The punctured euclidean space is simply connected (dim>2)

**CSM** · January 12th 2011, 04:46 AM

I'm looking for a proof of the statement that $\mathbb{R}^n\backslash \{0\}$ is simply connected.

I've already got a proof, but I'm trying to find another one (one that uses the fact that a product of simply connected space is again simply connected).
Proof: Any loop $\gamma$ in $\mathbb{R}^n\backslash \{0\}$
can be homotopied to a path on the unit sphere $S^{n-1}$ by $\gamma_t(s)=\gamma(s)/||\gamma(s)||^t$ . Paths on the sphere $S^{n-1}$ can be homotopied to a point for $n>2$

**Drexel28** · January 13th 2011, 11:00 PM

Originally Posted by CSM

I'm looking for a proof of the statement that $\mathbb{R}^n\backslash \{0\}$ is simply connected.

I've already got a proof, but I'm trying to find another one (one that uses the fact that a product of simply connected space is again simply connected).
Proof: Any loop $\gamma$ in $\mathbb{R}^n\backslash \{0\}$
can be homotopied to a path on the unit sphere $S^{n-1}$ by $\gamma_t(s)=\gamma(s)/||\gamma(s)||^t$ . Paths on the sphere $S^{n-1}$ can be homotopied to a point for $n>2$

What other proof could you hope for besides using the fundamental group?

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