1. Limited time only! Sign up for a free 30min personal 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!

Algebraic Topology Questions

  1. Nov 6, 2007 #1

    JasonRox

    User Avatar
    Homework Helper
    Gold Member

    I'm totally stuck on these two.

    The first is...

    Let A be a subset of X; suppose r:X->A is a continuous map from X to A such that r(a)=a for each a e A. If a_0 e A, show that...

    r* : Pi_1(X,a_0) -> Pi_1(A,a_0)

    ...is surjective.

    Note: Pi_1 is the first homotopy group and r* is the homomorphism induced by h.

    I can visually see in my mind why this is so, but I can't even think of how to write this down at all.

    I'm still thinking about it. No need to post anything right now.

    The way I'm thinking that is if [f] is in A then I need to show that there is a g in X such that g is path homotopic to f using probably r to create my path homotopy. (f is a loop around a_0 in A)

    Once I do that, then it should come out like... r([g]) = [r o g] = [f].

    I'm still thinking about this.
     
  2. jcsd
  3. Nov 7, 2007 #2
    Can't you use [f] itself for g? I mean, if [f] is a loop in A, then it is a loop in X, and since r fixes A, it should map [f] onto itself.
     
  4. Nov 7, 2007 #3

    JasonRox

    User Avatar
    Homework Helper
    Gold Member

    That's exactly what I was thinking too!

    But you must show that f is a loop in X by creating the path homotopy that I was speaking about from f to g. Isn't that right?
     
  5. Nov 7, 2007 #4
    f is a loop in X because it is a loop in A, and A is a subset of X.
     
  6. Nov 7, 2007 #5

    JasonRox

    User Avatar
    Homework Helper
    Gold Member

    So, to show it is surjective, we have...

    r([f]) = [r o f] = [f]

    If that is so, where does the properties of r even play a role in here?

    I feel like there is a pasting lemma in here or something.

    Is it really that simple?
     
  7. Nov 7, 2007 #6

    JasonRox

    User Avatar
    Homework Helper
    Gold Member

    I do understand this.

    For some reason, I feel like it's a little too simple.
     
  8. Nov 7, 2007 #7
    Well, if r did not fix a then we would not have r(f)=f.
     
  9. Nov 7, 2007 #8

    JasonRox

    User Avatar
    Homework Helper
    Gold Member

    That's true.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Algebraic Topology Questions
Loading...