When I think of mappings and double/surface integrals or mappings and triple/"volume" integrals, I think of a mapping going from the parameter domain to the domain of interest. So consider a surface integral over some surface $S$
$$ \iint_S f(x,y,z)\,dS$$
I might come up with a nice mapping $g(u,v) = ( x(u,v), y(u,v), z(u,v) )$, in which a simple region in the parameter domain maps to the complicated surface in the domain of interest. I can now create an integral over the parameter domain
$$\iint_{u,v} f(g(u,v)) \|\vec{g}{_u} \times \vec{g}{_v}\|\,du\,dv $$ Question
Consider the single integral $$ \int_a^b f(x)\, dx.$$
When I set up a $u$-substitution, I create some function $u = g(x)$, use a linear approximation which becomes a true equality in the limit $du = g'(x)\,dx$, and try to get everything in terms of $u$. However, this seems like a map going in the wrong direction? I'm just trying to reconcile how my textbook talks about mapping/multiple integrals and the simplest case of all of this, being $u$-substitution. My textbook is always presenting the theory using a map from the parameter domain to the domain of interest.
The type of mapping you do with double/triple integrals is what, in single-variable calculus, is sometimes called "inverse substitution". The most often-encountered example is probably trigonometric substitutions.
This is where you declare $x$ to be a function of some other variable, say $x=g(t)$, and so $dx = g'(t) \, dt$, and then the integral becomes:
$$ \int_a^b f(x) dx = \int_{g^{-1}(a)}^{g^{-1}(b)} f(g(t)) \, g'(t) \, dt $$
In the trig-sub case, you usually do something like $x = \sin(t)$, and $dx = \cos(t) \, dt$. In this way, the multi-variable change-of-variables formula is more like "inverse substitution".
Both types of substitution are essentially just the "chain rule":
$$ \frac{d}{dt} \, f(g(t)) = f'(g(t)) g'(t) $$
$u$-substitution goes from the right-hand side to the left, and inverse substitution goes from the left-hand side to the right.