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Suppose $M$ is a $n$-dim. smooth manifold and $f : M \to \mathbb{R}^n$ is a smooth embedding of $M$ into some Euclidean space. I have two related questions.

  1. I have read on this site that a smooth embedding is proper iff its image is closed, i.e. $f(M)$ is closed in $\mathbb{R}^n$. Is this a true statement (using my definitions below)?

  2. Now, a homeomorphism is both an open and a closed map, i.e. it maps open sets to open sets and closed sets to closed sets. When we think of $M$ as a topological space $M$ is both open and closed (provided that $M$ is connected only $\emptyset$ and $M$ are clopen). Since a smooth embedding is in particular a homeomorphism it seems to follow that therefore $f(M)$ is both open and closed in $\mathbb{R}^n$ (which is not possible unless $f(M)$ is equal to the empty set or $\mathbb{R}^n$). Obviously there must be an error in my deduction. If we additionally assume that $f$ is proper, how does this change the fact that $f(M)$ must be open and closed under the homeomorphism $f$?

My definitions (following Lee, Introduction to Smooth Manifolds):

  • $g : X \to Y$ is a proper map between topological spaces, if for any compact $K \subseteq Y$ the pre-image $g^{-1}(K)$ is again compact in $X$.
  • $f: M \to N$ is a smooth embedding, if $f$ is an injective immersion (immersion meaning that the differential ${\rm df}_p$ is injective everywhere) and a homeomorphism onto its image $f(M)$ in the subspace topology of $N$.
jazzinsilhouette
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    $f$ is only assumed to be a homeomorphism onto its image, so it takes open sets to open sets in the subspace topology on $f[M]$, not in the ambient topology of $\mathbb R^n$. – paul blart math cop Jun 15 '21 at 16:18
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    Here is an MSE answer in a more general context which is relevant to (1): https://math.stackexchange.com/a/1605659 – Alex K Jun 15 '21 at 16:18
  • @paulblartmathcop: This is a good point! So this means that $f(M)$ can be closed in the ambient topology of $\mathbb{R}^n$ while being open in the subspace topology induced on $f(M)$? – jazzinsilhouette Jun 15 '21 at 16:28
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    Yup that's exactly right. Imagine the inclusion $S^1 \subseteq \mathbb R^2$. It is closed in $\mathbb R^2$, so closed subsets of $S^1$ are closed in $\mathbb R^2$, but its interior in $\mathbb R^2$ is empty. – paul blart math cop Jun 15 '21 at 16:44

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The second question was answered by paul blart math cop in the comments, so I will tackle the first question. It's true, and here's the proof: First, suppose that we have $f : M \to N$ a proper smooth embedding. Let $x$ be a point in the closure of the image of $f$, and $B$ a closed ball around $x$. Then, $f^{-1}(B)$ is compact, hence $B \cap f(M)$ is compact because $f$ is an embedding, and therefore $B \cap f(M)$ is closed and must contain $x$. We conclude that $f(M)$ is closed.

Now, let suppose that $f : M \to N$ is a smooth embedding whose image is closed. Let $K \subseteq N$ be a compact subset, then $K \cap f(M)$ is a compact subset, and because $f$ is an embedding, $f^{-1}(K) = f^{-1}(K \cap f(M))$ is compact.

Observe that we didn't use the fact that we have smooth manifolds, but only topological manifolds.

Tourbon Kitsch
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  • Thank you for your answer. I have some problems understanding your proof though. First, in the first direction you are talking about a closed ball in $N$. Since $N$ is not generally a metric space i'm not sure what you mean by this. Why is its pre-image compact? – jazzinsilhouette Jun 16 '21 at 14:52
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    Any point has an open neighborhood homeomorphic to $\mathbb{R}^n$, so any point has a closed neighborhood homeomorphic to a closed ball in $\mathbb{R}^n$. So that's what I meant by a closed ball. Its preimage is compact by the definition of a proper map. – Tourbon Kitsch Jun 16 '21 at 17:26