boundary of image subset of image of boundary Notation: Let $A^o,\bar{A},\partial A, A^c$ denote the interior, closure, boundary & complement of $A$, resp. 
This is a question I had while reading Buck's Advanced Calculus (I'm on the corollary to Theorem 4, Sec 8.3)
Let $D\subseteq \mathbb{R}^n$ be compact.  If it's known $T(D^o) \subseteq T(D)^o$, why is it true that $\partial T(D) \subseteq T(\partial D)$? 
I'm confused b/c we know LHS=$T(D)\cap (T(D)^o)^c$, which is  $\subseteq T(D) \cap T(D^o)^c$ by the assumption. RHS=$T(D\cap (D^o)^c)$, which we only know is $\subseteq T(D) \cap T((D^o)^c)$, since we don't know T is 1-to-1. 
Here is more background for my question.  We consider $\Omega \subseteq \mathbb{R}^n$ open, $T:\Omega \to \mathbb{R}^n$, $T \in C'$ in $\Omega$.  Furthermore, its Jacobian $J_T \neq 0$ on $\Omega$.  We are trying to show, if $D \subseteq \Omega$ is compact and $\partial D$ has no volume, then $\partial T(D)$ has no volume either.  Buck has already convinced me that $T(D^o) \subseteq T(D)^o$, but then he asserts $\partial T(D) \subseteq T(\partial D)$, which is where I'm lost.
 A: Take $y \in \partial T(D)$, that means there is a sequence $(x_n)_n \subset D$ such that $T(x_n) \rightarrow y$. $D$ is compact so $(x_n)_n$ has a subsequence $(x_{n_k})$ who converge at $a \in D$, then $T(x_{n_k})$ converges to $T(a)$, but the sequence $T(x_n)$ is already converge to 7, either do $T(x_{n_k})$. So $T(a)=y$. The problem leads to prove $a \in \partial D$.
We have $D=D^0 \cup \partial D$ because D is compact then close, so if $a \in D^0$, $y=T(a) \in T(D^0) \subset T(D)^0$ (as you said that you ok with this), this yields a contradiction that we took $y \in \partial T(D)$, so $a \in \partial D$
A: Let $f:X\to Y$ be a set function between spaces. Let $C\subset X$ be closed. Then, as sets $C=\partial C\amalg \mathrm{int}C$. Suppose $fC\subset Y$ is also closed. Then as sets $f(C)=\partial f(C)\amalg \mathrm{int}f(C)$. Under these assumptions we have the following situation as sets.
$$\partial C\amalg \mathrm{int}C=C\longrightarrow f(C)=\partial f(C)\amalg \mathrm{int}f(C)$$
Suppose $f$ commutes with interiors. Then the second component of the above arrow - the one out of $\mathrm{int}C$ - lands in $\mathrm{int}f(C)$. Thus, if $\partial f(C)$ is nonempty, it's in the image of the first component. That is $$f(\partial C)\supset\partial f(C).$$
In your case, $f(C)$ is closed because $f$ is continuous and $C$ is compact. In your case, $f$ commutes with interiors because it's open as a local homeomorphism (by the inverse function theorem).
