I would like to ask if my thought process in answering the question in the Title is correct.
I know there are many posts on this general topic: e.g. Is the inclusion map always smooth?, When is an inclusion map smooth?, but my takeaway from reading those is that sometimes the arguments of (often knowledgeable) posters arise from different definitions of submanifold. That is, some allow immersed submanifolds, some embedded, some see a submanifold as the image of an injective immersion that may or may not be an embedding. I'd like to try to answer the question in the title by assuming the submanifold is defined by the use of submanifold charts. This is not homework but it is a question posed in a series of lecture notes I am studying and I want to test my understanding of the subject matter.
So, we have an $m$-dimensional manifold $M$ with a smooth atlas A $ \equiv (U_\alpha,\phi_\alpha)_\alpha$.
The $s$-dimensional submanifold $S \subset M$ has a smooth atlas A' $\equiv (U^\prime_\alpha,\phi^\prime_\alpha)_\alpha$;
here $U^\prime_\alpha = U_\alpha \bigcap S$, $\phi^\prime_\alpha = \pi_s \circ \phi_\alpha$ and $\pi_s$ is the projection onto $R^s$. I have skipped the submanifold charts because, from what I understand, they do not technically comprise an atlas for $S$ as a manifold "in its own right", these do, although of course the difference is trivial. However, I emphasize this point because proving the inclusion map $i: S \rightarrow M$ is smooth involves viewing $S$ and $M$ as separate manifolds, each with its own differentiable structure, so that the question does not become inessential.
I would therefore like to know if either of the following two are actual proofs: The key to both of them is the use of the smooth immersion $j: R^s \rightarrow R^m := j(x_1,\ldots,x_s) = (x_1,\ldots,x_s,0,\ldots,0)$.
First, since smoothness is a local requirement we can just concentrate on chart neighborhoods of each $p \in S$, and on each such neighborhood the inclusion $i$ is just equal to the composition $\phi_\alpha^{-1} \circ j \circ \phi^\prime_\alpha$. Note each chart map $\phi_\alpha,\phi_\alpha^\prime$ is smooth by the definition of a smooth structure.
Or second, to show that $i$ is smooth we need to show that, given any $p \in S$ and for every $U^\prime_\alpha$ containing $p$ and $U_\beta$ containing $i(p) = p$, the map: $\phi_\beta \circ i \circ (\phi^\prime_\alpha)^{-1}$ is smooth as a map from $R^s \rightarrow R^m$. But this map is just $j \circ \phi_\beta^\prime \circ (\phi^\prime_\alpha)^{-1}$ and $\phi_\beta^\prime \circ (\phi^\prime_\alpha)^{-1}$ is smooth by definition of the smooth structure on $S$. So $i$ is smooth.
