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Van Kampen's Theorem

  1. Nov 8, 2012 #1
    So I'm having a little trouble with the part of Van Kampen's theorem my professor presented to us. He called this the easy 1/2 of Van Kampen's theorem.

    Theorem (1/2 of Van Kampen's)
    - Let X,x=A,x U B,x (sets with basepoint x) where A and B are open in X and A[itex]\bigcap[/itex]B is path-connected. Then [itex]\pi[/itex]1(X) is generated by [itex]\pi[/itex]1(A) and [itex]\pi[/itex]1(B).

    [itex]\pi[/itex]1(A) and [itex]\pi[/itex]1(B) are not necessarily subsets of [itex]\pi[/itex]1(X), at least in general. So if anyone can enlighten me on exactly what the Professor meant. I would think he just means the embedding of [itex]\pi[/itex]1(A) and [itex]\pi[/itex]1(B) in [itex]\pi[/itex]1(X) but I don't think, at least in general, the homomorphism induced by the inclusion is injective.

    Thanks very much.
    Last edited: Nov 9, 2012
  2. jcsd
  3. Nov 9, 2012 #2


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    Your interpretation is correct on all points. When we talk of [itex]\pi_1(A)[/itex] and [itex]\pi_1(B)[/itex] "in [itex]\pi_1(X)[/itex]", we mean their image by the homomorphism induced by the canonical inclusion. And of course, this homomorphism is usually not injective (i.e. not an embedding). Consider for instance the homomophism induced by the inclusion of A:=R²\{0} into X:=R².
  4. Nov 9, 2012 #3
    Thanks very much! (I changed my original post because I felt I was being too talkative). So, even if A and B have nontrivial fundamental groups, their images under homomorphism induced by the inclusion may both be trivial, in which case they would "generate" the trivial group, right?
  5. Nov 9, 2012 #4


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    And in general, the easy 1/2 of V-K's thm states that any pointed loop [itex]\gamma[/itex] in X is homotopic to some concatenation of loops [itex]\sigma_1 * \ldots * \sigma_n[/itex] with each [itex]\sigma_i[/itex] lying entirely either in A or in B.
  6. Nov 9, 2012 #5
    Great! Thanks very much.
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