Is there a typo on this remark? Should it be $(\bar u,\bar v)$ instead of $(u,v)$? I'm working through a text that provides an introduction to differential geometry, and the author writes the following in one of the demonstrations:

(...) Let $X(u,v) = (x(u,v), y(u,v),z(u,v))$ be a regular parameterized surface and let $\overline{X}(\overline{u}, \overline{v}) = X \circ h=X(h(\overline{u}, \overline{v}))$ be a reparametrization of $X$, denoting by $(u,v) = h(\overline{u}, \overline{v})$ (...)

I believe there is a typo, and the following would be correct:

(...) Let $X(u,v) = (x(u,v), y(u,v),z(u,v))$ be a regular parameterized surface and let  $\overline{X}(\overline{u}, \overline{v}) = X \circ h=X(h(\overline{u}, \overline{v}))$ be a reparametrization of $X$, denoting by $\color{red}{(\overline{u}, \overline{v})} = h(\overline{u}, \overline{v})$ (...)

Since taking $(u,v) = h(\overline{u}, \overline{v})$ then $\overline{X}(\overline{u}, \overline{v}) = X \circ h = X(u,v)$, the same parametrization! And this would be rather pointless. I haven't provided further context because I don't think it's necessary (if it is I'll do so, though). 
 A: This is probably no typo, but using the notation in a different way as you might think. I am not a big fan of dropping arguments (e.g. writing $x$ instead of $x(t)$) and associating variable names with more than what they actually mean ($f(x)$ is the same function as $f(s)$, we just renamend the bound variable). Both can lead to confusion as seen here.
What the author probably wants to say with the line $(u,v)=h(\bar u,\bar v)$ is that $u$ and $v$ should be seen as functions of $\bar u$ and $\bar v$, hence
$$h(\bar u,\bar v)=(u(\bar u,\bar v),v(\bar u,\bar v)).$$
So we have
$$\bar X(\bar u,\bar v)=X\circ h(\bar u,\bar v) = X(u(\bar u,\bar v),v(\bar u,\bar v))\quad\not=\quad X(\bar u,\bar v).$$
The right side would be the same as $X(u,v)$ because renaming the variables does change exactly nothing. But since $u$ and $v$ are reused as functions in this context, we should not use them as names for bound variables.
Note that writing $(\bar u,\bar v)=h(\bar u,\bar v)$ makes $h$ to do nothing. This defines the identity and is the same as $h(u,v)=(u,v)$.
