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I want to confirm my answer for this question:

Calculate the fundamental group of $X=\mathbb{R}^3\setminus \{\text{union of n lines passing through origin}\}$.

My idea is that $X$ deformation retracts onto $S^2\setminus \{\text{union of $2n$ number of points}\}$, which will be homeomorphic to $\mathbb{R}^2\setminus\{\text{union of $(2n-1)$ number of points}\}$. Hence fundamental group of $X$ is same as fundamental group of wedge of $(2n-1)$'s circles, which is free group on $(2n-1)$'s symbols. Please confirm it.

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For some basic information about writing math at this site see e.g. here, here, here and here. –  Julian Kuelshammer Nov 30 '12 at 8:18
    
@ Julian : Is my editing fine now? –  Shraddha Srivastava Nov 30 '12 at 8:21
    
yes, only in the title, the spacing is somewhat different –  Julian Kuelshammer Nov 30 '12 at 8:23
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@user51266: your solution looks fine. –  user27126 Nov 30 '12 at 8:46
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+1 for the effort to improve the question. –  Rasmus Nov 30 '12 at 9:28
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Your proof works fine but a little bit more needs to be said. Now in reality $\Bbb{R}^3$ will not deformation retract onto $S^2$ but if you remove the origin then it will. Your deformation retract will then be done as follows. For any point $x$ in $\Bbb{R}^3$ if it is already on $S^2$ then do nothing, otherwise draw the line $l$ through the origin and $x$ and "slide" $x$ to the the point $l \cap S^2$.

For each line through the origin it will intersect $2$ points on $S^2$ and hence removing $n$ lines and then doing the deformation retract means that $X$ is homotopy equivalent to $S^2$ minus $2n$ points. Now if you remove a point then $S^2$ minus that point will be homeomorphic to $\Bbb{R}^2$. I visualize this by imagining the petals of flower when they're closed up (looking like $S^2$) and then opening up and flattening out to look like $\Bbb{R}^2$. Hence

$$X \simeq \Bbb{R}^2 - (\text{$2n-1$ points}).$$ We compute the fundamental group of $X$ by induction on the number of points that you remove. In the base case that $ n =1$, $\Bbb{R}^2$ minus a point is homemorphic IIRC via a variant of the exponential map to $S^1 \times \Bbb{R}$ and so we get that $\pi_1(\Bbb{R}^2 - {x_1}) = \Bbb{Z}$. For the inductive step suppose it is know for $n = k-1$ that the fundamental group of $\Bbb{R}^2 - (\text{$2k - 3$ points})$ is $\underbrace{\Bbb{Z} \ast \ldots \ast \Bbb{Z}}_{\text{ $2k - 3$ times}}$. Then in the case that $ n= k$, by a modified version of my argument given here I think you can choose a line that divides $\Bbb{R}^2$ into two regions $A$ and $B$, $A$ containing $2k - 3$ of those points and $B$ containing two points. By the inductive hypothesis the fundamental group of $A$ is $\Bbb{Z}$ free product $2k -3$ times, and it is easily seen that the fundamental group of $B$ is $\Bbb{Z} \ast \Bbb{Z}$. You can choose $A$ and $B$ so that their intersection contains none of those points, so that $\pi_1(A \cap B ) =0$. Now apply Van Kampen's theorem to get that

$$\begin{eqnarray*} \pi_1(X) &\cong& \underbrace{\pi_1(S^1) \ast \ldots \pi_1(S^1)}_{\text{$2k-3$ times}} \ast \Bbb{Z} \ast \Bbb{Z}\\ &\cong& \underbrace{\Bbb{Z} \ast \ldots \ast \Bbb{Z}}_{\text{$2n -1$ times}} .\end{eqnarray*} $$

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When you write $$\mathbb{R}^2 - (2n - 1 \mbox{ points}) = \cup_{i = 1}^{2n-1} \mathbb{R}^2 - \{x_i\}$$ did you mean to make a $\cap$ instead of a $\cup$, or am I missing something? –  andybenji Dec 7 '12 at 8:39
    
@andybenji Yes that was a mistake in my argument above and I have corrected it. –  user38268 Dec 23 '12 at 11:29
    
For the fundamental group of $\mathbb{R}^2$ minus a point, you can use the same argument from the first paragraph: $\mathbb{R}^2$ minus a point deformation retracts onto $S^1$ just like $\mathbb{R}^3$ minus a point deformation retracts onto $S^2$. –  Jason DeVito Dec 23 '12 at 13:52
    
@JasonDeVito Yes that is true, but I wanted to point out to the OP that we don't just have a homotopy equivalence between two spaces but something even stronger, a homeomorphism. –  user38268 Dec 26 '12 at 0:11
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