# Proving that two links in $\mathbb{R}^{n}$ are not equivalent

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.

• What is $\pi_1(\cdot,\cdot)$??? Sep 5, 2019 at 18:27
• Sep 5, 2019 at 18:39
• 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. Sep 5, 2019 at 18:48
• Thank you so much! Your suggestion is perfect. Sep 5, 2019 at 18:57

$$X$$ 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.
• Yes, I do unfortunately think so. Why must it be that the loop is homotopic to $x$ in $X_{1}$ simply because $x \in X_{1}$? Sep 5, 2019 at 18:59
• Because $X_1$ is simply connected. Sep 5, 2019 at 19:00
• 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? Sep 5, 2019 at 19:14