# Is the image of a proper smooth embedding always a closed set?

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 $$N$$.
• $f$ 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 $\mathbb R^n$. Commented Jun 15, 2021 at 16:18
• Here is an MSE answer in a more general context which is relevant to (1): math.stackexchange.com/a/1605659 Commented Jun 15, 2021 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 $f(M)$? Commented Jun 15, 2021 at 16:28
• 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. Commented Jun 15, 2021 at 16:44

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.
• 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? Commented Jun 16, 2021 at 14:52
• 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. Commented Jun 16, 2021 at 17:26