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I need to calculate the fundamental groups of the following spaces:

$X_1 = \{ (x, y, z) \in \mathbb{R}^3 | x \neq 0\} $

$X_2 = \mathbb{R}^3 \backslash \{ (x, y, z) | x = 0, y = 0, 0 \leq z \leq 1 \}$

$X_3 = \mathbb{R}^3 \backslash \{ (x, y, z) | x= 0, 0 \leq y \leq 1 \} $

I'm not sure at all how one would calculate these. I think that $X_1$ is still a convex space, so the fundamental group might be {1} but I'm really not at all sure...I need to calculate the fundamental groups of the following spaces:

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The fundamental group of X2=X1, and the first groups is the same of $\mathbb{S}^1$. –  checkmath Mar 17 '12 at 12:45
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Your first space is not path-connected, it doesn't make sense to speak of "the" fundamental group for it (even if the fundamental groups for any base point is the same in this case). –  Najib Idrissi Mar 17 '12 at 13:59
    
Contrary to chessmath's comment, $X_1$ is not $\mathbb{S}^1$, and because the first isn't even connected I don't like saying it has the same fundamental group as $X_2$ (and $X_2$ is, incidentally, deformation retractable to a sphere). –  mixedmath Mar 17 '12 at 16:51

1 Answer 1

Try to draw the spaces - the answer should be "obvious" from the drawings.

$X_1$ is just $\mathbb{R}^3$ sliced in two by a plane. Each part is convex. Loops are connected, so they lie in exactly one of the parts, and so can be contracted to a point. So $\pi_1(X_1)=0$.

$X_2$ is just $\mathbb{R}^3$ with a finite length line removed, so $X_2$ deformation retracts to $\mathbb{R^3}\backslash \{0\}$, which has trivial $\pi_1$. This is because $\mathbb{R}^3\backslash \{ 0\} \cong S^2 \times \mathbb{R}$ and therefore $\pi_1(X_2) \cong \pi_1(\mathbb{R}\times S^2) \cong 0$.

$X_3$ deformation retracts to $\mathbb{R}^3$ with a line removed. Now intuition tells you that this should have fundamental group isomorphic to $\mathbb{Z}$.

(and I think you should be able to find a deformation retract of $\mathbb{R}^3 \backslash \{\text{a line}\}$ to $\mathbb{R}^2\backslash \{ 0\}$, which has fundamental group $\mathbb{Z}$.)

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Dear Frederik, why is every loop in $\mathbb{R^3}\backslash\{0\}$ contained in a convex subspace? –  Georges Elencwajg Mar 17 '12 at 16:19
    
@Georges: It's not ;) I've corrected my answer. Thanks. –  Fredrik Meyer Mar 17 '12 at 16:56
    
It is not true that $\pi_1(X_1)=0$ because there are loops that can not be contracted into a point. For example, take a loop which turns around the x axis. –  user42912 Nov 10 '12 at 20:20
    
@user42912: I don't see how that's a counterexample. You are allowed to contract any such loop. (read my description of $X_1$ as a disjoint union of two half-spaces) –  Fredrik Meyer Nov 11 '12 at 0:11

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