Colimits of topological spaces

  • Thread starter galoiauss
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Can someone please explain to me what the following notation/objects are:

(Here X,Y are topological spaces)

colim(X-->Y<--X) where the first arrow is a map f, the second is a map g.
colim(X==>Y), where there are 2 maps f,g from X to Y (indicated by double lines, but couldn't draw 2 arrow heads)
 

Answers and Replies

  • #2
matt grime
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What definition have you got? It should be an object with certain morphisms, and a universality property.

The answer will be a (co)fibration in this kind of example.
 
  • #3
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the only thing I'm familiar with for colimits is if I have a diagram, the colimit in the diagram is the object through which everything factors. i'm not familiar with what a cofibration is; is there a more elementary way of interpreting the above examples?
 
  • #4
Hurkyl
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Incidentally, did you mean colim(X<--Y-->X)? Abstractly, do you know how that relates to the disjoint union X+X?
 
  • #5
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Incidentally, did you mean colim(X<--Y-->X)? Abstractly, do you know how that relates to the disjoint union X+X?
No, the original colim notation is accurate. I would appreciate knowing what the relationship is. Thank you.
 
  • #6
Hurkyl
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Well, the colimit of the diagram

X --> Y <-- X

is very easy: it's Y.


What do you know about colimits?
 
  • #7
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Well, the colimit of the diagram

X --> Y <-- X

is very easy: it's Y.


What do you know about colimits?
On a very informal level. I haven't learned category theory, but one of my teachers introduced some basic terminology and ideas. He mentioned something about if there's a commutative diagram, with a lot of things D going into the colimit, and for any other object Z that the D's map to, there is a unique map from the colimit to that object Z. I'm not really sure I'm getting this, and I'm confused about the 2 examples I listed because I don't see how the 2 are different (it seems like both cases have 2 maps from X to Y, and only the placement of the X is different). Can you explain why the colimits are what you say they are? Thanks.
 
  • #8
Hurkyl
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One difference between the diagrams X==>Y and X-->Y<--X is that the former only has two vertices, while the latter has three. In a cone from a diagram D to the object V, each vertex of the diagram gets its own map to V. So, in a cone from X==>Y, there is a single map originating from X. But in a cone from X-->Y<--X, there is a (possibly different) map from each X.
 
  • #9
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One difference between the diagrams X==>Y and X-->Y<--X is that the former only has two vertices, while the latter has three. In a cone from a diagram D to the object V, each vertex of the diagram gets its own map to V. So, in a cone from X==>Y, there is a single map originating from X. But in a cone from X-->Y<--X, there is a (possibly different) map from each X.
I'm sorry, I'm confused because the example said there were 2 maps in X==>Y, not 1.
 
  • #10
Hurkyl
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A cone from X==>Y to V is a commutative diagram:

Code:
X ===> Y
 \     |
  \    |
   \   |
    \  |
    _| \/
      
       V
I was referring to the fact there is only one map from X to V, as opposed to a cone from the other diagram in which X appears twice, so each X gets its own map to V.
 
  • #11
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I see, that makes sense. Thank you. So in general, if we're given a diagram (such as the examples above) how are we supposed to find the colim? For example, how did you determine that Y is colim for the first example? I want to try to work out the 2nd one myself and check with you if you don't mind.
 
  • #12
Hurkyl
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For X-->Y<--X, it was "obvious". The intuitive comment is that everything already points to Y, so Y must be a colimit of the diagram. (Try working out the proof)
 
  • #13
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Here's another example I just ran into:

If * represents a single point, then colim(X<--*-->Y)=XvY, where v represents the wedge product.

Where is XvY supposed to fit into the diagram X<--*-->Y? Is it just placed in there and we make arrows from every vertex pointing to it? How do we know it's XvY? I'm just lost as to how I'm even supposed to approach this kind of problem. Sorry for my lack of understanding.
 
  • #14
Hurkyl
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Is it just placed in there and we make arrows from every vertex pointing to it?
That's how you said the colimit is defined, right? Actually, it's often drawn like this:

Code:
* ---> X
|      |
|      |
V      V
Y --> XvY
and we don't explicitly draw the arrow from *, because it can be inferred from the diagram.

How do we know it's XvY? I'm just lost as to how I'm even supposed to approach this kind of problem.
Grind through the definition!

I think maybe you're in the wrong mindset at the moment? Maybe you're wondering "How would I calculate that?" -- but for now you might be better off thinking "How can I prove that's the answer?" (Once you get some practice, and learn some tricks, then you are much better equipped to try and answer "How would I calculate that?")

But, either way you look at it, your only real approach at this point is to simply grind through the definition of a colimit.

Code:
* ---> X
|      |\
|      | \
V      V  \
Y --> XvY  \
 \         |
  \        V
   \-----> Z
What if you had a diagram like this? Can you find one map from XvY to Z that makes the diagram commute? Can you show there aren't two?

(Again, I didn't draw the arrows from *. Can you see why I can ignore them?)
 
  • #15
matt grime
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Here's a rough hint as to 'how' to make a colimit. You look at the diagram and then you put anything there that will accept arrows from all the objects. Then you try to think 'how to make it smaller' by removing duplicated information.
 

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