I would be grateful if someone takes a look at my solution to the following problem to see if it is correct. Thanks a lot for helping people even during the holidays.
Sorry, I have to delete the question due to some moral stuff.
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I hope you know that your whole question can be still seen in the edit-history. – Daniel Bernoulli Dec 30 '16 at 08:26
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@DanielBernoulli, the OP is not searchable which is good. The question was once on a comprehensive exam, and I thought that I shouldn't share the solution with others unless it is not searchable. – Aleph-null Dec 30 '16 at 19:13
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I assume that you want to get a comment on your proof.
Here it is: Your argument looks solid, but I don't understand, why you take $\pi_1(U)=\langle \alpha\rangle$ instead of $\pi_1(U)=\langle \beta\rangle$, because if you take away the point $x_0$ where $A,B,C$ are identified, an obvious deformation retraction in my opinion would retract $Z\setminus\{x_0\}$ to $D$. Then for $\phi\colon \pi_1(U\cap V)\to \pi_1(U)$ the image $\phi(\gamma)=\alpha^3$ is really evident by this deformation retraction.
All in all your solution is plausible.
Off topic: The knot $K$ should be a torus knot. :)
Daniel Bernoulli
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@DanelBernoulli, thanks man, I will edit my post. +1 for your clarification. – Aleph-null Dec 28 '16 at 19:08