Hi, I've been trying to prove this statement for a while now but haven't made much progress:(adsbygoogle = window.adsbygoogle || []).push({});

Suppose [tex]f:[a,b]\to\mathbb{R}[/tex] has the property that for each [tex]y[/tex] in the image of [tex]f[/tex], there are EXACTLY two distinct points [tex]x_1,x_2\in [a,b][/tex] that map to it. Then, [tex]f[/tex] 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 [tex]f[/tex] is continuous. Construct a function [tex]g:[a,b]\to[a,b][/tex] as follows. Each [tex]x\in [a,b][/tex] has a unique partner [tex]x'\in[a,b][/tex] such that [tex]f(x)=f(x')[/tex] via the two-to-one property of [tex]f[/tex]. So, we can have [tex]g:x\mapsto x' [/tex] and [tex]g[/tex] is also bijective. Now, here's the problem. I have an instict that [tex]g[/tex] is continuous because of [tex]f[/tex] 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:)

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# Two-to-One Function

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