Understanding 1/2 of Van Kampen's Theorem

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The discussion centers on the easy half of Van Kampen's theorem, which states that the fundamental group π1(X) is generated by the fundamental groups π1(A) and π1(B) when A and B are open in X and their intersection is path-connected. It clarifies that π1(A) and π1(B) are not necessarily subsets of π1(X), as the induced homomorphism from inclusion is typically not injective. An example is provided with the inclusion of A = R²\{0} into X = R², illustrating that images of nontrivial fundamental groups can be trivial in π1(X). The theorem also asserts that any pointed loop in X can be homotopically represented as a concatenation of loops from A or B. Understanding these concepts is crucial for applying Van Kampen's theorem effectively.
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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\bigcapB is path-connected. Then \pi1(X) is generated by \pi1(A) and \pi1(B).

\pi1(A) and \pi1(B) are not necessarily subsets of \pi1(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 \pi1(A) and \pi1(B) in \pi1(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 \pi_1(A) and \pi_1(B) "in \pi_1(X)", 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 \pi_1(A) and \pi_1(B) "in \pi_1(X)", 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 \gamma in X is homotopic to some concatenation of loops \sigma_1 * \ldots * \sigma_n with each \sigma_i lying entirely either in A or in B.
 
Great! Thanks very much.
 

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