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Dec 20, 2019 at 15:14 comment added HJRW Or, to put it another way, Hatcher doesn't use the assumption that the vertex spaces are aspherical when proving that the inclusion-induced maps are injective. I hope that's some help.
Dec 20, 2019 at 15:11 comment added HJRW @Blade, the above discussion is more or less equivalent to Theorem 1B.11 from Hatcher's Algebraic topology. The only difference is that Hatcher assumes that his vertex and edge spaces are apsherical. But you can make your vertex and edge spaces aspherical by attaching $n$ cells for $n\geq 3$. Since attaching such $n$-cells doesn't change the fundamental group, the result that you want follows.
Dec 19, 2019 at 13:41 comment added Yaniv Ganor The most general form of Van-Kampen I am comfortable with is Ronnie Brown's version with the groupoids and base points, (Although I am aware there is a more general version saying that $\Pi\left(\operatorname{hocolim} X_i\right)$ = $\operatorname{colim} \Pi\left(X_i\right)$ where $\Pi$ is the fundemental groupoid.) but still, I don't know how to use even that to show the claim about the presentation as fundamental group as graph of groups, or how to transition from knowing something about fundamental groupoids back to fundamental group. Is there an easy argument that I am missing? Thanks
Dec 19, 2019 at 13:40 comment added Yaniv Ganor Dear HJRW, Thank you for your answer. Coming back to it now, after having done some reading about Van-Kampen and about graphs of groups (the latter I read from Hatcher), I realize I don't understand why Van-Kampen's theorem implies that $\pi_1(M)$ is presented as the fundamental group of the graph of groups. Now I should say that I do not come from geometric group theory and have little experience with its notions. 1/2
Oct 21, 2019 at 13:02 history edited HJRW CC BY-SA 4.0
Another typo.
Oct 20, 2019 at 15:29 history edited HJRW CC BY-SA 4.0
Corrected typo.
Oct 18, 2019 at 12:16 vote accept Yaniv Ganor
Oct 18, 2019 at 9:35 history answered HJRW CC BY-SA 4.0