Let $f: [0,1] \to \mathbb{R}$ be absolutely continuous and bijective. Is $f^{-1}$ absolutely continuous? Let $f: [0,1] \to \mathbb{R}$ be absolutely continuous and bijective. Is $f^{-1}$ absolutely continuous?
If we had the additional condition $f'>0$ then it is true that $f^{-1}$ is absolutely continuous (exercise 5.8.52 from this book). I suspect that it will not hold without this condition, but I haven't been able to come up with an example.
 A: [Method: Start with a known example that is not a.c. (I will write "a.c." for "absolutely continuous"), and work backward.]  
Let $c : [0,1]\to[0,1]$ be the Cantor function; $c$ is not a.c.  
Define $g(x) := x+c(x)$.  Now $g$ is continuous with $g(0) = 0, g(1) = 2$.  So $g$ is bijective and strictly increasing from $[0,1]$ onto $[0,2]$.  And $g$ is not a.c.  Indeed, it has $g'(x) = 1 + c'(x) = 1 + 0 = 1$ for almost all $x$, but
$$
g(1)-g(0) = 2 \ne 1 = \int_0^1 g'(x)\;dx .
$$
Remark. For $0 \le x \le y \le 1$ we have $g(y)-g(x) \ge y-x$.  Indeed,
$c$ is nondecreasing, so $g(y)-g(x) = y-x+c(y)-c(x) \ge y-x$.
Now let $f :=  g^{-1}$.  Then $f$ is continuous, bijective from $[0,2]$ onto $[0,1]$.  We claim $f$ is a.c.
From the Remark above: for $0 \le u \le v \le 2$ we have $f(v)-f(u) \le v-u$.  This Lipschitz property implies that $f$ is a.c.  Indeed, let $\delta > 0$ be given.  Let $(u_i,v_i):1 \le i \le n$ be disjoint intervals with $\sum(u_i - v_i) < \delta$.  Then
$$
\sum |f(v_i) - f(u_i)| = \sum (f(v_i) - f(u_i)) \le \sum (v_i - u_i) < \delta.
$$
This completes the proof. 

As the OP notes, we cannot have $f'(x) > 0$ for all $x$.  Here we have $f'(x) = 1$ on a dense open set of measure $1$ (the domain of $f$ has measure $2$), but probably $f'(x) = 0$ for almost all other points.
