An embedded submanifold is an immersed submanifold for which the inclusion map is a topological embedding. A properly embedded submaniold is one which is embedded and the inclusion map is proper. There are many classical examples of one-to-one immersions which are not emeddings e.g. a line of irrational slope on the 2-torus. The assiciated inclusion map in this case is obviously not proper, but I am having trouble thinking of an embedded submanifold which is not properly embedded. There are examples when the ambient manifold is not Hausdorff, for instance, but can someone think of an example when both manifolds are well behaved?
An example of an embedding which is not proper
differential-geometrydifferential-topologygeneral-topology
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What you write is true, only you have to be a bit more careful with the definitions and the arguments. The inclusion mapping doesn't always have to be a smooth immersion and the statement that "the derivative of the inclusion is the identity transformation" and this doesn't make sense in general. Let me state a definition and then provide some examples:
Let $(M,\tau_M, \mathcal{A}_M)$ be a smooth manifold. An immersed submanifold of $M$ is a triple $(X,\tau_X, \mathcal{A}_X)$ where:
- The set $X$ is a subset of $M$.
- The set $\tau_X$ is a topology on $X$ making $(X,\tau_X)$ a topological manifold.
- The set $\mathcal{A}_X$ is a collection of charts on $(X,\tau_X)$ which defines a smooth structure on $X$.
such that the inclusion map $i \colon (X,\tau_x,\mathcal{A}_X) \rightarrow (M,\tau_M,\mathcal{A}_M)$ is a smooth immersion (and in particular, it is continuous).
Consider the following "artificial" examples:
- Let $M = \mathbb{R}$ with the standard topology and smooth structure. Choose some subset $X \subseteq M$ for which there exists a bijection $\varphi \colon X \rightarrow \mathbb{R}^2$ (as sets!). Use the map $\varphi$ to endow $X$ with a topology $\tau_X$ and a smooth structure $\mathcal{A}_X$ that turns $\varphi$ into a diffeomorphism. Then $(X,\tau_X, \mathcal{A}_X)$ satisfies the first three properties above but the inclusion map is not continuous and in particular can't be a smooth immersion.
- Let $M = \mathbb{R}^2$ with the standard topology and smooth structure and let $X = \{ (x, |x|) \, | \, |x| < 1 \}$. Let $\tau_X$ be the subspace topology on $X$ and consider the map $\varphi \colon \mathbb{R} \rightarrow \mathbb{R}^2$ given by $$ \varphi(x) = \begin{cases} \left( -e^{-\frac{1}{x^2}}, e^{-\frac{1}{x^2}} \right) & -\infty < x < 0, \\ (0,0) & x = 0, \\ \left( e^{-\frac{1}{x^2}}, e^{-\frac{1}{x^2}} \right). & 0 < x < \infty \end{cases}$$ You can verify that $\varphi$ is a smooth map with $\varphi(\mathbb{R}) = X$ and that $\varphi$ is a homeomorphism onto $(X,\tau_X)$ . Use the map $\varphi$ to endow $(X,\tau_X)$ with a smooth structure $\mathcal{A}_X$ that turns $\varphi$ into a diffeomorphism. Then $(X,\tau_X,\mathcal{A}_X)$ satisfies the first three properties and the inclusion is a smooth homeomorphism onto the image but it is not an immersion at $(0,0)$. In fact, Lee shows that the set $X$ cannot be endowed with a topology $\tau_X$ and smooth structure $\mathcal{A}_X$ making $(X,\tau_X,\mathcal{A}_X)$ into an immersed submanifold.
- The notion of a local diffeomorphism makes sense only if the domain and the range are smooth manifolds. If the image of a map happens to be a smooth submanifold of the target manifold, one can say " $f$ is a local diffeomorphism onto its image" by restricting the range. Any other use is just made-up (by various MSE users, it seems) and should be avoided (at least until you are very comfortable with the subject). Instead you can simply say:
...The image of a map $f: X\to Y$ is a smooth submanifold and $f: X\to f(X)$ is a local diffeomorphism.
You may also sometimes encounter the following, describing what an immersion is:
A map $f: X\to Y$ of smooth manifolds is an immersion if and only if locally, it is a diffeomorphism to its image, meaning that $\forall x\in X \exists$ a neighborhood $U$ of $x$ such that $f(U)$ is a smooth submanifold of $Y$ and $f: U\to f(U)$ is a diffeomorphism.
But, again, given ambiguity of the language, it is better to avoid using this terminology in the beginning. The ambiguity comes from the word "image": It can either mean the image of the original map or the image of the map with the restricted domain.
- Everything that you wrote up to the line "However, the first and third posts..." is correct and proofs are very straightforward.
However: I did not check your guesses on how it may or may not be related to various MSE posts.
One thing, you should not repeat ad nauseum "with dimension". (Every manifold has dimension and, except for the empty set, its dimension as a smooth manifold equals its dimension as a topological space. As for the empty set: For every $n\ge 0$, the empty set is a manifold of dimension $n$. At the same time, from the general topology viewpoint, the empty set has dimension $-1$.)
- As for what various MSE users meant in their answers and comments, I prefer not to discuss: Frequently, there is no consistency in their use of mathematical terminology. (Many are only beginners, many have trouble with English, etc.)
Addendum. I am not sure who came up with the idea of allowing manifolds to have variable dimension on different connected components, but I wish this never happened as this just leads to a confusion. I checked several sources in geometry and topology, and the only author allowing manifolds to have variable dimension is Lang.
Related Question
- What’s an example where the inclusion map $\iota: A \to B$ is smooth and a topological embedding but not an immersion
- Are all smooth embeddings proper
- Are manifold subsets, now with subspace topology, that are immersed submanifolds (regular/embedded) submanifolds
- Every smooth $n$-manifold is diffeomorphic to a properly embedded submanifold of $\mathbb{R}^{2n+1}$.
Best Answer
Think of the embedding $\mathbb R^n\to\mathbb S^n$, which image is $\mathbb S^n\setminus \{N\}$. Then, for any compact neighborhood $C$ of $N$ in $\mathbb S^n$, the inverse image of $C$ is not a compact space.