How is the inclusion map both an Immersion and Submersion between Manifolds Hi i am trying to figure out why an inclusion map is both and immersion and submersion. This is what i have tried so far.Let $S$ be an open subset of a manifold $M$. Now the inclusion mapping  is $\iota:S\to M$. Now to prove immersion/submersion we have to show that each of its differentials $\iota_{*,p}$, for $p \in S$, is, resp, injective/surjective. By definition of differential of a smooth map $F$,we have $(F_{*,p}(X_p))f=X_p(f\circ F)$ where $F_{*,p}:T_pM\to T_{F(p)}N$ where $M$ and $N$ are manifolds.Now, taking $F$ as the inclusion map $\iota$, we get  $(\iota_{*,p}(X_p))f=X_p(f\circ \iota)=X_p(f)$ where $\iota_{*,p}:T_pS\to T_{\iota(p)=p}M$.
My questions are as follows:
How do we know from this that each $\iota_{*,p}$ is injective and surjective so that $\iota$ is both an immersion and a submersion?

Say that $M$ is an $n$-dimensional smooth manifold, that $S \subset M$ is an open subset, that $i: S \to M$ is the inclusion and choose $p \in S$. Since $S$ is an open subset of $M$, $S$ is also an $n$-dimensional smooth manifold (this is true because the topology on $M$ is induced from the coordinate charts on $M$, which have target $\mathbb{R}^n$). Choose a coordinate system $\phi: U \to \mathbb R^n$ for $M$, centered at $p$. By replacing $U$ with $U\cap S$ if necessary (and using the fact that $S$ is open) we may assume that $U \subset S$. We see that $\phi$ is therefore also a coordinate system for $S$ centered at $p$. Representing the inclusion map in these coordinates gives
$$\phi \circ i\circ\phi^{-1} = Id,$$
where $Id:\mathbb R^n \to \mathbb R^n$ is the identity map. The differential of $Id$ at $0$ ($0$ since $\phi$ is centred at $p$) is
$$Id_{*,0} :T_0\mathbb R^n \to T_0\mathbb R^n,$$
which is also the identity map.
In particular, this shows that each differential $i_{*,p}$ is both injective and surjective.
