Continuous function, mapping of a set to itself

Question

Let $f: [0,1] \rightarrow [0,1]$ be continuous. Any idea on how we can prove that it is not possible for $f$ to map $[0,1]$ onto $[0,1]$ exactly two-to-one. That is, there is no continuous $f$ as above such that for each $y \in [0,1]$, there are exactly two values $x_1$ and $x_2$ such that

$$y = f(x_1) = f(x_2)$$

before going to a function, you should know about a relation. if a relation is many-one , that can not be a function. — mahavir, Mar 01 '14 at 06:53
@Paramanand Singh Thank you, but in this question f is assumed to be continuous, correct? — Sam, Mar 01 '14 at 07:14
The linked question in my comment says that if $f$ is two-to-one then it can't be continuous. Both your version and my version are saying the same thing in different words. — Paramanand Singh, Mar 01 '14 at 07:15
@Paramanand Singh I see what you are saying, helpful hint, I appreciate it — Sam, Mar 01 '14 at 07:17

BigM · Accepted Answer · 2014-03-01T08:14:27.543

0

Suppose such a function does exist. It attains its max and min in [0,1].so there exists $x_1, x_2\in [0,1]$ such that $f(x_1)=f(x_2)=1$ Lets assume $0<x_1<x_2<1$.similarly we can find $0<y_1<y_2<1$ Such that $f(y_1)=f(y_2)=0$. Then by intermediate value property you can find atleast three points $a,b,c$ with $f(a)=f(b)=f(c)$ but this is contradiction.(to be more precise we should split into cases according to the location of $x_i,y_i, i=1,2$ )

edited Mar 01 '14 at 08:14

answered Mar 01 '14 at 08:06

BigM

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Thank you for your answer, so helpful :) – Sam Mar 01 '14 at 23:06

Continuous function, mapping of a set to itself

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