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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|| $.

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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 $V$.

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    $\begingroup$ In the $m>n$ case, you have to say a bit more, because the continuous injection might not be a homeomorphism. One thing you can do is write $U$ as a countable union of open subsets whose closures in $U$ are compact, and then by the Baire category theorem, the image of one of these sets must have nonempty interior. By compactness, our continuous injection is actually a homeomorphism onto an image when restricted to this set. $\endgroup$ – Eric Wofsey Feb 13 '16 at 4:25
  • $\begingroup$ Ah yeah, good point, and nice argument $\endgroup$ – Michael Harrison Feb 13 '16 at 4:35
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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.)

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    $\begingroup$ Doesn't the embedding renders the real line to be a closed set? ($\mathbb{R}$ is a closed subset of $\mathbb{R}^2$) $\endgroup$ – NeerajKumar Feb 6 '16 at 14:04
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    $\begingroup$ A set being open is relative to the space in which it is embedded. When defining the embedding as above, it is defined on $\mathbb{R}$ which is open in $\mathbb{R}$. If we were seeing the domain as a closed subset of $\mathbb{R}^2$, we have already embedded it and there is no need of such a function. $\endgroup$ – ChaPi Feb 6 '16 at 21:56
  • $\begingroup$ @ChandrahasPiduri - That's true but the claim by OP that this will satisfy the 1-1 case is false. The embedding maps $\mathbb{R}$, which is by itself open, to $\mathbb{R}^2$ which is closed. But where are the interior points getting mapped to the boundary. There is no boundary of $\mathbb{R}^2$, unless see it as a subset of $\mathbb{R}^3$, in which case the closed set $V=\mathbb{R}^2$ has all the point as boundary point but then it has empty interior! The assumption about non-empty interior precisely removes all such cases. $\endgroup$ – NeerajKumar Feb 7 '16 at 9:45
  • $\begingroup$ Oh, you are right. The image does not have non-empty interior... $\endgroup$ – Paul K Feb 7 '16 at 9:47

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