The norm on $\mathbb{R}^2$ is a quotient map

Question

I think this is a simple example of a quotient map: $$ f:\mathbb{R}^2 \rightarrow \mathbb{R}_{\geq 0} \text{ defined by } f\left(x\right) = \left|x\right|$$ where $\mathbb{R}^2$ and $\mathbb{R}_{\geq 0}$ have the usual Euclidean bases.

A map $f:X\rightarrow Y$ between topological spaces $X$ and $Y$ is called a quotient map if it is surjective and $V \subset Y$ open $\iff$ $f^{-1}\left(V\right) \subset X$ open.

I'm trying to show that this map satisfies those properties. It is obviously surjective. If $V \subset Y$ is open, then let $x\in f^{-1}\left(V\right),$ so that $f\left(x\right)\in V$. Since $V$ open, there is a ball neighborhood of $f\left(x\right)$ contained in $V$, and --- without going through the details --- this pulls back under $f$ to either an open disc or an open annulus centered at $\left(0,0\right)$, either of which is open. Hence, $V\subset \mathbb{R}_{\geq 0}$ open $\implies f^{-1}\left(V\right) \subset \mathbb{R}^2$ open.

I'm struggling with the other direction. If $f^{-1}\left(V\right) \subset \mathbb{R}^2$ is an open set, I want to show that $V$ is open.

My work

Let $y \in V$. Then $f^{-1}\left(y\right) \subset f^{-1}\left(V\right)$, and $f^{-1}\left(V\right)$ is a circle of radius $\left|y\right|$. For every point $x \in f^{-1}\left(y\right)$, there is a ball $B_{r_x}\left(x\right) \subset f^{-1}\left(V\right)$ since the latter set is open in $\mathbb{R}^2$. Hence, $f^{-1}\left(y\right) \subset \cup_{x \in f^{-1}\left(y\right)} B_{r_x}\left(x\right) =: B$ which is an open set. Then $f\left(B\right) \subset V$ is a set containing $y$ in $V$. But I can't see how I could conclude that $B$ is open?

Second attempt

Let $f^{-1}\left(V\right) \subset \mathbb{R}^2$ open. Then for $\left(x,y\right) \neq \left(0,0\right) \in f^{-1}\left(V\right)$, there is a ball neighborhood $B_r \left(\left(x,y\right)\right) \subset V$. But we have $\left(x',y'\right) \in B_{r}\left(\left(x,y\right)\right)$ if any only if we have $$ d -r < \left|\left(x',y'\right)\right| < d + r $$ where we've written $\left|\left(x,y\right)\right| = d$. Hence, $$ f\left(B_r\left(\left(x,y\right)\right)\right) = \left(d-r, d+r\right) \subset V $$ which is an open neighborhood of $f\left(\left(x,y\right)\right)$ in $V$. Thus $V$ is open.

If $\left(x,y\right) = \left(0,0\right)$, the same logic applies except we end up with an interval of the form $\left[0,r\right) \subset V$.

Answer 1

Here is another approach. It is rather long but mainly consists of checking certain facts.

Let $X = \mathbb{R}^2$, $Y = [0, \infty) \subseteq \mathbb{R}$, and define the function $d: X \to Y$ by $d (x \times y) = \sqrt{x^2 + y^2}$ corresponding to the standard Euclidean norm. We want to show that $d$ is a quotient map.

Surjectivity of $d$ is indeed simple to show: for any $y \in Y$, the point $y \times 0 \in X$ is in $d^{-1} (y)$, as is easily checked.

Suppose $V$ is open in $Y$; we want to show that $d^{-1} (V)$ is open in $X$. So let $x \times y \in d^{-1} (V)$. Equivalently, $d (x \times y) \in V$. Since $V$ is open, there is a neighborhood $W$ of $d (x \times y)$ that is contained in $V$. If $d (x \times y) = 0$, then clearly $x \times y = 0 \times 0$. $W$ must contain the interval $[0, \varepsilon)$ for some $\varepsilon > 0$. In this case, $d^{-1} (W) = B_d (0 \times 0, \varepsilon)$. Otherwise, if $d (x \times y) > 0$, then $W$ must contain the interval $(d (x \times y) - \varepsilon, d (x \times y) + \varepsilon)$ for some $\varepsilon > 0$. Note that $B_d (x \times y, \varepsilon) \subseteq d^{-1} (W)$, which can be verified using the triangle inequality for $X$.

Now suppose $d^{-1} (V)$ is open in $X$; we want to show that $V$ is open in $Y$. So let $z \in V$. Consider the set defined by

$$U = \bigcup_{x \times y \in d^{-1} (z)} B_d (x \times y, \varepsilon_{x \times y})$$

where for each $x \times y\in d^{-1} (z)$, $\varepsilon_{x \times y}$ is chosen such that $B_d (x \times y, \varepsilon_{x \times y}) \subseteq d^{-1} (V)$. Construction of $U$ is possible if you allow the axiom of choice. We claim that $d (U)$ is a neighborhood of $z$ that is contained in $V$.

That $z \in d(U)$ is easy to check: take an arbitrary $x_0 \times y_0 \in d^{-1} (z)$; it is contained in $B_d (x_0 \times y_0, \varepsilon_{x_0 \times y_0}) \subseteq U$, which shows that $z = d (x_0 \times y_0) \in d (U)$.

That $d (U) \subseteq V$ is also not hard to show. We have that $B_d (x \times y, \varepsilon_{x \times y}) \subseteq d^{-1} (V) \implies d (B_d (x \times y, \varepsilon_{x \times y})) \subseteq V$ for each $x \times y \in d^{-1} (z)$. Since

$$d (U) = \bigcup_{x \times y \in d^{-1} (z)} d (B_d (x \times y, \varepsilon_{x \times y}))$$

then clearly $d (U) \subseteq V$.

To show that $d (U)$ is open in $Y$, it is sufficient to check that $d (B_d (x \times y, \varepsilon_{x \times y}))$ is open for each $x \times y \in d^{-1} (z)$. You can easily verify that each of these images is an interval $[0, b)$ (if $0 \times 0 \in B_d (x \times y, \varepsilon_{x \times y})$) or $(a, b)$ (if $0 \times 0 \notin B_d (x \times y, \varepsilon_{x \times y})$), both of which are open in $Y$. This completes the proof.

Answer 2

Take $(x,y)\in f^{-1}(V)$. Since $f^{-1}(V)$ is an open set, there is some $\varepsilon>0$ such that$$(x-\varepsilon,x+\varepsilon)\times(y-\varepsilon,y+\varepsilon)\subset f^{-1}(V).$$If $x\ne0$, you can assume, without loss of generality, that $\varepsilon<|x|$. Then, since $V=f\bigl(f^{-1}(V)\bigr)$:

• if $x=0$, $V\supset[0,\varepsilon)$;
• if $x>0$, $V\supset(x-\varepsilon,x+\varepsilon)$;
• if $x<0$, $V\supset(-x-\varepsilon,-x+\varepsilon)$.

Therefore, $V$ is a neighborhood of each of its points, and therefore it is an open set.

Answer 3

Here is an easy way to see $f$ is a quotient map provided you know what proper and perfect maps are and their properties.

Let $K\subseteq \mathbb{R}_+$ be a compact set. Choose $n$ such that $K\subseteq [0, n]$. Then $$f^{-1}(K)\subseteq \{x : |x|\leq n\}$$ so the closed set $f^{-1}(K)$ is compact. Since $f$ is a proper map between metric spaces, its perfect. Thus $f$ is closed. Since $f$ is closed, continuous and surjective, it follows that $f$ is a quotient map.