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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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    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).2012-03-17

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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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    @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)2012-11-11