# Two-to-One Function

1. Jun 10, 2010

### dmuthuk

Hi, I've been trying to prove this statement for a while now but haven't made much progress:

Suppose $$f:[a,b]\to\mathbb{R}$$ has the property that for each $$y$$ in the image of $$f$$, there are EXACTLY two distinct points $$x_1,x_2\in [a,b]$$ that map to it. Then, $$f$$ is not continuous.

Well, I considered two approaches. I figure it has to do with the intermediate value theorem somehow, but I could be wrong. The argument that I'm hoping would do the trick goes as follows and it is a proof by contradiction approach:

We shall assume that $$f$$ is continuous. Construct a function $$g:[a,b]\to[a,b]$$ as follows. Each $$x\in [a,b]$$ has a unique partner $$x'\in[a,b]$$ such that $$f(x)=f(x')$$ via the two-to-one property of $$f$$. So, we can have $$g:x\mapsto x'$$ and $$g$$ is also bijective. Now, here's the problem. I have an instict that $$g$$ is continuous because of $$f$$ but I don't know how to prove it. If this is indeed true, then we're done because of the fixed point theorem.

Any suggestions would be appreciated, thanks:)

Last edited: Jun 11, 2010
2. Jun 11, 2010

### mathman

Suggestion: [a,b] can be divided into two intervals L and U as follows. For each pair (x1,x2) that have the same image and x1 < x2, put x1 in L and x2 in U. I suspect all points in L are then < all points in U (you will need to use the fact that there are exactly two points going into the image). Since they must make up the entire interval [a,b], L and U will meet at some point x. There will be no other point x' such that f(x)=f(x').