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Algebraic topology

  1. Jul 11, 2007 #1
    Please read the following problem first:

    Suppose n > 1 and let S^n be the n-sphere in R^{n+1}. Let e be the unit-coordinate vector (1,0,...,0) on S^n. Prove that the fundamental group pi_1(S^n;e) is the trivial group.


    Okay, now my question is what does the notation "pi_1(S^n;e)" mean?

    I understand the fundamental group but I don't understand the "semi-colon and then an element"-part.

    Thank you! :confused:
     
  2. jcsd
  3. Jul 11, 2007 #2
    As pi_1(S^n) is the group of equivalence classes of loops in S^n, pi_1(S^n;e) is the group of equivalence classes of loops in S^n that start and end at the point e.
     
  4. Jul 11, 2007 #3
    Thank you so much jimmysnyder for clearing that up for me. I looked through Allen Hatcher's book but it wasn't helpful.

    So for our example above, is pi_1(S^n) isomorphic to pi_1(S^n;e)? Do you think I need to show that they are isomorphic or can I assume this as "obvious"?
     
  5. Jul 11, 2007 #4
    It's in his book entitled "Algebraic Topology". It's hidden on page 26, just above Proposition 1.3 with no indication that it is a definition.

    I don't know. But if a space X is path connected, then there is no difference between [itex]\pi_1(X)[/itex] and [itex]\pi_1(X; x)[/itex] for all [itex]x \in X[/itex].
     
    Last edited: Jul 11, 2007
  6. Jul 11, 2007 #5
    Hi jimmysnyder, I think what was confusing (at first) is that the notation on these practice problems used a semi-colon, instead of a comma. Semi-colon can mean Homology with Coefficients, which I didn't know.

    Yes, I agree that there is no difference between [itex]\pi_1(X; x)[/itex] and [itex]\pi_1(X; e)[/itex] but I think I need to prove that these two groups are really isomorphic by using basepoint change homomorphism (Propositon 1.5 page 28 of Hatcher), don't you think so? And after doing that, re-prove that [itex]\pi_1(X; x)=0[/itex] (Prop 1.14 page 35)?

    This is a lot of work but I'm not sure what we're allowed to assume.

    Also, there's another problem that I came across:

    Let [itex]I = [0,1][/itex]. Let [itex]X[/itex] be a space, and let [itex]p[/itex] and [itex]q[/itex] be two points of [itex]X[/itex]. Give an example of a connected space [itex]X[/itex] and points [itex]p[/itex] and [itex]q[/itex] such that [itex]\pi_1(X;p)[/itex] is not isomorphic to [itex]\pi_1(X;q)[/itex].

    So [itex]\pi_1(X)[/itex] does depend on the basepoint!
     
    Last edited: Jul 11, 2007
  7. Jul 12, 2007 #6
    I don't have time right now to look at it. There is a difference between connected and path connected. This is the red herring rule in mathematics. A red herring is neither red, nor is it a herring.
     
    Last edited: Jul 12, 2007
  8. Jul 12, 2007 #7
    It is easy to see that for any space X, [itex]\pi_1(X; x) \subset \pi_1(X)[/itex] since every loop that begins and ends at x is a loop. Since by Proposition 1.14 (page 35), [itex]\pi_1(S^n) = 0[/itex] for [itex]n\ge2[/itex], most of the work is done. Note that [itex]S^0[/itex] is not path connected, and [itex]\pi_1(S^0;1) \ne \pi_1(S^0)[/itex]. That is why in the statement of the problem, n > 1.

    I don't have a formal proof that [itex]\pi_1(X;x) = \pi_1(X) \ \forall x \in X[/itex] when X is path connected. But informally, given a loop in X, there is a loop that starts at x, goes to the start point of the loop, loops back to the start point and then goes back to x (because X is path connected). so [itex]\pi_1(X) \subset \pi_1(X; x)[/itex].

    It seems you have no more to do than to prove that [itex]S^1[/itex] is path connected to show that [itex]\pi_1(S^1;1) = \pi_1(S^1) = Z[/itex]. So that group is not trivial.
     
    Last edited: Jul 12, 2007
  9. Jul 12, 2007 #8
    Thank you so much for your help!
     
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