 2006 #15

    mathwonk

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    oh, I guess, F plays no role at all, as a direct limit construction makes sense for all directed systems, and I am defining the direct limit construction, on all direcetd nsystems at once, i.e. on all functors from I to C.

    apparently an adjoint functor for my c above would be a functor from the category of all functors from I to C, back to C. so a functor from I to C is like an indexed system of elements of C, and the direct limit is an element of C.

    lets take I to be a directed set, so that Fun(I,C) is the category of all directed systems of elements of C. then the constant functor c takes an element X of C to the constant direcetd system with all indexes assigned to X.

    A direct limit construction takes a directed system F to an element X of C such that the maps out of X are the same as directed systems of maps out of the system F. i.e. the same as a map in the functor category, from F to c(X).


    so if we denote the limit construction by lim:Fun(I,C)--->C, then for all elements F of Fun(I,C), i.e. all directed systems {Xi} of elements of C, and for all elements X of C, the adjointness condition says that

    Hom(limF,X) = Hom(F,c(X)), i.e. in old fashioend language,

    that Hom(lim{Xi},X) = Hom({Xi}I,{X}I), where {X}I denotes the constant family.

    I think this is right.
     
    Last edited: Aug 22, 2006
  17. Aug 22, 2006 #16

    mathwonk

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    this is a good example of the difference between a definition and a construction or existence theorem this isa definition. there is no assertion that direct limits, i.e. adjoints of cthe costnat functor, always exist.

    but if C is sets and I is a directed set, they do.
     
  18. Aug 22, 2006 #17

    mathwonk

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    continuing matts dewscription:

    "Thus given a directed graph and a category we can assign to the graphs vertices some of the objects of the category and to each arrow a map. The colimit C of the diagram is then, if it exists, an object with a map from every object in the diagram to C such that all possible triangles commute. "

    im going to say by analogy a direct sum is an object with maps from every one of the indexed objects into it (now no triangles to commute) and am going topa dd that maps out of the direct sum should correspond to arbitrary indexed families of maps out of the indexed objects.

    hence a colimit of a functor F:I-->C, is an object limF, together with some maps into limF. now this is what it means to have an element of

    Hom(F,c(limF)), i.e. for each element i of I, we have a map from F(i) to limF, and the fact these are maps of functors says all triangles commute. the additional statement about the mapping property says that for all X in C, we have Hom(F,c(X)) i.e. indexed systems of maps, corresponds to

    Hom(limF,X) i.e. to single maps out of limF.


    moreover, if we take the identity map in Hom(limF,limF) we get a special systems of maps in Hom(F,c(limF)) and this is the special system that a limit comes equipped with.

    then given any map limF-->X, since c is a functor, we get a map c(limF)-->c(X), and if we compose with the special system of maps just obtained from F-->c(limF), we get a system of maps F-->c(X), and in fact this is the system that yields the one map limF--->X.


    i.e. the correpondfence between systems of maps F--->c(X) and single maps limF--->X, is given by composition with the special system F--->c(limF) corresponding as above to the identity map.


    which says you do not need all this mumbo jumbo to describe it, just take the indexed system of maps matt described from {F(i)}--->limF, and then the adjoint functor correspondence is given by composing with these.

    i.e. the correspondence taking a map limF--->X to the system of compositions

    {F(i)}--->limF--->X, defines the isomorphism between


    Hom(limF,X) and Hom(F,c(X)) for all X, which shows that lim and c are adjoint functors.


    whew! my brain feels full of smoke.
     
    Last edited: Aug 22, 2006
  19. Aug 22, 2006 #18

    mathwonk

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    i rest my case on the matter of illustrating how category theorists can easily make things so abstract that few people can understand them, including those who state them, like me.

    on the other hand i admit to liking it, and enjoying contemplating how efficiently it encapsulates so much apparently complex data in a few words or symbols.

    i.e. this nonsense i just laid out does sort of make me say to myself: gee is that all there is to it?
     
  20. Aug 22, 2006 #19

    mathwonk

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    this brings me to the topic of stacks, a concept discussed in a paper i just downloaded, where we are told that a stack can be thiough of as a "sheaf of groupoids", thanks a bunch.


    such nonsense should be treated i dont know how. come on guys, a stack is presumably just the theory that developed to understand the local structure of a moduli space of objects that have automorphisms, so is some kind of space plus a finite cover of a nbhd of each point, corresponding to the action of the automorphism group of that object. now i admit i have never even read the definition of a stack, but this is what people were studying, so this must be what it means.


    of course i could be wrong. this is after all the luddite point of view. but try to keep your feet on the ground until they are lifted off by the force of the ideas you are learning.
     
  21. Aug 23, 2006 #20
    Thanks guys

    Matt: I meant functor not function, sorry typo, I know enough to tell them apart. I'm from a small university and so I'm categorically self taught. I haven't seen the notion that categories are diagrams but it seems rather interesting. Can directed graphs be large enough to accomadate categories? I seem to remember directed graphs being defined set theorectically and not with something potentially as large as a class.

    Could you recommend an introduction to more of these ideas (thanks for the website).

    Once again I appreciate all of your help.
     
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