- If $B$ is a manifold, and $A\subseteq B$ is a regular submanifold, then the inclusion map $i:A\to B$ is an embedding and thus smooth.
That $i$ is an embedding seems a bit strong. Is there another way to do this?
I seem to recall so far in learning differential geometry that inclusions are smooth specifically when doing compositions, but I can't find it upon looking through my notes. In general, for any manifold subset $A$ (a subset that is also a manifold but not necessarily a regular submanifold or immersed submanifold or neat submanifold, etc (I think not all irregular submanifolds are manifolds anyway)) of a manifold $B$, I know $i:A \to B$ is continuous (assuming subspace topology), but when is $i:A \to B$ smooth?
Upon second look of my notes, I found that it was true for the 'inclusion' $i_b: A \to A \times B, i_b(a)=a \times b$ for fixed $b$. That's different from what I usually think of inclusion and instead more like this 'inclusion'.
Upon third look, I think many of those subset manifolds $A$ were open in $B$. Since open implies regular submanifold, all those compositions were safe, but we didn't have submanifolds then. I ask about this in another question.
Thanks in advance!
The inclusion map is automatically an embedding if it is smooth unless...
1stly, talking about an inclusion map as smooth means both the objects in question are manifolds.
The inclusion map is an embedding if and only if the domain is a regular/an embedded submanifold of the range. It's possible this isn't the case if the domain is given a manifold (or even topological!) structure s.t. the domain is not a regular/an embedded submanifold of the range. See here: Are manifold subsets submanifolds?