Construction of a continuous function which maps some point in the interior of an open set to the boundary of the Range

Question

I was studying the Inverse function theorem when I came across the following problems :

(Let the closed set $V$ i.e the range have non-empty interior)

1. Does there exist a continuous onto function from an open set $U$ in $\mathbb{R}^n $ to a closed set $V$ in $\mathbb{R}^m$ such that some points in the interior of $U$ get mapped to the boundary of $V$?

2. Does there exist a continuous $1-1$ map from an open set $U$ in $\mathbb{R}^n $ to a closed set $V$ in $\mathbb{R}^m$ such that some points in the interior of $U$ get mapped to the boundary of $V$?

If there are examples in $C(\mathbb{R})$ i.e continuous functions from $\mathbb{R}$ to $\mathbb{R}$, then that would be great too! Though I do need some example in the general case too.

Simpler examples will be really appreciated.

Thanks in advance.

Edit:

The case (1) can be dealt with using any "cut-off" function. e.g let $U,V$ two balls around $0$ in $\mathbb{R}^n $ with radius $r(>1)$ and $1$, and be open and closed respectively.

Let $f: U \rightarrow V $ such that $x \in V \implies f(x)=x$ and $x \in U-V \implies f(x)= x/||x|| $.

Answer 1

I am assuming you want $V$ to actually be the image of $U$. In this case, there is no such map satisfying your second condition.

If $m = n$, this follows from invariance of domain, since the image of $U$ will necessarily be open.

If $m < n$, there is no continuous injective map from $\mathbb{R}^n$ to $\mathbb{R}^m$ (let alone to a closed subset). You can find some more elementary arguments here, or you can again apply invariance of domain. In particular, if $f : U \to V$ is continuous and injective, and $\iota : \mathbb{R}^m \to \mathbb{R}^n$ is an inclusion map, then $\iota \circ f : U \to \mathbb{R}^n$ is an open map, but the image of $U$ is not open in $\mathbb{R}^n$.

For $m > n$, similar logic: we cannot have a continuous injective map from $U$ onto a set with non-empty interior in $\mathbb{R}^m$ (since $\mathbb{R}^n$ and $\mathbb{R}^m$ are not homeomorphic).

Since we didn't even use the fact that an interior point maps to the boundary of $V$, I suspect there is an easier argument. (Maybe take an interior point $u$ which maps to the boundary, restrict to a small compact neighborhood of $u$, and use the fact that the map on the compact neighborhood is a homeomorphism and must preserve the boundary).

Also, I guess it's possible that you never intended for $V$ to actually be the image of $U$. In this case, we still cannot find such a function when $m \leq n$, based on the same arguments as above, but we can for $m > n$. For example, take $V$ to be the unit disk in $\mathbb{R}^2$ and consider the map $(-1/2,1/2) \to \mathbb{R}^2 : x \mapsto (x,1-4x^2)$, or something similar. This maps $0$ to the boundary of $V$.

Answer 2

The sine/cosine functions map $\mathbb{R}$ to the closed interval $[-1,1]$.

(THIS IS WRONG: Moreover, the embedding $\mathbb{R} \to \mathbb{R}^2, x \mapsto (x, 0)$ satisfies both assertions.)