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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.

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    $\begingroup$ There is a question whether a Klein bottle can be immersed into 3-space (it cannot be embedded). That question has the answer "yes" if immersions are defined as pairs consisting of the map of manifold $A$ into manifold $B$, together with the immersing map $f:A \to B.$ Here there is a danger of confusing the "inclusion map" of $A$ (Klein bottle) into $B$ (3-space) with the embedding map $f$. [The latter involves "looping" the Klein bottle around itself and having it "pass through itself" to re-connect, as available pictures show.] So the language used in the post re. immersions seems vague. $\endgroup$ – coffeemath Nov 7 '16 at 13:20
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    $\begingroup$ Above I should have written more clearly "pairs $(A,f)$ where $f$ is the proposed immersing map from $A$ into $B.$ $\endgroup$ – coffeemath Nov 7 '16 at 13:32
  • $\begingroup$ @coffeemath I agree, although I am even more confused about what is meant by the inclusion. Lee's definition is the following: "smooth immersion" a smooth map $F: M \to N$ such that its differential is injective at every point (rank F = dim M), and a smooth embedding is an injective smooth immersion that is a homeomorphism onto its image in the subspace topology. Let $M$ be a smooth manifold with or without boundary. An embedded submanifold is a subset $S \subseteq M$ that is a manifold (without boundary) in the subspace topology, endowed with a smooth structure with respect to which the $\endgroup$ – Chill2Macht Nov 7 '16 at 13:47
  • $\begingroup$ inclusion map $S \hookrightarrow M$ is a smooth embedding. Let $M$ be a smooth manifold with or without boundary. An immersed submanifold of $M$ is a subset $S \subseteq M$ endowed with a topology (not necessarily the subspace topology) with respect to which it is a topological manifold (without boundary), and a smooth structure with respect to which the inclusion map $S \hookrightarrow M$ is a smooth immersion. $\endgroup$ – Chill2Macht Nov 7 '16 at 14:21
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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:

  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.
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  • $\begingroup$ What did I write that was true? Also these examples are really helpful -- you're right that I need to be more careful with the definitions, especially regarding the smooth structures. How do you calculate the derivative/differential of the inclusion map in general, i.e. accounting for the information provided by the atlases? I see that assuming it is always the identity doesn't always work out very well, but I am unsure of how to calculate it, since I'm not sure how to calculate it. I'm also even wondering whether the inclusion is always the restriction of the identity or always injective. $\endgroup$ – Chill2Macht Nov 7 '16 at 21:05
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    $\begingroup$ @William: An immersed submanifold $(X,\tau_X,\mathcal{A}_X)$ will be an embedded submanifold if and only if $\tau_X$ is the subspace topology on $X$ and this follows immediately from the definitions. To calculate the differential of a map, compose it with local coordinates and calculate there. If $i \colon X \rightarrow M$ is the inclusion, this means you need to look at $\phi \circ i \circ \varphi^{-1}$ as a map between open sets in $\mathbb{R}^n / \mathbb{R}^m$. This won't necessarily be the inclusion map (or a restriction of such). $\endgroup$ – levap Nov 7 '16 at 21:17
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    $\begingroup$ @William: If $X$ is an embedded submanifold, then you can actually pick charts in which this composition will look like an inclusion (such coordinates charts are called slice charts). $\endgroup$ – levap Nov 7 '16 at 21:19
  • $\begingroup$ Ahh OK yes this makes a lot more sense now -- I agree it is definitely much clearer to understand how either $\phi \circ i \circ \varphi^{-1}$ or its derivative need not necessarily be the identity. Thank you also for mentioning slice charts, since that also explains their motivation better for me also (slice charts were the second-next thing I was going to look at after this). Since $i$ is a function with manifolds as its domain and codomain, its derivative isn't even defined directly in most cases. $\endgroup$ – Chill2Macht Nov 7 '16 at 21:21
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    $\begingroup$ @William: In fact, any immersion looks locally like an inclusion (this follows from the constant rank theorem) so this holds also for immersed submanifolds but a slice chart around a point $p$ of an embedded submanifold gives you something stronger. $\endgroup$ – levap Nov 7 '16 at 21:31

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