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This is from p.156 of Topology, Munkres: The unit sphere $S^{n-1}$ in $\mathbb{R}^n$ is path-connected, since it is the continuous image of the surjective function $g: \mathbb{R}^n -0\to S^{n-1}$ by $g(x)=x/|x|$. (Note that the punctured euclidean space $\mathbb{R}^n -0$ is path-connected for $n>1$.)

But how can I show that $g$ is continuous? I have an approach but this looks not cool: Since the domain and codomain are metric spaces, we can use epsilon-delta method. So $$\begin{align*} \left|\frac x {|x|} - \frac y {|y|}\right| &= \frac1{|x||y|}\Big|\big(x|y|-|x|y\big)\Big|\\ &\le \frac1{|x||y|}\Big(\Big|\big(x|y|-|y|y\big)\Big|+\Big|\big(|y|y-|x|y\big)\Big|\Big)\\ &\le \frac1{|x||y|}\Big(|y||x-y|+||y|-|x|||y|\Big)\\ &\le \dfrac{|x-y|}{|x|}+\dfrac{|y-x|}{|x|}\\ &<2\dfrac{\delta}{|x|}\;. \end{align*}$$

So at first by choosing delta to be smaller than $2\dfrac{\delta}{|x|}<\epsilon$, this function is continuous.

  1. Is it right?
  2. Are there any cool method?
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6  
It's easier the split this sort of thing up into steps. (1) The function $x$ is continuous. (2) The function $|x|$ is continuous. (3) If $f$ is a continuous function which doesn't vanish, then $1/f$ is continuous. (4) If $f$ and $g$ are continuous, then $fg$ is continuous. –  froggie May 10 '12 at 1:29
    
Never mind go with froggie –  john w. May 10 '12 at 1:30
    
@froggie I thought your comments apply only to functions to real numbers. How can we multiply a function to $\mathbb{R}^n$ with a function to $\mathbb{R}$ in (4)? –  Gobi May 10 '12 at 1:39
1  
@Gobi: It's still true that if $f$ is a continuous function to $\mathbb{R}^n$ and $g$ is a continuous function to $\mathbb{R}$, then $fg$ is a continuous function to $\mathbb{R}^n$, but I guess this requires proof. Of course, you could just split $f$ into coordinate functions $f = (f_1,\ldots, f_n)$ so that $gf = (gf_1,\ldots, gf_n)$. Then you can use the fact that $gf_i$ is continuous for each $i$. –  froggie May 10 '12 at 1:45
    
@froggie Got it, great! –  Gobi May 10 '12 at 2:03

1 Answer 1

up vote 2 down vote accepted

The map $g$ is the projection map when the unit sphere is taken to be the quotient space of $\mathbb{R}^n-0$ under the equivalence relation that identifies point on the same ray from the origin.

In fact, you can prove that $g$ is continuous directly from the definition using open sets: A basis for the induced topology of the sphere is the intersection of balls from the space with the sphere. The inverse images under $g$ of these sets are open cones in the space.

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Sure it is, but to use this to prove the continuity of $g$ you need to show that the topology of $S^{n-1}$ is the same as the quotient topology. Can you do that without proving in the process that $g$ is continuous? –  Mariano Suárez-Alvarez May 10 '12 at 2:17
    
@Mariano, a basis for the induced topology of the sphere is the intersection of balls from the space with the sphere and these are also the images of open cones under $g$. Are the details too hard? Or have I just proved that $g$ is continuous? –  lhf May 10 '12 at 2:19
    
Well, that observation is exactly the same as «the preimage of the open sets of a basis of the topology of the sphere under $g$ are open in $\mathbb R^n$» :) –  Mariano Suárez-Alvarez May 10 '12 at 2:22
    
The answer was a bit abstract, but with the comments it helped me. –  Gobi May 10 '12 at 2:30
    
@Gobi, I've edited my answer. I hope it's clearer now. –  lhf May 10 '12 at 2:50

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