One can achieve a 1-1 correspondence from (0, 1) onto R easily by taking tan(pi * x - pi/2).

When I first looked at this problem, I came up with this. I'm curious if it works too.

Let f(x) be given by f(x) = 1/(e ^ x - 1) on {0 < x <= ln((e + 1)/2)}

f(x) = 1/(e - e ^ x) on {ln((e + 1)/2) < x < 1}