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In my lecture we proofed the "easy" seifert van kampen theorem namly:

If $X=U\cup V$ U,V open and U,V,$U\cap V$ path connected. Let $x_0\in U\cap V$ then $\pi_1(X,x_0)\cong\pi_1(U,x_0)*_{\pi_1(U\cap V,x_0)}\pi_1(V,x_0)$. where the amalgamation is induced by the inclusions.

In Hatchers book the theorem is:

If $X=\bigcup_\alpha U_\alpha$ is an open covering such that all $U_\alpha$ , all $U_\alpha \cap U_\beta$ and all $U_\alpha\cap U_\beta \cap U_\gamma$ are path-connected and all intersection contain $x_o$. then $\pi_1(X,x_0)\cong *_\alpha\pi_1(U_\alpha,x_0)/N$ where N is generated by the elements $i_{\alpha\beta}(\omega)i_{\beta\alpha}(\omega)^{-1}$ for $\omega\in\pi_1(A_\alpha\cap A_\beta)$.

My question is can I proof the second statment with the first. I tried to do it inductivly by taking $U_\beta,\bigcup_{\alpha\neq\beta}U_\alpha$ as a covering in the sense of the first statment but then I dont really know how the Amalgamation goes on if I divide $\bigcup_{\alpha\neq\beta}U_\alpha$ again in to two open sets.

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  • $\begingroup$ Why would you assume $\alpha$ only ranges over a finite index? $\endgroup$ – PVAL-inactive May 10 '17 at 18:17
  • $\begingroup$ There is a more general theorem referenced at math.stackexchange.com/questions/198348/… using the fundamental groupoid on a set of base points, not assuming path connectivity, and proved by verifying the universal property, which makes the general case as easy as the 2-fold covering case. $\endgroup$ – Ronnie Brown May 11 '17 at 7:02

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