As stated i need to prove that the fundamental of the upper half of $R^3$ minus the line $y=0$ and the line segment $ 0 \leq z \leq 1 $ has a trivial fundamental group.Im just started in these kind of things. So i dont know how to start. I know i have to prove that given a random loop on $A$ with a base point $a_0$there exist a path-homotopy to a constant function. But also i have to prove that all fundamental groups are isomorphic since i get different groups on different base points. Which means i must also prove $A$ is path connected .My main questions is as to what strategies are available ? Why cant i just take the straight-line homotopy between any loop? and if i can how do i show that its an ok homotopy to use?
Proof: Can i just say the only loops in $A$ not shrinkable to a point are the loops around the segment $ 0 \leq z \leq 1$ since the loops aroun the axis-y are not in $A$ so no problem for those. Now for those around the segment i just use straight-line homotopy to the same loop just above them. say with a base point (0,0,10) .And those are homotopic to a constant. Now only thing left is THat $A$ is path connected. FOr any 2 points in $A$ just connect them with a line. If that line passes through the line segment $ 0 \leq z \leq 1$ or $y=0$ just go around it following a circle with centre the point that the initial line "struck" the unwanted point. So there we have a path for every point in $A$. Is this a correct proof?
Say that $\Omega \subset \Bbb R^n$ is star-shaped if there is $p \in \Omega$ with the following property : for all $ q \in \Omega$, the segment $[p,q] \subset \Omega$.
The following proposition is clear :
Now we just need to verify that $A$ is star-shaped, I claim that this is the case with for example $p = (0,0,2)$.