[Math] Condition for immersed but not embedded submanifold

Question

Claim: an immersed submanifold is not an embedded submanifold if and only if its manifold topology does not agree with the subspace topology.

Why I suspect the claim is true: the inclusion map is always injective by definition. Also, since it is a restriction of the identity, its derivative everywhere is just the identity transformation, and thus clearly injective. Thus the inclusion is always a smooth immersion.

Therefore the only way the inclusion map could fail to be a smooth embedding is if its image was not homeomorphic to the manifold itself. Since its image always has the subspace topology, the only way it could fail to be a homeomorphism is if the manifold topology were different from the subspace topology.

For the other direction we proceed by contraposition. If its subspace topology agrees with its manifold topology, then the inclusion map has to be a homeomorphism.

Is this correct? This claim is never made (I think) in Lee's Introduction to Smooth Manifolds, so I am doubtful it is correct, because it seems a lot simpler than his discussion of the topic.

Also I was struggling to understand why the inclusion map wouldn't always be an embedding, and the only reason I could think of was that it might not be a homeomorphism onto its image if the topology of the manifold as an "independent space" was different from its topology as a subspace. But I don't know if that's actually correct.

Namely I don't even know if the claim that the inclusion map is always a smooth immersion is correct (it definitely has to be injective).

Definitions: An immersed submanifold is a subset of another manifold that is a topological manifold, and the inclusion map is an injective smooth immersion, where a smooth immersion is a smooth map whose derivative is injective at every point.

An embedded submanifold is a subset of another manifold which is a topological manifold and for which the inclusion map is a smooth embedding, which is an injective smooth immersion that is also a homeomorphism onto its image.

Answer 1

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:

1. The set $X$ is a subset of $M$.
2. The set $\tau_X$ is a topology on $X$ making $(X,\tau_X)$ a topological manifold.
3. 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:

1. 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.
2. 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.