$f(\partial K)\subset\partial f(K)$ and $f(\operatorname{Int}K)\subset\operatorname{Int}f(K)$ for a well-behaved $f:\mathbb R^2\to\mathbb R^2$?

Question

Let $f:\mathbb{R}^2\to\mathbb{R}^2$ be a continuously differentiable bijection with nonzero Jacobian. Let $K\subset\mathbb{R}^2$ be a compact subset. Then $f(K)$ is compact. Does the boundary of $K$ map into the boundary of $f(K)$ ? Does the interior of $K$ map into the interior of $f(K)$ ? That is, are $$f(\partial K)\subset\partial f(K)\qquad\mbox{and}\qquad f(\operatorname{Int}K)\subset\operatorname{Int}f(K)$$ true? I was wondering if $C^1$ condition, bijectivity, and nonzero Jacobian are enough to ensure that these are true.

Answer 1

Yes, it is true. Indeed, we only need $f$ to be a continuous injection.

We first prove that $f(\partial K)\subset \partial f(K)$. If not, there exists $x\in\partial K$ such that $f(x)\notin \partial f(K)$. Hence $f(x)\in \text{Int }f(K)$ or $f(x)\in \text{Int }\left(\mathbb R^2\setminus f(K)\right)$. If $y=f(x)\in \text{Int }f(K)$, then there exists $r>0$ such that $B=B(y,r)\subset f(K)$. Since $f$ is injective, we have $f^{-1}(B)\subset K$. The continuity of $f$ implies that $f^{-1}(B)$ is an open set containing $x$, hence $x\in \text{Int }K$, contradicting to $x\in\partial K$. Similarly, we can handle the case where $f(x)\in \text{Int }\left(\mathbb R^2\setminus f(K)\right)$. If $y=f(x)\in \text{Int }\left(\mathbb R^2\setminus f(K)\right)$, then there is an open ball $B=B(y,r)\subset \text{Int }\left(\mathbb R^2\setminus f(K)\right)$. Since $f$ is continuous and injective, we have $f^{-1}(B)\subset \mathbb R^2\setminus K$, and $f^{-1}(B)$ is an open set containing $x$, hence $x\in\text{Int }\left(\mathbb R^2\setminus K\right)$, contradicting to $x\in\partial K$.

Now, we prove that $f(\operatorname{Int}K)\subset\operatorname{Int}f(K)$. By invariance of domain, $f(\operatorname{Int}K)$ is an open subset of $\mathbb R^2$ contained in $f(K)$, hence $f(\operatorname{Int}K)\subset\operatorname{Int}f(K)$.

Note that here we don't need $K$ to be a closed set. In fact, $K$ can be any set. For completeness, we prove that, indeed, we also have $f(\operatorname{Int}K)\supset\operatorname{Int}f(K)$ and $f(\partial K)\supset \partial f(K)$.

Since $f$ is continuous and injective, we know that $f^{-1}(\operatorname{Int}f(K))$ is an open subset of $\mathbb R^2$ contained in $K$, so $f^{-1}(\operatorname{Int}f(K))\subset \operatorname{Int}K$, implying that $\operatorname{Int}f(K)\subset f(\operatorname{Int}K)$.

To prove $\partial f(K)\subset f(\partial K)$, we first prove $f^{-1}(\partial f(K))\subset \partial K$. Since $f(\operatorname{Int}K)\subset\operatorname{Int}f(K)$, we have $f^{-1}(\partial f(K))\cap \operatorname{Int}K=\emptyset$. Using the invariance of domain we can also prove that $f(\operatorname{Int}\left(\mathbb R^2\setminus K\right))\subset\operatorname{Int}\left(\mathbb R^2\setminus f(K)\right)$, hence $f^{-1}(\partial f(K))\cap \operatorname{Int}\left(\mathbb R^2\setminus K\right)=\emptyset$. So, $f^{-1}(\partial f(K))\subset \partial K$. But this time, we cannot conclude that $\partial f(K)\subset f(\partial K)$, because $f^{-1}(\partial f(K))$ could be empty set (I believe that there should be some examples.) If we require $f$ to be surjective, then everything will be OK.

Therefore, we've proved that

If $f:\mathbb R^2\to\mathbb R^2$ is a continuous bijection and $X\subset\mathbb R^2$ is a subset, then $$f(\partial X)= \partial f(X),\qquad f(\operatorname{Int}X)=\operatorname{Int}f(X).$$

Disclaimer: I'm not very good at geometry or topology. It is very likely that I've made some silly mistakes in proving the reverse inclusion. So, please read the proof with a critical eye. If there is something wrong, let me know, please.