Is $\varphi : \mathbb{R^2} -\{0\}\rightarrow S^1$ defined by $\varphi (x)=\frac{x}{|x|}$ a quotient map?

Question

$\textbf{My problem:}$ Prove that $\varphi : \mathbb{R^2} -\{0\}\rightarrow S^1$ defined by $\varphi (x)=\frac{x}{|x|}$ is a quotient map.

$\textbf{My attempt:}$ I need to prove that $V\subset S^1$ is open if and only if $\varphi ^{-1}(V)$ is open in $\mathbb{R^2} -\{0\}$. I know $\varphi $ is a coontinuous function because us the product by two continuous functions. Thus, I have the first implication. However, I have problems with to prove the second implication...any idea?...somebody have another way to resolve this problem?...

Answer 1

Exercise: If $f : X \to Y$ is a continuous map with the property that there exists a continuous map $g : Y \to X$ such that $f \circ g = \operatorname{id}_Y$, then $f$ is a quotient map.

Now note that $\varphi \circ i = \operatorname{id}_{S^1}$, where $i : S^1 \to \mathbb R^2 \setminus \{0\}$ is given by $i(x)=x$.