When is a boundary mapped to a boundary? In multiple integration exercises one often wants to change the domain of integration using a well behaved function. Say you want to evaluate $\int_D f$, a mapping $\varphi$ transforms $D$ into another domain $\varphi(D)$. From what I've seen, "most of the time", the boundary $\partial D$ is mapped into the boundary $\partial \varphi(D)$. What are the requirements for that statement to hold? That is, when do we have $\varphi(\overline{D} \setminus \mathring{D})=\overline{\varphi(D)} \setminus \mathring{\varphi(D)}$?
 A: Your statement surely holds when $D \subseteq \mathbb{R}^n$ is simply connected and $\varphi$: $D \to \text{Im} \ \varphi(D)$ is an homeomorphism. The proof of this is by absurd: 
Let $x_0 \in \partial D$ and $\varphi (x) \notin \partial \varphi{D}$. 
Then $\left. \varphi(x) \right|_{x_0}$ is an homeomorphism and it induces an isomorphism between the two homology chains: $$ \varphi : D \to \text{Im}(D)$$ $$ H_q(\varphi): H_q(D) \to H_q(\text{Im}(D))$$
By hypothesis $H_q(D) = \lbrace 0 \rbrace \ \forall q \in \mathbb{Z}$. (*)
The second homology sequence contains, for some $i \in \mathbb{Z}$ (**), $H_{i} \cong \mathbb{Z}$, so we get a contradiction.
I think this statement it's valid for every topological variety with a finite number of holes. Using homology you are formalizing the idea that the number of holes is a invariant under homeomorphisms and in fact you are adding an hole with that missing inner point 
(*)  (even without that point, I can retract the space to a point, or in other words, there aren't "holes" inside it)
(**)for example in a $2$-dimensional topological variety with an hole inside it the loop around the hole isn't trivial (= can't be retracted to the constant path). To generalize this reasoning you have to consider (using singular homology maybe), the $n-1$ cycle around the hole and note that it's equivalence class is not trivial.
