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Apr 18 at 7:36 comment added HJRW Without the hypothesis that $\pi_1(S)$ embeds into $\pi_1(M)$ and $\pi_1(N)$, it's not even a free product with amalgamation. Also, you the OP may need to understand HNN extensions if $S$ is non-separating in $M$ or $N$.
Apr 17 at 21:02 comment added Sam Nead Ah, one more thing - Fernando Muro was being a bit unclear (or just wrong) when they referred to $\partial S$. After all, $S$ is closed, so $\partial S$ is empty...
Apr 17 at 20:57 comment added Sam Nead You still need to specify how you glue. That induces maps on the fundamental groups... this is not a standard part of the notation, unfortunately! (By the way, your usage of $\pi_1(\partial S)$ is not right... you want the boundary of the normal bundle... anyway.)
Apr 17 at 19:26 comment added Jeremy Yes, the normal bundles are isomorphic in this case. So, will the fundamental group $\pi_1(M\#_S N)$ just be the amalgamated free product $\pi_1(M) \ast_{\pi_1(\partial S)} \pi_1(N)$ due to the Van Kampen theorem?
Apr 17 at 17:49 history answered Sam Nead CC BY-SA 4.0