Why is an inclusion map from an open subset smooth?

Question

If $B$ is a manifold and $A\subseteq B$ is a regular submanifold of $B$, then the inclusion map $i:A\to B$ is an embedding and thus smooth.

If $B$ is a manifold, and $A\subseteq B$ is an open subset of $B$, then how do we justify that the inclusion map $i:A\to B$ smooth without using that $i$ is an embedding (in the sense of differential geometry; 'embedding' in the sense of elementary topology is within scope) or $A$ is a regular submanifold? Also, do not use germs or tangent space please.

Thanks in advance!

Here's my answer:

Let $i: A \to B$ be inclusion for $A$ an open subset of $B$. To show $i$ is smooth, we must show $i$ is continuous (known from elementary topology) and that for all $a \in A$, there are charts $(C,\gamma)$ about $i(a)=a \in B$ and $(D,\delta)$ about $a \in A$ for which $\gamma \circ i \circ \delta^{-1}: \gamma(i^{-1}(C) \cap D) = \gamma(C \cap D)$ $\to \gamma(C)$ is smooth

For any $a \in A$, choose $C=D=A$ and $\gamma=\delta=\text{id}_A$ identity function on $A$. Then $\gamma \circ i \circ \delta^{-1}: \gamma(i^{-1}(C) \cap D) = \gamma(C \cap D)$ $\to \gamma(C)$ becomes $i: A \to A$. We know that the inclusion map, with its range restricted to its image which happens to be its domain, becomes the identity map on its domain. Since the identity map on any manifold is smooth (I think), $i: A \to A$ is smooth. This satisfies the definition for $i: A \to B$ to be smooth.

This is trivial. By definition, the inclusion map is always an embedding with the subspace topology, and with the induced smooth structure it is obviously smooth. — Matt Samuel, Jan 19 '19 at 17:13
@MattSamuel I said without using embedding, but perhaps you are referring to a embedding of topological manifolds instead of smooth manifolds? — Ekhin Taylor R. Wilson, Jan 19 '19 at 17:15
What I mean is that the inclusion map from $A$ is trivially an embedding. — Matt Samuel, Jan 19 '19 at 17:17
@MattSamuel Do you mean embedding in the sense of elementary topology (a map that becomes a homeomorphism when its range is restricted to its image) instead of in the sense of differential geometry (injective immersion whose image in the subspace topology is homeomorphic to its domain) ? — Ekhin Taylor R. Wilson, Jan 19 '19 at 17:27
How do you define the smooth structure on an open subset? If you look at the definition you'll see it's automatically a smooth embedding for an open subset. — Matt Samuel, Jan 19 '19 at 17:28
@MattSamuel Okay, I had a second look at the definition, and I think I just proved it. I don't think this is obvious for a beginner however. — Ekhin Taylor R. Wilson, Jan 19 '19 at 17:42
@MattSamuel Thanks. I tried answering it, but I realised I actually don't know why identity on a manifold is smooth, if it is. Is it? — Ekhin Taylor R. Wilson, Jan 19 '19 at 18:02
@MattSamuel Does some diffeomorphism $f$ exist on any smooth manifold $B$ by definition of the differentiable structure on $B$? — Ekhin Taylor R. Wilson, Jan 20 '19 at 09:05
I think my proof is more of an intuitive one. You really should prove the it is a diffeomorphism directly. But if it isn't, the manifold isn't diffeomorphic to any manifold, even itself, and diffeomorphism wouldn't be an equivalence relation. — Matt Samuel, Jan 20 '19 at 11:32
@MattSamuel is my answer here relevant maybe? When is an inclusion map smooth? — BCLC, Apr 26 '21 at 03:46

user20672 · Answer 1 · 2021-03-01T04:08:49.633

1

If you have a smooth structure $\{u_a, \phi_a\}$, it induces a natural smooth structure in the subspace topology. Overload the notation for simplicity

$$ \phi_a \circ i \circ \phi^{-1}_b = \phi_a \circ \phi^{-1}_b $$

which must be smooth by the definition of a smooth structure, since we require that all transition functions be smooth.

The fact that it's open just guarantees that it remains locally homeomorphic to balls in $\mathbb{R}^{n}$

edited Mar 01 '21 at 04:08

answered Mar 01 '21 at 03:51

user20672

626

is my answer here relevant maybe? When is an inclusion map smooth? – BCLC Apr 26 '21 at 03:46

Why is an inclusion map from an open subset smooth?

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