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I'm having trouble demonstrating this map is a quotient map, $f: \Bbb R^2 \to \Bbb R^2$ defined by $(x,y)\mapsto (x \cos(y),x\sin(y))$ with $x \neq 0$. Showing the map is surjective isn't difficult but I'm stuck with the other requirements for quotient maps.

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Doesn't a little open box map to a little open annular sector? –  user66345 Apr 10 '13 at 3:21
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@user66345: Unfortunately, $f$ is neither open nor closed. It is open whenever $x\neq 0$. –  Stefan Hamcke Apr 10 '13 at 11:48
    
okay so assuming $x\neq 0$ –  user71839 Apr 10 '13 at 12:10
    
No no, that's not what I meant. I was referring to @user66345's comment which suggests that the map is simply an open map. But it isn't. It can still be a quotient map, though. –  Stefan Hamcke Apr 10 '13 at 12:57
    
@StefanH. Thanks, you're right. –  user66345 Apr 10 '13 at 14:00

1 Answer 1

up vote 2 down vote accepted

This could be helpful: A map $f:X\to Y$ is called pseudo-open if for each $y\in Y$ and every open neighbourhood $U$ of $f^{-1}(y)$, the interior int$f(U)$ of the image contains $y$. One can show that open or closed surjections are pseudo-open and that pseudo-open maps are quotient maps.

Now show that $f$ is pseudo-open.
Where $x\neq0$ the function is open anyway. Show that a small basis neighbourhood $(x-\epsilon,x+\epsilon)\times(y-\epsilon,y+\epsilon)$ around $(x,y)$ has an image which contains a small enough ball $B_\delta(f(x,y))$. I'm not sure, but you could try $\delta=$ min$(\epsilon/2,\ x\epsilon/2)$. You'll need the addition theorem for cosine.
For $0$ the fiber $f^{-1}(0)$ is just the $y$-axis. Show that an open set containing the $y$-axis has an open image using a compactness argument. Ask if you don't know how.

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