# Why does $\{(x,y,z): z \ge 0\}-\{(x,y,z): y=0,0\leq z \leq 1\}$ have trivial fundamental group?

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.

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I don't know if you are allowed to use the following technique, but I'll say it anyway. If you have a loop in $X$ you can homotop it to lie in the subspace where $z\geq 1$ by just floating it upward. (Increasing the z-coordinate). Then you can contract it to a point inside the $z\geq 1$ subspace which is convex. (This assumes the basepoint has $z\geq 1$.) –  Grumpy Parsnip Apr 5 '11 at 14:40

You know that convex subsets of $\mathbb{R}^n$ have trivial fundamental group. Check the proof to see that it proves also this: if $X\subset \mathbb{R}^n$ and $a\in X$ are such that for every $b\in X$ the segment connecting $a$ and $b$ is in $X$ (i.e. if $X$ is "star-shaped") then $\pi_1(X,a)$ is trivial. (Since $X$ is then contractible, along the segments going to $a$.)

In your case $a=(0,0,2)$.

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+1 Star-shaped is the canonical way to prove this. –  Sam Nead Apr 5 '11 at 15:02
Ah.. Thank you. My textbook use a homotopy like $F(x,t)=(1-t)f(x)+tg(x),t\in[0,1]$(f and g are maps and F is the homotopy) to prove convex subset of $\mathbb{R}^n$ has trivial fundamental group. I thought that such a homotopy always calls for convexity. Now I know since g is a constant loop, only star-shaped is needed. Also I forgot to use the property that fundamental group is independent of choice of base point. I can figure out the proof now. Thanks again. –  Roun Apr 5 '11 at 15:12
Actually, I don't know if you are assuming this, but the fundamental group is independent of the choice of point only when your space is path-connected; the isomorphism is constructed by using a path going between the two points, which "pulls back" (or pushes forward) the loops based at , e.g., x to those at y. –  gary Jul 13 '11 at 22:31

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.

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Suggestion - break $H$ (your set) into two pieces using the half plane $P = \{ (x,y,z) : y = 0, z > 1 \}$. Show that the two pieces $K^\pm$ of $H - P$ each have trivial fundamental group, as does $P$. Now assemble.

Two more suggestions: 1. If you want to talk about stuff, it helps to give the stuff names. 2. Look up the Seifert-van Kampen theorem.

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Thank you for your suggestions. Uh..are there more basic proofs? I looked up the Seifert-van Kampen theorem and found that I still have 9 sections to go before I reach there in my textbook... –  Roun Apr 5 '11 at 14:49
Oh, I think user8268's proof is more understandable for me. Thank you all the same. –  Roun Apr 5 '11 at 14:59