1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Eeeeaaargh! (homotopy, what else)

  1. Apr 16, 2007 #1

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    1. The problem statement, all variables and given/known data
    Question in my 3 hours final from 10 minutes ago that I thought about for 2 hours:

    Let f,g,h:X-->X be continuous fonctions. If f o h ~ g o h, is it true that f ~ g.


    3. The attempt at a solution
    Nothing convincing.
     
  2. jcsd
  3. Apr 16, 2007 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Of course not. Take h to be a "constant" function: h(x)= p for all x. Then
    f o h= g o h for all f and h.
     
  4. Apr 16, 2007 #3

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Observe that what you just wrote is the definition of "h is a epimorphism". (in the category of spaces and maps modulo homotopy)
     
    Last edited: Apr 16, 2007
  5. Apr 16, 2007 #4

    AKG

    User Avatar
    Science Advisor
    Homework Helper

    What does the ~ stand for?
     
  6. Apr 16, 2007 #5

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    Is homotopic to (as the title of the post implies).
     
  7. Apr 16, 2007 #6

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You meant "for all f and g"?

    And how is this possible. Let f(p)=q and g(p)=r.

    And finally, even if you're right, what does this prove??? :confused:
     
    Last edited: Apr 16, 2007
  8. Apr 16, 2007 #7

    StatusX

    User Avatar
    Homework Helper

    He meant f and g, and this means f and g need not have any relation to each other (other than that f(p) can be joined by a path to g(p)) so if you can find any two that aren't homotopic, you'll have a counterexample.

    This is the same kind of confusion in the other thread, namely, if you are wondering whether A=>B, but you see you can get A in an almost completely generic situation, then it almost always follows that this implication is false, you just need to find one example that satisfies B and one that doesn't (and which both satisfy A, which wll follow from the genericness).
     
  9. Apr 16, 2007 #8

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    What do you mean what does it prove? It is a counter example. Since fh=gh for all f and g, you certainly can't cancel on the right, unless you think all maps are homotopic.
     
  10. Apr 16, 2007 #9

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    In my exam, I said

    Let X=[0,1]u[2,3], let h be the constant function h(x)=0, let f be the identity and g(x)=x if x is in [0,1] and g(x)=x-2 if x is in [2,3].

    Then (f o h)(x) = (g o h)(x) = 0 for all x, so f o h ~ g o h. And then I said, not half convinced, that f is not ~ to g.

    Is that any good? It kind of ressembles what Halls of Ivy said, except he said (f o h)(x) = (g o h)(x) no matter f and g which I can't make any sense of even if my life depended on it.
     
  11. Apr 16, 2007 #10

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    Actually, I mispoke. If h is constant, the fh is constant, and gh is constant for any f,g. Let's suppose these are homotopic, i.e. the points lie in the same path component, and fh~gh. Then are you saying that you think *all* maps must be homotopic? Clearly this can't be: let f be the identity map and g be *any* map not homtopic to the identity. Surely you can think of spaces that are pathconnected (so all constant maps are homotopic), and that have two non-homotopic maps.
     
  12. Apr 16, 2007 #11

    AKG

    User Avatar
    Science Advisor
    Homework Helper

    Yeah, that sounds good.
    What he should have said is this (which is exactly what matt grime also just said):

    Let h be constant, with value c. If f(c) and g(c) lie in the same path-connected component of X, then f o h ~ g o h. This holds for ANY two functions f and g such that f(c) and g(c) lie in the same p-c component of X. This is a very weak condition to impose on f and g, so chances are that there are many function f and g which aren't homotopic to one another, even though f(c) and g(c) can be path-connected.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Eeeeaaargh! (homotopy, what else)
  1. Homotopy Problem (Replies: 6)

  2. Homotopy equivalence (Replies: 0)

  3. Homotopy Classes (Replies: 2)

Loading...