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Trouble understanding the path-homotopy equivalence class

  1. Apr 1, 2007 #1

    JasonRox

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    Ok, I'm having trouble understanding the path-homotopy equivalence class. It's kind of blurry when they apply the operation...

    [f]*[g] = [f*g]

    ...where [f] is the path-homotopy equivalence class of f.

    I can see that an element in [f]*[g] is in [f*g], but not the other way around.

    For example, if f is a path from x0 to x1 and g from x1 to x2, then f*g is a path from x0 to x2. This means that [f*g] is the equivalence class of f*g with paths from x0 to x2.

    But take a look at the picture I got. We have a path on the bottom from x0 to x2 without crossing x1, but how does that fit into [f]*[g]?

    What's going on here?
     

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  3. Apr 1, 2007 #2
    I think you're misinterpreting the * sign and mixing up different things here.

    Now, what's the point of defining all these abstract equivalence class things? You want to give it some more structure. By f*g one denotes the composition of the two. However, one requires (in almost all cases I have seen) that the end point of f be the same as the starting point of g, so that f*g is continuous (iff f and g are). A priori there is no product of equivalence classes. However, one can define one (denoted here by °) by setting [f]°[g]:=[f*g], which makes sense (topologically), if end point of f=starting point of g. If you want to equip the whole thing with more structure you choose a "base point", which serves as starting and end point for all paths you consider, i.e. you only consider closed ones. Then the definition of ° makes the set of all equivalence classes with a certain base point a group (the "fundamental group"), as is easily verified.

    Notice that (using your notation) [f]*[g] is per definition [f*g].

    Is it clear(er) now?
     
  4. Apr 1, 2007 #3

    JasonRox

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    I know and I use this in my picture.

    I want to understand why my picture is wrong (if it is).

    I understood most of what you're saying, but I don't see how that applies to my picture. If it does, then I'm completely lost.

    Remember, in my picture, I'm saying that the first graph [f]*[g] consists of all f's with end points x1 and g's with starting points x1. So, they all look something like the first graph on my picture.

    But the equivalence class [f*g], we have an element f*g with starting point x0 and ending point x2, so that implies that [f*g] must include all those functions path-homotopic to f*g. So, here comes my second graph, which is a path from x0 to x2 that is path-homotopic to f*g (which has points x0 to x2). But the graph I have does not contain x1, so it can't be in [f]*[g].

    If you didn't look at the picture, it's probably best to look at it.

    Thanks for your help though. I really need to understand this operation before I'll feel comfortable actually using it.
     
    Last edited: Apr 1, 2007
  5. Apr 1, 2007 #4

    StatusX

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    Just because two paths are homotopic doesn't mean they pass through the same point (except the endpoints, assuming we're looking at homotopies relative to them). The homotopy class [f]*[g] does not consist of just paths passing through x_1. It is exactly equal (by definition) to [f*g]. There is no other interpretation to check consistincy with.
     
  6. Apr 1, 2007 #5
    What part don't you understand?

    I think you didn't quite catch the drift, although I have to admit I haven't had a look at your picture because it's still pending approval.

    Tell me, how did you define [f]*[g]? Because that is not a natural thing, there isn't simply one way of composing equivalence classes! If you define [f]*[g] as the equivalence classes of all paths starting at [itex]x_0[/itex], ending at [itex]x_2[/itex] and going through [itex]x_1[/itex], that's certainly not what [f*g] amounts to!
     
  7. Apr 1, 2007 #6
    O wait, I just really noticed this:
    If that's really your definition of [f]*[g] you're dealing with free homotopy, where the end and starting points don't have to be fixed. But again then, all of your composing of equiv. classes is complete nonsense and certainly far from true.
     
  8. Apr 1, 2007 #7

    JasonRox

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    No, it does contain a fixed starting point for f and end point for g.

    I guess you have to see the picture first. Hopefully, it comes up soon.

    Thanks for everything though!

    Thanks for letting me know what's coming too.:wink:

     
  9. Apr 2, 2007 #8

    matt grime

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    Something in [f*g] is by definition homotopic to f*g, which is obviously in [f]*[g].

    Why should something homotopic to [f*g] have to go through the end point of f/start point of g? This is homotopy - you can deform things around at will. In particular in the picture you can deform the one that doesn't go through x_1 into one that does.
     
  10. Apr 2, 2007 #9

    JasonRox

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    So, [f]*[g] includes the second curve?

    I thought you were suppose to take a curve from [f] say f1 and g1 from [g] and perform the operation f1*g1 to get a curve from [f]*[g]. But f1 and g1 always have the point x1 in it, but the second graph does not.
     
  11. Apr 2, 2007 #10
    You can take some function f1 in [f] representing [f] and some functino g1 in [g] representing [g] and compose them to get a curve in [f]°[g]. But of course, you obtain all functions in [f]°[g] as the homotopy equivalence class of this particular f1*g1, which will give you all functions homotopic to f1*g1, which will also include those which don't go through x1.

    I don't mean to be rude, but did you read what I wrote?
     
  12. Apr 2, 2007 #11

    JasonRox

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    Of course I read it.

    I understand now. I actually made this more complicated than it really was. :grumpy:
     
  13. Apr 2, 2007 #12

    matt grime

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    *UP TO HOMOTOPY* that is what you do. You pick two composable curves in the two homotopy classes you want to compose, and take the homotopy class of the composition. If no such choice is possible, then you cannot compose them (i.e. if the end points lie in different path-connected components, but let's assume that the space is path connected).
     
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