If $v = |u|^{p-1}\mathrm{sign}\, u$, how can we explicitly write $u$ as a function of $u$?
If we didn't have the sign function there, we could just write $$u = v^{m},$$ with $m= 1/(p-1)$. The motivation is the following. In the PDE $$(|u|^{p-2}u)_t - \mathrm{div}(|Du|^{p-2}Du)=0, \quad 1 < p < \infty,$$ can we use a change of variables to rewrite it in the form $$v_t - \mathrm{div}(|Dv^m|^{p-2}Dv^m)=0$$ for some $m$?
OK, we have that:
$v = |u|^{p-1}\mathrm{sign}\, u$
From here, it's obvious that $u$ and $v$ have the same sign. $(*)$
It's worth noting that: $sign (u) = u / |u|$, for $u \ne 0$
1) If $p=1$ we obviously cannot solve for u, so we have no inverse function.
Why? Well, because $v=1$ always (for any $u$).
2) If $p \ne 1$ we get:
2.1.) $u \ge 0$
In this case $v = u^{(p-1)}$ and so: $u = v^{1/(p-1)}$
2.2.) $u \lt 0$
In this case we get:
$v = (-u)^{(p-1)} \cdot (-1)$
$-v = (-u)^{(p-1)}$
$-u = (-v)^{1/(p-1)}$
$u = - ( (-v)^{1/(p-1)} )$
So using $(*)$ we can write 2.1.) and 2.2.) as follows:
$ u = \left\{ \begin{array}{ll} v ^ {1/(p-1)} & \mbox{if } v \geq 0 \\ - ( (-v) ^ {1/(p-1)} ) & \mbox{if } v \lt 0 \end{array} \right. $
And now, we can luckily unify the two cases/branches:
$ u = sign(v) \cdot |v|^{1/(p-1)}$
This formula is true for $p \ne 1$ and for all values of $v$.