I read in Munkres's topology about Seifert Van-Kampen , classical verson . Assume the hypotheses of the preceding theorem ( Seifert Van-Kampen ) . Let $j : \pi_{1}(U,x_{0}) * \pi_{1}(V , x_{0}) \to \pi_{1}(X,x_{0})$ . But i'm not sure $\pi(U)*\pi(V)$ is well-defined . Anybody explain it for me ? Thanks .
Seifert Van Kampen theorem
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general-topology
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0It is well defined. You can take the free product of any collection of groups. May be you are thinking the map is not well defined or it is not one-one? – 2017-01-23
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0We can define free product if any pair in family group intersect at 1-point and the representation of $1$ by empty word is unique ? Maybe I confuse ? – 2017-01-23
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0Are you sure that there are no more conditions on $U$ and $V$? They intersect at at least one point, which is $x_0$, so the 1-point intersection is done over with. I do not understand the representation of $1$ part. – 2017-01-23
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0$U,V,U \cap V$ are path-connected and $U,V$ are open in $X$ . The representation of $1$ is empty word . You can define free product of a family group $G_{\alpha}$ if reduce word of identity element is unique . – 2017-01-23
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0You can assume the reduced word of identity element is unique and proceed. – 2017-01-23
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0It's matter . I just know that the map induced by inclusion map from fundamental group of $U,V$ to $X$ is generated $\pi(X)$ . But I'm not sure any alternating production of some paths is unique . – 2017-01-23
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0Yes, but this page : https://en.wikipedia.org/wiki/Normal_form_for_free_groups_and_free_product_of_groups says that every word, including identity word, has unique reduced form. – 2017-01-23
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0I think we have a problem such that Munkres's definition is intrinsic definition but by the usual way we can always define free products . Ok , I understood and have no question more . – 2017-01-23
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0Ok great. Always happy to help. – 2017-01-23