Understanding 1/2 of Van Kampen's Theorem

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Discussion Overview

The discussion revolves around the interpretation and implications of the first half of Van Kampen's theorem, particularly focusing on the fundamental groups of open sets and their relationship to the fundamental group of a larger space. Participants explore the conditions under which these groups interact and the nature of the induced homomorphisms.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant expresses confusion regarding the professor's explanation of the theorem, particularly about the relationship between the fundamental groups of the open sets A and B and the larger space X.
  • Another participant confirms the interpretation that the groups π1(A) and π1(B) are considered in terms of their images under the homomorphism induced by the inclusion into π1(X), noting that this homomorphism is generally not injective.
  • A later reply suggests that even if A and B have nontrivial fundamental groups, their images under the inclusion homomorphism could both be trivial, leading to the generation of the trivial group.
  • Another participant adds that any pointed loop in X can be homotopically represented as a concatenation of loops that lie entirely within A or B, which is a key aspect of the theorem.

Areas of Agreement / Disagreement

Participants generally agree on the interpretation of the theorem and the nature of the induced homomorphisms, but there remains some uncertainty regarding the implications of these relationships, particularly concerning the injectivity of the homomorphisms and the generation of groups.

Contextual Notes

There are limitations regarding the assumptions made about the fundamental groups and the conditions under which the theorem applies, particularly in relation to the injectivity of the induced homomorphisms.

sammycaps
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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.
 
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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.
 
quasar987 said:
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.

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?
 
Right.

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.
 
Great! Thanks very much.
 

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