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Algebraic Topology Questions

  1. Nov 6, 2007 #1

    JasonRox

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    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

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    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

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    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

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    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

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    That's true.
     
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