Are manifold subsets, now with subspace topology, that are immersed submanifolds (regular/embedded) submanifolds

differential-geometrydifferential-topologygeneral-topologymanifoldssubmanifold

Same as this but now with subspace topology.

From here, we know manifold subsets are not only not necessarily regular/embedded submanifolds but also not necessarily immersed submanifolds.

Now I ask:

Let $A$ and $B$ be sets with $A \subseteq B$. Let $B$ become a smooth $b$-manifold and $A$ become a smooth $a$-manifold, but $A$ is not necessarily a smooth regular/embedded $k$-submanifold of $B$. (I believe $k$ not equal $a$, if ever.)

Additionally now: assume $A$ has the subspace topology.

If it somehow makes sense to say $A$ is a smooth immersed submanifold of $B$, then is $A$ a smooth regular/embedded $k$-submanifold of $B$?

Okay so about the manifold structures:

  1. '$A$ a regular/an embedded $k$-submanifold $B$' –> As I recall: given a smooth manifold structure on $B$ and a subset $A \subseteq B$, there is exactly 1 smooth manifold structure on $A$ s.t. this holds. So I guess no issue here.

  2. '$A$ is a smooth immersed submanifold of smooth $b$-manifold $B$ and $A$ is a smooth $a$-manifold' –> This may be kind of weird, like maybe it doesn't make sense to talk about $A$ as a smooth immersed submanifold of $B$ if it doesn't automatically upgrade from smooth immersed to smooth regular/embedded as soon as $A$ actually is a smooth manifold, in w/c case prove this please. But I think it should make sense because I think an immersed submanifold of a manifold could be a manifold under a different manifold/topological structure.

Note: In Tu's An Introduction to Manifolds, it says

'If the underlying set of an immersed submanifold is given the subspace topology, then the resulting space need not be a manifold at all!'

In this case, I'm asking: If the resulting space is actually a manifold with the subspace topology, does immersed upgrade/promote to embedded/regular?

Edit: Further explanation: My new question is about whether or not an immersed submanifold upgrades/promotes to an embedded/a regular submanifold assuming the underlying topological space can become manifold on its own like not necessarily concerning the ambient space/parent manifold. this question differs from the previous one where i take into account only the underlying set rather than the underlying topological space. if underlying set only, then we can follow Lee Prop 5.2 and Prop 5.18 or we can simply do this.

Remark: Based on answer: As it turns out, the immersion $f:A \to B$, if is a local diffeomorphism onto image, i.e. $f(A)$ is a smooth embedded/regular $a$-submanifold of $B$ and then $f: A \to f(A)$ is a local diffeomorphism…but wait i think $C$ in the answer is equal to the $A$ in my question, sooo…a different $A$ i guess…

Best Answer

To the best of my understanding of your question, you are asking the following:

Question 1. Suppose that $A, B$ are smooth manifolds and $f: A\to B$ is an immersion such that $f(A)$ happens to be a smooth submanifold of $B$ (when equipped with the subspace topology, which is a default in this setting). Does it follow that $f$ is an embedding?

The answer to this question is negative. The simplest example is $B=S^1\subset {\mathbb C}$, $A={\mathbb R}$ and $f(t)=e^{it}$.

It is possible, however, that what you have in mind is different:

Question 2. Suppose that $A, B$ are smooth manifolds and $f: A\to B$ is an injective immersion such that $C=f(A)$ happens to be a smooth submanifold of $B$ (when equipped with the subspace topology). Does it follow that $f$ is an embedding?

This question has positive answer, in fact, $f: A\to C$ is a diffeomorphism in this situation (as follows from the inverse mapping theorem).

Lastly, my suggestion is to avoid the terminology "immersed submanifold" when you are just learning Differential Topology. Instead, talk about "immersions"and "embeddings" of smooth manifolds, as well as "submanifolds." Tu is really doing his readers disservice by introducing the terminology "an immersed submanifold" at the early stage. But this is just my opinion.

Edit. It seems that the correct reading of the question is:

Question 3. Suppose that $A, B$ are smooth manifolds, that $f: A\to B$ is an immersion and that $f$'s image $C=f(A)$ (with the subspace topology) is a topological manifold. Is it true that $C$ is a smooth submanifold of $B$?

This one has positive answer too and the proof is similar to one in the case of Question 2.

Step 1. The topological manifold $C$ has the same dimension $a$ as $A$.

Proof. Suppose not. Let $U_j, j\in J$ be the open subsets of $A$ such that $f|U_j$ is an embedding $U_j\to B$ for each $j\in J$, where $J$ is a countable index set. (My definition of manifolds requires that they are 2nd countable.)

In particular, $f|U_j$ is 1-1. Thus, by the invariance of domain theorem, for each $j$, $f(U_j)$ is nowhere dense in $C$. Thus, $C$ is a union of countably many nowhere dense subsets, contradicting Baire's Theorem.

Step 2. $C$ is a smooth $a$-dimensional submanifold of $B$.

Proof. For each $U_j$ as above, again, by the invariance of domain theorem, $f(U_j)$ is open in $C$. But each $f(U_j)$ is a smooth submanifold of $B$. Hence, for each $x\in f(U_j)$ there is a neighborhood $W_x$ of $x$ in $B$ and a diffeomorphism $h: W_x\to R^b$ ($b$ is the dimension of $B$) sending $W_x\cap U_j$ to an open subset of $R^a\subset R^b$. Thus, $C$ is a smooth submanifold of $B$. qed

Remark: As it turns out, $f$ is a local diffeomorphism onto image, i.e. $f: A \to C$ is a local diffeomorphism.

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