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I want to compute the $\pi_1(X)$ where $$X=\mathbb{R}^2-(([-1,1]\times \{0\})\cup (\{0\}\times [-1,1]))$$ my only tools at the moment are the basic definitions and the fundamental group of a circle, I think this should be the same group $\mathbb{Z}$ but I don't know how to calculate it explicitly.

Smurf
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    Do you mean $[-1,1]$? Remember that the fundamental group is a homotopy invariant, so if two spaces are homotopy equivalent, then they have the same fundamental group. – John Gowers Jan 26 '16 at 18:15
  • I am trying to follow a book to study this subject by myself, and that concept appears later so I tink I am not suposed to use it yet. – Smurf Jan 26 '16 at 18:19
  • $X$ is homotopy equivalent to $\mathbb{R}^2-(0,0).$ – Brad A.M. Jan 26 '16 at 18:20
  • Which book are you using? – John Gowers Jan 26 '16 at 18:20
  • Is a small book written by a professor in my university "Topología algebraica, Vicente Muñoz" – Smurf Jan 26 '16 at 18:23
  • Since $X$ is homotopy-equivalent to $S^1$, an explicit calculation--without utilizing helper theorems--would be fairly difficult for a beginning student. Are you sure you're not supposed to rely on other results? Perhaps "explicit" means: "Identify an explicit generator for the group"? – PeterJL Jan 26 '16 at 18:27
  • where can I find an explanation for what $\mathbb{R}^2-(([-1,1]\times 0)\cup (0\times [-1,1]))$ would mean ? – reuns Jan 26 '16 at 18:31
  • Well I would be glad if I really need more concepts to solve it because I was getting frustrated. I will ask the author in case this is an errate. – Smurf Jan 26 '16 at 18:42
  • would be possible to show that $\mathbb{R}^2-{(0,0)}$ is homeomorphic to $X$? – Smurf Jan 27 '16 at 14:06

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$X$ is homotopy equivalent to $\mathbb{R}^2-(0,0).$ To see this retract the $(([−1,1]×{0})∪({0}×[−1,1]))$ to the origin. Then consider loops that contain the origin and loops that do not contain the origin.

Brad A.M.
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The space X in questions is homotopy equivalent to $R^{2}\setminus (0,0)$ because you can immage to extend its hole in the two coordinate direction.

Now $\pi_{1}(X)\cong \pi_{1}(R^{2}\setminus (0,0))$ and the last one is $\mathbb{Z}$ because $R^{2}\setminus (0,0)\cong \mathbb{R}\times S^{1}$ (for example see this: Prove that $\Bbb R^2 - \{0\}$ is homeomorphic to $S^1 \times \Bbb R$. ).

Finally $\pi_{1}(\mathbb{R}\times S^{1})\cong \pi_{1}(\mathbb{R})\times \pi_{1}(S^{1}) \cong 0\times \mathbb{Z}$.

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The Number Theorist
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  • The OP says in the comments that they're not allowed to use the fact that fundamental group is a homotopy invariant. – John Gowers Jan 27 '16 at 00:41