TITLE: Proof regarding n-connectedness
QUESTION [0 upvotes]: If one has to prove that $R^n - \{0\}$ is not $(n-1)$-connected, is it necessary to prove formally that there exists a non contractible $(n-1)$-sphere or can that simply be stated. If one must formally prove this, how may that be done? Can one, for example, consider the intersection of the $(n-1)$-sphere and $R^n$ with an arbitrary 2D plane and then say (for the sake of contradiction by the definition of 1-connectedness) that the given sphere is contractible if and only if the circle that corresponds to the intersection of that sphere is contractible in the subspace topology defined by the intersection of the plane and $R^n$?

REPLY [1 votes]: This is not immanent in the definitions, so there is need for a proof.
Your approach, however, is not sufficient. In general, there could be a contraction of the $n-1$-sphere outside the plane, without the existence of a contraction inside the plane. I think your approach would say something about relative homotopy groups.
The standard approach (at least I don't know an easier way) to show that $R^n\setminus \{0\}$ is not $n-1$-connected goes by first showing that $R^n \setminus \{0\} \simeq S^{n-1}$ (homotopy equivalence). Therefore, all homotopy groups will be isomorphic.
So it remains to show that $S^{n-1}$ is not $n-1$-connected. It is well known that $S^{n-1}$ is $n-2$-connected, so by the Hurewicz theorem, $\pi_{n-1}(S^{n-1}) \cong H_{n-1}(S^{n-1})$. Homology is not so hard to calculate, in particular $H_{n-1}(S^{n-1})\cong \mathbb{Z}\neq 0$, so $S^{n-1}$ is not ${n-1}$-connected. 
See also Wikipedia for this.
{"set_name": "stack_exchange", "score": 0, "question_id": 3061415}