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I was asked a few "challenge problems". Maybe it's not that hard, but i don't know how to solve them.

1) What's the fundamental group of $R^3 \setminus \{ \{z\text{-axis}\} \cup \{ x^2 + y^2 =1\}\}$?

2) What's the fundamental group of $(S^1 \times S^1) \setminus \{\text{a point}\}$?

I know that the fundamental group of $(S^1 \times S^1)$ is isomorphic to $\mathbb{Z} \times \mathbb{Z}$. but take out a point?

3) What's the fundamental group of $R^n \setminus \{m\text{ distinct points}\}$ $(n \ge 2)$?

I have a feeling that I need to use induction on this?

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In 3) it it easy to see that you can always move each loop a bit to omit removed points and then contract this loop. So the fundamental groups is trivial. – Damian Sobota Dec 12 '11 at 1:54
For 2), more generally, what happens when I have a cell complex and remove a point from one of the top-dimensional cells? – Thomas Belulovich Dec 12 '11 at 1:56
And for 1), can you find a suitable deformation-retract of $\mathbb{R}^3-({z-axis}\cup {x^2 + y^2 = 1})$ onto a nice space? – Neal Dec 12 '11 at 2:15
What Damian said is only true when $n > 2$. When $n = 2$ you get a different answer. – Jason DeVito Dec 12 '11 at 2:29
@AlexJ. and Neal : When you write $z-axiz$ without using \text{}, then not only is proper spacing respected, but also the word is improperly italicized and the hyphen actually looks like a minus sign instead of a hyphen. If you put z\text{-axis} inside TeX, then it looks like this: $z\text{-axis}$. That's the way to do it. – Michael Hardy Dec 12 '11 at 4:01
up vote 7 down vote accepted

For $2)$, a torus with a point deleted should deformation retract to a wedge sum of two circles (note a square with an inside point deleted deforms to its edges and use the induced retraction on the quotient space by identifying the edges of the square). So the fundamental group is isomorphic to $\mathbb{Z}*\mathbb{Z}$.

For $3)$, when $n=2$, it should be a free product of $m$ copies of $\mathbb{Z}$ (I don't really know how to make the statement truly rigorous,but the conclusion is true). For $n > 2$, argue by induction to show that deleting m points has the same fundamental group as deleting $m-1$ points. Let $P_1,\dots,P_m$ be the deleted points. $X$ be the space with $\{P_1,\dots,P_{m-1}\}$ points deleted, $U$ be a nbhd of of $P_m$ in $X$, $V=\mathbb{R}^n-\{P_1,\dots,P_m\}$. $U\cap V$ is homeomorphic to $S^{n-1}$(thus having a trivial fundamental group). And use the van Kampen theorem to conclude $\pi_1(V) \cong \pi_1(X)$. Thus the fundamental group is trivial for $n>3$. (I believe this is an excercise in Hatcher).

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for 2) shouldn't it be a figure eight? – Alex J. Dec 12 '11 at 2:50
In 3), in case of n=2, one can write a deformation retraction of the plane without those $m$ points onto a wedge of $m$ circles. And this proves that the fundamental group is a free product of $\mathbb{Z}$'s. – Damian Sobota Dec 12 '11 at 2:53
@AlexJ. I don't really see the difference between a figure eight and a wedge of two circles – Michael Luo Dec 12 '11 at 2:57
i take it back. I wasn't reading it carefully. – Alex J. Dec 12 '11 at 2:59
for 3) n=2, $\pi_1 (\mathbb{R}^n) \setminus (a \ point)=\mathbb{Z}$ and is trivial when n>2. So the same holds with the similar reasoning. Got it! – Alex J. Dec 12 '11 at 3:02

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