TITLE: Why does $\{(x,y,z): z \ge 0\}-\{(x,y,z): y=0,0\leq z \leq 1\}$ have trivial fundamental group?
QUESTION [8 upvotes]: I have just begun to learn about the fundamental group.
An exercise asks me to prove that $$X=\{(x,y,z): z \ge 0\}-\{(x,y,z): y=0,0\leq z \leq 1\}$$ has trivial fundamental group.
What I know is:
1) the definition of the fundamental group.
2) X  has trivial fundamental group iff any loop in X can be shrunk into a constant loop at the base point.
3) Homeomorphic (path-connected) spaces have isomorphic fundamental groups.
4) Any convex subset of $\mathbb{E}^n$ and $S^m,m\ge 2$ has trivial fundamental group.
I tried to construct a homeomorphism from X to a convex subset of $\mathbb{E}^3$ such as an area like this:
$$\{(x,y,z): -1\leq y \leq 1,z>0\}$$
But I failed.
Can you please help me? Thank you!

REPLY [2 votes]: If $T$ is your space, then $T$ has the property that if $(x,y,z)\in T$ and $t >0$ then $(x,y,z+t)\in T$.
This lets you find a homotopy between any loop in $T$ to a loop in $T_0=\{(x,y,z): z>=1\}$  But $T_0$ is convex, so it is simply connected.