# Prove that $f$ is surjection.

$f:\mathbb{R}^n\to \mathbb{R}$ be a continuous map. $f$ not be a surjection if for all $c>0$, $f^{-1}(c)$ bounded.

My idea: I guess I can use "if $A$ is a connected set, then $f(A)$ is connected." But I can't prove $f$ is a surjection.

• @MikePierce the preimage of a point is not bounded – Andres Mejia Aug 18 '18 at 22:16
• Typing tips: A typographical "function", like \mathbb or \sqrt does not require brace-brackets if you want to apply it to a single keyboard character... And \Bbb is identical to \mathbb. So to get $\Bbb R^2$ you can type \Bbb R^2. – DanielWainfleet Aug 19 '18 at 1:12

Assume that the distance from the origin to any point from $f^{-1}([-1, 1])$ is at most $R$. Let $B_R$ be the ball of radious $R$ with the center at the origin. In particular $f^{-1}([-1, 1]) \subset B_R$. Since $f$ is continuous, this means that $f(B_R)$ is bounded. Let $C > 0$ be such that absolute value of any point in $f(B_R)$ is smaller than $C$. Since $f$ is surjective, there are two points, $a, b\in\mathbb{R}^2$ such that $f(a) = C, f(b) = -C$. These two points ($a$ and $b$) are outside $B_R$. Connect this to points by a continious path which does not intersect $B_R$. There is a point on this path, call it $c$, such that $f(c) = 0$. This is a contradiction, because $c\notin B_R$.
This even means that if $f:\mathbb{R}^2\to\mathbb{R}$ is continuous and surjective, then $f^{-1}(0)$ is unbounded.
• Is this question limited for functions $f:\mathbb{R}^2\to\mathbb{R}$? I do not see, where this is needed in particular. So does it hold in general for $f:\mathbb{R}^n\to\mathbb{R}$? – Cornman Aug 18 '18 at 23:20
• @Cornman Yes, that's true for every $n\ge 2$. The point is that on a plane (or in $\mathbb{R}^n$ for $n\ge 2$) if you have two points which are outside a given ball, then you can connect these two points by a continuous path which does not intersect this ball, while the same is false for a line – Sasha Kozachinskiy Aug 18 '18 at 23:27
• I edited the Q to say $f:\Bbb R^2 \to \Bbb R$ from the original$f:\Bbb R\to \Bbb R,$ which didn't make sense and was surely a typo. – DanielWainfleet Aug 19 '18 at 1:07