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?

  • $\begingroup$ Sorry my previous comment was wrong. Just translate your $\gamma$ until it is in a upper half plane, so it will be clear that you can contract it to zero. For the second part, make a picture and then try to prove formally that $A$ is connected. $\endgroup$ – Nicolas Hemelsoet Jan 3 '18 at 12:12
  • $\begingroup$ @NicolasHemelsoet i sketched a proof can you check it? $\endgroup$ – Manolis Lyviakis Jan 3 '18 at 12:16
  • $\begingroup$ I didnt justify the argument that " i only need to solve my problem for the loops around tha line segment" which is just an intuitevly geometric approach. How would i justify it clearer? $\endgroup$ – Manolis Lyviakis Jan 3 '18 at 12:21
  • $\begingroup$ Your argument is not very clear for "simply connected". For connected, your argument works. There is a cleanest argument for both statement (see my answer). $\endgroup$ – Nicolas Hemelsoet Jan 3 '18 at 12:57

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 :

If $\Omega$ is star-shaped, then $\Omega$ is contractible, that is homotopy equivalent to a point. In particular, $\Omega$ is connected and simply-connected.

Now we just need to verify that $A$ is star-shaped, I claim that this is the case with for example $p = (0,0,2)$.


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