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New question: Can manifold subsets always be made into submanifolds?


My book is An Introduction to Manifolds by Loring W. Tu.

A. Regular/embedded submanifolds are manifolds. My question is about the converse.

In algebra:

  • B. Subset groups are equivalent to subgroups (at least with the same law, but I believe "same identity" is not required because they'll just turn out to have the same identity anyway).

  • C. Rings not so much: For (commutative unital) rings, if $B$ is a ring and if $A \subseteq B$ and $A$ is a subring of $B$, then $A$ is a ring (with the same laws and identity as $B$ because this is how subring is defined anyway). However conversely, if they are both rings (NOT necessarily with the same laws or identity), then $A$ is not necessarily a subring of $B$.

    • D. For example, $B$ has an idempotent element $e$ besides identity, and $A$ is the principal ideal generated by $e$, where we have $e$ as the identity of $A$ but not of $B$ (Algebra by Michael Artin Proposition 11.6.2). I think the laws of $A=(e)$ are the same as the one of $B$, and the only thing lacking for $A$ to be a subring of $B$ is that $A$ has a different identity from $B$ (I understand that $A=(e)$ has a different identity from $B$ if and only if $A$ doesn't contain the identity of $B$).
  • E. Based on what I think is the issue in (D) and based on my guess that manifolds have no such analogue for "identity", I expect manifold subsets to be regular/embedded submanifolds.

    • Update: Based on Eric Wofsey's answer, I guess since there are indeed ways, that subset rings are not subrings, besides not sharing identity. I guess the ways are to do with the laws $+$ and $\times$ differing between $A$ and $B$, kind of like in the above parenthetical remark for groups.

Question: Let $A$ and $B$ be manifolds with respective dimensions $a$ and $b$. If $A \subseteq B$ (given the subspace topology because apparently people don't just assume this), then is $A$ a regular/an embedded $a$-submanifold of $B$?

I'll just attempt to prove embedded (I won't prove regular directly). Please verify.

$A$ is the image of the inclusion map $\iota: A \to B$. I will show $\iota$ is an embedding, with this definition (Using this equivalent definition would be circular since such definition says "smooth submanifold" and not "smooth manifold"):

  1. Smooth: An inclusion between two smooth manifolds is smooth.

    • Edit: I guess this is the problem. I can't quite use Theorem 11.14, but i think one can somehow modify the proof of Theorem 11.14 to prove "If N is a (smooth) manifold subset of M, then the inclusion $i: N \to M, i(p) = p$, is an embedding"
  2. Immersion: Inclusions are the prototype of immersions.

    • Edit: Oh, at least for Euclidean spaces.
  3. Topological embedding: The restriction $\tilde{\iota}: A \to \iota(A)=A$ is identity on $A$, a homeomorphism of $A$ (because of subspace topology).

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    $\begingroup$ Do you assume any relationship between the smooth manifold structures on $A$ and $B$ at all? If not, then merely knowing that $A$ is a subset of $B$ tells you extremely little. Any set of cardinality $2^{\aleph_0}$ can be given a smooth manifold structure of any positive dimension... $\endgroup$ – Eric Wofsey Jul 20 at 5:26
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    $\begingroup$ In any case, you have not given any justification for most of the claims in your "proof". $\endgroup$ – Eric Wofsey Jul 20 at 5:29
  • $\begingroup$ @EricWofsey Thanks. I added a link for (2). For (3), I think this is known from elementary topology. For (1), I guess this is the main issue. Is it related to one of the issues here? $\endgroup$ – Selene Auckland Jul 20 at 5:43
  • $\begingroup$ @EricWofsey Can you give a counter example please? $\endgroup$ – Selene Auckland Jul 20 at 5:44
  • $\begingroup$ @EricWofsey Thanks for the edit! I cannot believe I put algebraic-geometry instead of differential-geometry. $\endgroup$ – Selene Auckland Jul 20 at 5:50
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No, this is very very false. For instance, let $B$ be $\mathbb{R}$ with its usual smooth manifold structure, and let $A$ be $\mathbb{R}$ with a smooth manifold structure given by picking a bijection $\mathbb{R}\to\mathbb{R}^2$ and pulling back the usual smooth manifold structure on $\mathbb{R}^2$. Then $A$ is certainly not an embedded submanifold of $B$, since it has larger dimension. Indeed, the inclusion map $A\to B$ cannot even be continuous.

Even if you assume $A$ has the subspace topology, it is still very false. For instance, in the example above, you can instead pick a homeomorphism $\mathbb{R}\to\mathbb{R}$ that is not a diffeomorphism and pull back the usual smooth manifold structure of $\mathbb{R}$ to a new one and call it $A$. Then the inclusion map $A\to B$ will be a homeomorphism but not a diffeomorphism.

The key thing to understand here is that being a manifold is not a property of a set. It's an additional structure you can put on a set. All that that $A\subseteq B$ tells you is that every element of $A$ happens to be an element of $B$; it tells you nothing at all about their manifold structures, which could be totally unrelated. (The same thing happens with rings: if $A$ and $B$ are rings with $A\subseteq B$, then there is no reason at all to think that $A$ is a subring of $B$, because the ring operations of $A$ are probably totally different from those of $B$.) Being a smooth manifold is similarly not a property of a topological space, but an extra structure you can put on it.

As for your proposed proof, all three of your claims are wrong as shown by the example above. You gave no justification for claim 1 or claim 2 ("inclusions are the prototype of immersions" is just a vague slogan that has no meaning in a proof). For claim 3, to prove $\iota$ is an embedding you need to prove it is a homeomorphism from $A$ to $\iota(A)$ with the subspace topology from $B$, and you have no reason to believe that topology is the same as the given topology on $A$.

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    $\begingroup$ Yes, claim 3 becomes true if you assume $A$ has the subspace topology. This is generally implied if you are just starting with a topological space and taking a subset of it, but that's not what's going on in your question: you started with two separate manifolds $A$ and $B$, and then added the assumption that $A$ is a subset of $B$. Given that you are explicitly not assuming $A$ is a submanifold of $B$, I would think most readers would think you aren't assuming it is a subspace either. $\endgroup$ – Eric Wofsey Jul 20 at 5:55
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    $\begingroup$ It is not true that the differential of an inclusion map of smooth manifolds is an inclusion (for one thing, it certainly isn't literally an inclusion at least for most ways of defining tangent spaces; at best you might hope for it to be an injection). $\endgroup$ – Eric Wofsey Jul 20 at 6:01
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    $\begingroup$ The question you linked seemed to be tacitly assuming that $X$ is a smooth submanifold of $Y$ (and the answer also makes this assumption). $\endgroup$ – Eric Wofsey Jul 20 at 6:03
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    $\begingroup$ No, consider $f:\mathbb{R}\to\mathbb{R}$ given by $f(x)=x^3$. It is smooth and injective but not an immersion. (It is not literally an inclusion, but any injection is "isomorphic" to an inclusion since you can transport the structures on the domain to a structure on its image.) $\endgroup$ – Eric Wofsey Jul 20 at 6:08
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    $\begingroup$ Let $A$ denote the image of $f$ (i.e., the set $\mathbb{R}$), equipped with the unique smooth structure that maps $f:\mathbb{R}\to A$ a diffeomorphism (push the smooth structure of $\mathbb{R}$ forward along $A$). Now the inclusion map $A\to\mathbb{R}$ is smooth and not an immersion (and it is basically the same as $f$; we have just renamed the elements of its domain along $f$ to make it an inclusion). $\endgroup$ – Eric Wofsey Jul 20 at 6:14

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