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:

  • $\begingroup$ The fundamental group of X2=X1, and the first groups is the same of $\mathbb{S}^1$. $\endgroup$ – checkmath Mar 17 '12 at 12:45
  • 1
    $\begingroup$ 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). $\endgroup$ – Najib Idrissi Mar 17 '12 at 13:59
  • $\begingroup$ 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). $\endgroup$ – davidlowryduda Mar 17 '12 at 16:51

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 $\mathbb{Z}$.)

  • $\begingroup$ Dear Frederik, why is every loop in $\mathbb{R^3}\backslash\{0\}$ contained in a convex subspace? $\endgroup$ – Georges Elencwajg Mar 17 '12 at 16:19
  • $\begingroup$ @Georges: It's not ;) I've corrected my answer. Thanks. $\endgroup$ – Fredrik Meyer Mar 17 '12 at 16:56
  • $\begingroup$ 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. $\endgroup$ – user42912 Nov 10 '12 at 20:20
  • $\begingroup$ @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) $\endgroup$ – Fredrik Meyer Nov 11 '12 at 0:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.