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I am trying to use S-vK Theorem in reverse; what I know are as follows:

  1. $U$ and $V$ satisfy the requirements (open, path-connected), $U\cup V = X$, $U \cap V = N$
  2. $\pi_1(N) = \langle c,d| cd=dc\rangle$
  3. $\pi_1(U) = ??$
  4. $\pi_1(V) = \langle d\rangle$
  5. $\pi_1(X) = \langle a,b|a^p=b^q\rangle$, where $p,q \in \mathbb{Z}$
  6. When $c$ and $d$ are injected into $X$, they become identity and $b$ respectively.

In fact does it work this way? Any other help? Thank you very much.

wilsonw
  • 1,016

1 Answers1

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I know this is an old question, but here's an answer anyway: let $\pi_1(U) = \langle a,d | a^p = d^q \rangle$ and the map $f: \langle c,d | cd = dc \rangle \to \pi_1(U)$ be given by $f(c) = e, f(d) = d$. The pushout of $\pi_1(U)$ and $\pi_1(V)$ over $\pi_1(N)$ is isomorphic to the $\pi_1(X)$ above.

Alternatively $\pi_1(U) = \langle a,d,c | a^p = b^q, cd = dc \rangle$ and $f(c) = c, f(d) = d$. This gives the same pushout but is not isomorphic to the previous $\pi_1(U)$.