Show that If $f : \mathbb{R}^2 \to \mathbb{R}$ is continuous, then $f$ is not injective.
My idea is to use the fact that continuous functions send connected in connected and if I take the origin to $ \mathbb{R} ^ 2 $ its connection is maintained while $ \mathbb{R} $ loses its connection
Prove by contradiction. Suppose that $f$ is injective. Let $g:[0,2\pi]\rightarrow\mathbb{R}^{2}$ be defined by $g(x)=(\cos x,\sin x)$. Note that $g$ is continuous. Moreover, $g\mid_{[0,2\pi)}$ is injective. Therefore, $\theta:=f\circ g$ is continuous and $\theta_{[0,2\pi)}$ is injective. Denote $a=\theta(0)$ and $b=\theta(\pi)$. Note that $a\neq b$. Without loss of generality, assume that $a<b$. Let $c=\frac{1}{2}(a+b)$. Since $g(x)\rightarrow g(0)$ as $x\rightarrow2\pi^{-}$ and $f$ is continuous, we have $\theta(x)\rightarrow a<c$ as $x\rightarrow2\pi^{-}$. Hence, there exists $x_{0}\in[0,2\pi)$, sufficiently close to $2\pi$, such that $\theta(x_{0})<c$.
Observe that $\theta(0)<c<\theta(\pi)$ and $\theta(\pi)>c>\theta(x_{0})$ By intermediate value theorem, there exist $x_{1}\in(0,\pi)$ and $x_{2}\in(\pi,x_{0})$ such that $\theta(x_{1})=c$ and $\theta(x_{2})=c$. However, this contradicts to the fact that $\theta\mid_{[0,2\pi)}$ is injective.
