Let $U$ be a star-shaped domain in $\mathbb{C}$. Prove that the subset obtained from $U$ by removing a finite number of points is not simply connected.
Usually when we want to show something is simply connected we show that any two curves with fixed endpoints are homotopic by finding a suitable homotopy. However, what can you do if you need to show that the domain is not simply connected? Can we somehow show that there doesn't exist a homotopy?
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With no technology available whatsoever this is an unreasonable question to ask a new student. It is exactly as difficult as the following theorem.
The punctured plane $\Bbb C \setminus \{0\}$ is not simply connected. In fact, the loop $\gamma_r(\theta) = (r\cos \theta, r \sin \theta)$ is not null-homotopic for any $r>0$.
If you know this fact, you can use it to prove your result.
If not, you are not ready yet to prove the stated claim. This fact is non-trivial and the proof is non-obvious.
If $\Omega$ is simply connected then $Ind_{\gamma} (z)=0$ for any closed path $\gamma $ in $\Omega$ and any point $z $ not in $\Omega$.
If $z$ is one of the points removed from your region there exists a closed disk $\overline {D} (z,r)$ around $z$ cantained in $U$ and the index of $z$ w.r.t. the circle of radius $r$ around $z$ is not zero. Hence, the region is not simply connected.