This is a purely theoretic problem which doesn't involve any specific calculation.
Suppose we have two coordinate systems for $\mathbb R^2$ and they are $xy$-system and $uv$-system respectively.
Suppose we have $x=g(u,v)$ and $y=h(u,v)$ and we know both $g$ and $h$ are continuously differentiable, i.e. $C^1$.
Suppose the mapping $T(u,v):=(g(u,v),h(u,v))=(x,y)$ is injective.
Suppose we have a small rectangle $S$ in the $uv$-plane. The general idea to prove the formula for change of variables is approximating the image of $S$ under $T$ in the $xy$-plane by a parallelogram in the $xy$-plane. People use the image of the boundary of $S$ under $T$ to calculate the area of the parallelogram I quoted. By doing this, people use the fact that $T$ preserves boundaries, i.e. $T(\partial S)=\partial [T(S)].$
My question is, how can we assure that $T(\partial S)=\partial [T(S)]$ ?
If the equality is not true, then how do we approximate the parallelogram in the $xy$-plane?
I know the equality is true if $T$ is a homeomorphism, but under our assumption, $T$ may not have a continuous inverse(though $T^{-1}$ is always defined on $T(S)$ due to our injective condition).
Edit: I think I missed an important condition that the Jacobian determinant $\frac{\partial (x,y)}{\partial (u,v)}$ is nonzero in the interior of the integral domain in the $uv$-plane. If we have this condition, then we can use the Inverse Function Theorem to guarantee the existence of local homeomorphism, and then our operation is valid.(Am I right?)
Second Edit: In my calculus book, the formula for change of variables doesn't require the Jacobian determinant is nonzero. I am stuck here again.
Third Edit: Intuitively I feel like the Jacobian must be nonzero and so I am trying to prove it now. Suppose the Jacobian is zero on a set with positive measure, which is denoted by $E$. The key equation in the proof is that $$dx\,dy= \frac{\partial (x,y)}{\partial (u,v)} du\,dv.$$ On $E$, we can have some $du,dv$ s.t. $du>0$ and $dv>0$. Since the Jacobian is zero on $E$, we have $dx\,dy=0$. So a rectangle with area of $du\,dv(>0)$ is mapped to a set with zero measure by an injective continuous function $T$, where $T$ is defined as above. Intuitively I feel like this statement is wrong, but I can not prove it. Any help with deriving a contradiction will be appreciate.
Best Answer
If $S$ is bounded then you can find a compact subset $K$ which contains an open neighbourhood of $S\cup\partial S$. Now $T$ restricts to a continuous bijective mapping $K\to T(K)$ where $K$ is compact and $T(K)$ is Hausdorff, hence a homeomorphism.
If $S$ is not bounded then $\partial(TS)=T(\partial S)$ can fail. Take the vertical strip $S=\{(x,y)\mid 0\leq x\leq 1\}$ and consider $T:\mathbb R^2\to\mathbb R^2$ given by $T(x,y)=(x,\arctan y)$.