Is it possible to construct a function f(x) which maps the closed interval [0,1] to [0,1] such that any...

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Problem

Is it possible to construct a function f(x) which maps the closed
interval [0,1] to [0,1] such that any element in the range is the image
of exactly TWO elements in the domain?

Pictorially, any horizontal line crosses the graph of y=f(x) either 0 or
2 times.  In this problem, f need not be continuous.

Alternatively, can anyone give me a proof that such a function is
impossible?

Show solution

I would suggest that it is not possible to do. If a horizontal line
crosses y=f(x) twice then y=f(x) must have 'come back' on itself at some
point, which I think means that dy/dx = 0 for some value of x. At this
point the horizontal line will only cross the line once.

-- Julian Cracknell <jcrackne@mic.dundee.ac.uk>

Which proves the function cannot be differentiable at at least one
point, but not that it can't exist.

I think, though, that you're right-- it isn't possible.  It is
possible on [0,1), [0,1).

f(x) = (2x, 0<=x<0.5)
f(x) = (2(x-0.5), 0.5<=x<1)

-- Matthew T. Russotto <russotto@wanda.phl.pond.com>

Consider the function f: [0,1] -> [0,1] given by

        f(x) = 2 * x - 1,        if x >= 1/2,
        f(x) = 4 * x,                if x = 2^(-k) for some integer k > 1,
        f(x) = 2 * x,                 otherwise,

I claim f is the required 2-to-1 function.  In fact, f maps each of the
intervals [0,1/2) and [1/2,1] onto [0,1], and f is 1-to-1 on each
subinterval.

-- Dave Seaman <ags@seaman.cc.purdue.edu>