Let $$M$$ be a smooth manifold of dimension $$n$$.

My notes say

Theorem: A subset $$S$$ of $$M$$ could be given a structure of smooth manifold of dimension $$k$$ such that $$S$$ is an embedded submanifold of $$M$$ (i.e. the inclusione map $$\iota:S\hookrightarrow M$$ is a smooth embedding) if and only if for each point $$p$$ in $$M$$ there exists a smooth chart $$(U,\phi)$$ for $$M$$ centered at $$p$$ and $$k$$-adapted to $$S$$ (i.e. $$U\cap S=\emptyset$$ or $$\phi(U\cap S)=\{x\in \phi(U):x^{k+1}=\dots=x^n=0\}$$)(Or equivalently: there exists a smooth atlas for $$M$$ which is $$k$$-adapted to $$S$$, meaning that for each smooth chart $$(U,\phi)$$ in that altas, I have $$(U,\phi)$$ is $$k$$ adapted to $$S$$).

Now, in the proof we only show that if I take a point $$p$$ in $$S$$ (not, in general, in $$M$$ as stated above!) then exists a smooth chart $$(U_p,\phi_p)$$ for $$M$$ centered at $$p$$ and $$k$$-adapted to $$S$$. But now if I consider $$\{(U_p,\phi_p)\}_{p\in S}$$ I could not have an atlas for $$M$$. (I only know that $$S\subseteq\bigcup_{p\in S}U_p$$ but not that $$M=\bigcup_{p\in S}U_p$$).

So, the statement of the above theorem is not very correct or am I missing something? How could I complete (if I can!) the set $$\{ (U_p,\phi_p)\}_{p\in S}$$ to obtain a smooth atlas for $$M$$ such that each chart is $$k$$-adapted to $$S$$ ?

Or should I modify the statement in

Theorem: A subset $$S$$ of $$M$$ could be given a structure of smooth manifold of dimension $$k$$ such that $$S$$ is an embedded submanifold of $$M$$ if and only if for each point $$p$$ in $$S$$ there exists a smooth chart $$(U,\phi)$$ for $$M$$ centered at $$p$$ and $$k$$-adapted to $$S$$ ?

• I am having a really hard time following your question. If $p \not \in S$, then choose a small enough chart which does not intersect $S$ at all. That is a chart "adapted to $S$". That gives you the rest of your charts. – user98602 Dec 28 '18 at 17:01
• Is it always possible? If for example $S$ is dense in $M$, then it is not possible to choose a "small enough chart which does not intersect $S$ at all". – Minato Dec 28 '18 at 17:28
• That is not an embedded submanifold, and in that case you won't get any adapted charts whatsoever. An embedded submanifold has the subspace topology with respect to the ambient space. – user98602 Dec 28 '18 at 17:30
• Shurely something stupid is confusing me, I'm very sorry for this, but in a topological space, a subset could well be dense in the ambient space and as a space in its own right have the subspace topology. (E.g. Q is dense in R and has the subspace topology). Now, according to my definition of embedded submanifold, why it is not possibile for $S$ to be dense in $M$? – Minato Dec 28 '18 at 17:43
• Can you explicitly construct a chart in $p$ which does not intersect $S$ at all? – Minato Dec 28 '18 at 17:45

The correct statement is that $$S \subset M$$ is a closed, embedded submanifold if and only if $$M$$ has an atlas of adapted charts. If I were being careful / if $$S$$ wasn't embedded a better way to phrase this would be to start by writing $$f: S \to M$$, which I state is a proper injective immersion.

The essential input into constructing such an atlas is the following.

1) Given any $$x \in S$$, choose a small chart on $$S$$ with domain $$V$$; then there is an open subset $$U \subset M$$ containing $$x$$ so that $$U \cap S \subset V$$. (That is, "far away points of $$S$$" from the perspective of the manifold topology on $$S$$ do not accumulate towards some fixed point $$f(x)$$ in the image.)

2) Given any $$x \in M \setminus S$$, there should be a neighborhood $$U \subset M$$ containing $$X$$ so that $$U \cap X = \varnothing$$.

The first is necessary so that $$f: S \to f(S) \subset M$$ is a homeomorphism, where we consider $$f(S)$$ with the subspace topology (which is guaranteed by the existence of adapted charts near points of $$S$$). The second is necessary to get adapted charts away from $$S$$.

(2) is equivalent to saying that $$S$$ is a closed subset of $$M$$.

(1) is harder: it says that if there is a sequence $$x_n \in S$$ and $$x \in S$$ so that $$f(x_n) \to f(x)$$, then in fact we must have $$x_n$$ near $$x$$; that is, there is a subsequence of $$x_n$$ which converges to $$x$$. This is saying that the embedding map is "proper": the inverse image of compact sets, like $$\{f(x_1), f(x_2), \cdots, f(x)\}$$, is compact.

But if $$f$$ is a proper embedding (= proper injective immersion = injective immersion with closed image), then one may indeed find adapted charts.

The definitions already almost tell us the proof. If $$p \in S$$, we know that we may find some chart $$V \subset S$$ of $$p$$ and some chart $$U \subset M$$ of $$f(p)$$ so that on these charts, the map $$f: V \cap U \to U$$ is given by the inclusion of $$\Bbb R^k$$ into $$\Bbb R^n$$, by the implicit function theorem. What we don't know is that $$U \cap S = V \cap S$$, so that $$S$$ "never otherwise appears in $$U$$". This is where your proof needs the properness assumption. This gives an adapted chart to any point $$p \in S$$.

If $$p \not \in S$$, then by assumption we may choose a chart with domain $$U$$ having trivial intersection with $$S$$. If you already had a chart with domain $$U'$$, then the new chart is the same map on the subdomain $$U = U' \cap (M \setminus S)$$, which is again open.