# Showing a map is surjective. It seems kind of tricky…

I have a map which I want to show is surjective. (It is only one small part of a much larger problem involving constructing a homeomorphism from the space $[([0,1]\times S^{n-1})/(\{1\}\times S^{n-1})]/(\{0\}\times S^{n-1})$.

If that notation is unfamiliar that's OK because it is irrelevant to my problem below:

I want to show that the map $g:[0,1]\times S^{n-1}\to S^n$ defined by $g(t,x) = \cos(\pi t)e_{n+1} + \sin(\pi t)x$ is surjective.

Here, $S^{n-1} = \{x\in \mathbb{R}^{n} : |x| = 1\}$ and $S^{n}$ is defined in the only consistent way. Also, by $e_{n}$ I mean the $n$th standard basis vector for Euclidean space.

The beginning of my approach is this:

Let $y\in S^{n}$. Then write $y = \Sigma_{i=1}^{n+1}y_{i}e_{i}$. By linear independence, I know immediately that if I want $g(t,x) = y$, I need to have $t = \cos^{-1}(y_{n+1})/\pi$ (chosen from $[0,\pi]$).

After typing this out I'm sort of having an idea, but since I've already typed the entire question out now I might as well post it. I'll post back with any updates. Thanks for any hints or suggestions you can give!

Update: I got it. Take $x = (y - y_{n+1}e_{n+1})$ and scale appropriately.

Sorry to trouble the crowd with this. I guess I just needed to organize the thinking process to figure it out!

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I was "upping" your comment/ replying to it but it disappeared. The vector you pointed out, which is $y - y_{n+1}e_{n+1}$ needed to be scaled by a factor of $\frac{1}{\sqrt{1 - y_{n+1}^2}}$ and then it worked out perfectly and still had norm $\leq 1$. – roo Jan 19 '12 at 4:46
Wait... it needs to be $= 1$. Hmmm... back to the drawing board? – roo Jan 19 '12 at 4:46
I decided to slightly expand it and make it an answer. – Jonas Meyer Jan 19 '12 at 4:47
Do you mean $y_ne_{n+1}$ and $y-y_{n+1}e_{n+1}$ are othogonal? – Jonas Meyer Jan 19 '12 at 4:48
The nice thing is that $\sin(\pi t)$ is always nonnegative for $t\in[0,1]$ so you don't really need to worry about its sign. You just need the fact that if $z\in \mathbb R^n$ has norm $a\geq 0$, then $z=ax$ for some $x\in S^{n-1}$. – Jonas Meyer Jan 19 '12 at 5:02

I would start by noting that $t\mapsto \cos(\pi t)$ maps $[0,1]$ onto $[-1,1]$, and therefore there exists $t\in[0,1]$ such that $y_{n+1}=\cos(\pi t)$. Then you can use the Pythagorean identity for $\cos$ and $\sin$ to find the norm of $\sum_1^n y_ie_i$.