A question regarding continuous curves.

Question

Say $f:\Bbb{R^2}\to\Bbb{R}$ is a continuous map. Now take the fibre of $a\in\Bbb{R}$, which is $f^{-1}(a)$. Will it always be a continuous curve in $\Bbb{R^2}$?

I tried constructing examples. Clearly $(x,y)\to x^2+y^2$ satisfies this condition, as $f^{-1}(a)$ is a circle for every $a\in\Bbb{R}$. However, I don't know how to prove this for the general case.

It is easy to make a fibre that consists of a finite number of points, or curves. — André Nicolas, Jun 19 '14 at 05:37
For $f(x,y) = x^2+y^2$, the fibre $f^{-1}(0)$ is a single point and $f^{-1}(a)$ for $a < 0$ is the empty set. — JimmyK4542, Jun 19 '14 at 05:43
@JimmyK4542- Can you map $(0,1)$ to an empty set? Is such a mapping defined? — freebird, Jun 19 '14 at 05:45

score 1 · Answer 1 · answered Jun 19 '14 at 05:40

1

Consider $f(x,y)=xy$ and think about $f^{-1}(0)$.

answered Jun 19 '14 at 05:40

Anurag A

41,067

@Anurag- How do you prove that the union of the X-Y axes is not a curve? – freebird Jun 19 '14 at 05:43

score 0 · Answer 2 · answered Jun 19 '14 at 05:58

0

Look at $f(x,y)=x^2(x-1)^2(x-2)^2+y^2$. Then $f^{-1}(0)$ consists of three points.

answered Jun 19 '14 at 05:58

André Nicolas

507,029

Thank you! This is exactly what I needed. – freebird Jun 19 '14 at 06:08
You are welcome. One can play more elaborate versions of the same game. – André Nicolas Jun 19 '14 at 06:10

A question regarding continuous curves.

2 Answers2