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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?

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    $\begingroup$ 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. $\endgroup$ – logarithm May 8 at 18:32
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    $\begingroup$ 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 $x, y \in S^1 \subseteq S^{n-1}.$ $\endgroup$ – Robert Shore May 8 at 18:33
  • $\begingroup$ Maybe can I use induction? $\endgroup$ – Mateus Rocha May 8 at 18:35
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    $\begingroup$ You'll only need to write more angles $x=(1,\theta_1,\theta_2,...,\theta_{n-1})$ and $y=(1,\phi_1,\phi_2,...,\phi_{n-1})$. $\endgroup$ – logarithm May 8 at 18:45
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    $\begingroup$ It is easier to prove that the punctured euclidean space is path connected. $\endgroup$ – Randall May 8 at 18:52
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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 $y$.

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