Let the set $S$ be defined as: $$S = \{(a_{1},a_{2} \dots, a_{n}) \mid \Sigma^{n}_{1}a^{2}_{i} = 1, a_{n} = 0\}$$ Then, let $A = \mathbb{R}^{n} - S$. Intuitively, it seems clear to me that $\pi_{1}(A,c) \cong \mathbb{Z}$, where $c$ is an arbitrary element of $A$. If this is true, how can I distinguish between curves in $A$ that are not homotopic to each other? Is there something like the crossing number that can generalize to higher dimensions that I can use for this?

For the sake of reference, my intuitions are based primarily on the arguments offered page 43 of hatcher, under the section on linking circles.

  • $\begingroup$ What is $\pi_1(\cdot,\cdot)$??? $\endgroup$
    – amsmath
    Sep 5, 2019 at 18:27
  • $\begingroup$ The fundamental group. $\endgroup$ Sep 5, 2019 at 18:39
  • $\begingroup$ Lightly thought out suggestion: Perhaps you could consider an equivalence relation such that $x \sim y$ iff $x_n = y_n$ and $\|x\|=\|y\|$? This reduces $S$ to a single point and $X$ to half a plane missing the $S$ point. $\endgroup$
    – copper.hat
    Sep 5, 2019 at 18:48
  • $\begingroup$ Thank you so much! Your suggestion is perfect. $\endgroup$ Sep 5, 2019 at 18:57

1 Answer 1


$X$ is the union of two simply connected open subspaces $X_1,X_2$. WLOG we can suppose $x \in X_1$. A loop with $x$ for base point is included in $X_1$ and homotopic to the constant loop equal to $x$. Therefore $\pi_1(X,x)$ is the trivial group.

Am I missing something?

  • $\begingroup$ Yes, I do unfortunately think so. Why must it be that the loop is homotopic to $x$ in $X_{1}$ simply because $x \in X_{1}$? $\endgroup$ Sep 5, 2019 at 18:59
  • $\begingroup$ Because $X_1$ is simply connected. $\endgroup$ Sep 5, 2019 at 19:00
  • $\begingroup$ Yes, but it is not necessary that the curve only lies in $X_{1}$, right? It can be that it consists of one path in $X_{1}$, followed by a path in $X_{2}$, followed by a path back to $x$, right? Or am I the one missing something here? $\endgroup$ Sep 5, 2019 at 19:14
  • $\begingroup$ How can a continuous curve have parts in two disconnected opens? $\endgroup$ Sep 5, 2019 at 19:17
  • $\begingroup$ The endpoint of one can approach the starting point of the other, right? $\endgroup$ Sep 5, 2019 at 19:18

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