Proof of path connectedness of $S^{n-1}$

I need to proof that $$S^{n-1}:=\{x\in\mathbb{R}^{n}\, :\, ||x||=1\}$$, $$n>1$$ is path connected.

So, for all $$x,y\in S^{n-1}$$, I need to show a function $$f:[a,b]\rightarrow S^{n-1}$$ such that $$f$$ is continuous, $$f(a)=x$$ and $$f(b)=y$$.

My proof: Let $$x,y\in S^{n-1}$$, and write it in polar coordinates, i.e., $$x=(1,\theta_{1},\phi_{1} )\quad\textrm{and}\quad y=(1,\theta_{2},\phi_{2}).$$

So, define a function $$f:[0,1]\rightarrow S^{n-1}$$ by $$f(t)=\left(1,\theta_{2}t+(1-t)\theta_{1},\phi_{2}t+(1-t)\phi_{1}\right)$$

So, $$f$$ is continuous, $$f([0,1])\subset S^{n-1}$$, $$f(0)=x$$ and $$f(1)=y$$.

Is my proof correct?

• Your polar coordinates have too few angles for $n$ large or too many for $n=2$, or you need to explain the notation. Otherwise, yes, that idea works. – logarithm May 8 at 18:32
• I think you've sort of got the right idea, but since you're working in arbitrary finite dimensions, it's not immediately obvious that you can write $x$ and $y$ in polar coordinates. To fill in this gap in your proof, you can prove that $S^1$ is path-connected and then prove that any two points $x$ and $y$ are on a copy of $S^1$ such that $x, y \in S^1 \subseteq S^{n-1}.$ – Robert Shore May 8 at 18:33
• Maybe can I use induction? – Mateus Rocha May 8 at 18:35
• You'll only need to write more angles $x=(1,\theta_1,\theta_2,...,\theta_{n-1})$ and $y=(1,\phi_1,\phi_2,...,\phi_{n-1})$. – logarithm May 8 at 18:45
• It is easier to prove that the punctured euclidean space is path connected. – Randall May 8 at 18:52

Your idea works for $$n=2$$. One way to extend to higher dimensions is to show first that antipodal points can be connected by a path. Then for any pair of non-antipodal points $$x,y\in S^{n-1}$$, let $$\alpha(t)=(1-t)x+ty$$ be the straight line connecting $$x$$ and $$y$$, and let $$\tilde\alpha(t)=\frac{\alpha(t)}{\|\alpha(t)\|}$$. Then $$\tilde\alpha:[0,1]\to S^{n-1}$$ is a path connecting $$x$$ and $$y$$.