Let $f:\Bbb R^2 \to [0,\infty)$ be defined as $f(x,y)=x^2+y^2$. Show that $f$ is a quotient map. 
Let $f:\Bbb R^2 \to [0,\infty)$ be defined as $f(x,y)=x^2+y^2$. Show that $f$ is a quotient map.

The map $f$ is surjective, but I don't know how to show that $U \subset [0,\infty)$ is open if and only if $f^{-1}(U)$ is open.
If $U$ is open in $[0,\infty)$, then $f^{-1}(U)=\{(x,y) \mid x^2+y^2 \in U\}$ i.e. the set of points in the plane for which the sum of the components is in $U$. How can I show that $f^{-1}(U)$ is open?
And conversely if $f^{-1}(V)$ is open for some subset $V$ of $[0,\infty)$ how can we conclude that $V$ is also open?
 A: $f$ is continous and surjective so it's enough to show that f is an open map.
Remember that $\{B(x,r)|x\in \mathbb R^2 ,r\in [0,\infty)\}$ is a base to the topology on $\mathbb R^2$ so it's enough to show that the image of this sets is open.
And indeed let $B(x,r)$

*

*If $|x|\ge r$: we have thah $f(B(x,r))=((|x|-r)^2,(|x|+r)^2)$
[
Indeed, Let $a \in f(B(x,r))$ thus there exist $y \in B(x,r)$ such that $a = f(y)$ now from the defnition of $B(x,r)$ we have that $|y-x|<r$ and thus from the $|y| \le |x|+|y-x| < |x| + r$ and thus $f(y)=|y|^2<(|x|+r)^2$ , in the same way we have $|x| \le |y|+|x-y|<|y|+r$ and thus $|y|>|x|-r$ and thus $f(y)=|y|^2>(|x|-r)^2$ qnd thus $a=f(y) \in ((|x|-r)^2,(|x|+r)^2)$
In the converse, let $a\in  ((|x|-r)^2,(|x|+r)^2)$ thus $\sqrt a\frac{x}{|x|} \in B(x,r)$ as $|x-\sqrt a\frac{x}{|x|}|=|x(1-\frac{\sqrt a}{|x|})|=|x-\sqrt a|<r$
and $f(\sqrt a\frac{x}{|x|})=a$
]
which is open in $[0,\infty)$


*if $r > |x|$: we have that $f(B(x,r)) = [0,(|x|+r)^2)$ which is open in $[0,\infty)$
And thus $f$ is an open continous and surjective map and thus a quotient map
A: Suppose $S$ is open and saturated with respect to $f$ in $\mathbb R^2$ and let $0\neq p\in S.$ Rotate $S$ so that without loss of generality, $p=(0,b).$ Then, $f^{-1}(f(\{p\}))=\{x^2+y^2=b^2\}\subseteq S$ (because $S$ is saturated.) And since $S$ is open, there is an $0<\epsilon<b$ such that the strip $\overline {(0,b-\epsilon), (0,b+\epsilon)}$ lies in $S.$ It follows that $f(p)\in ((b-\epsilon)^2,(b+\epsilon)^2)\subseteq f(S).$ That is, $f(S)$ is open.
The case $p=0$ is even easier.
We have shown that $f$ maps saturated open sets to open sets, which proves that $f$ is a quotient map.
