Each chart of the canonical structure of a submanifold S is locally a submanifold chart I am bugged about some technical details in the following proof regarding submanifolds. I am reading a book that is not in english , so I'll try to translate here a couple of terms in the best possible way (privileged chart and submanifold chart)  but I am not sure thouse are the english names, in case  you know them, please tell me.
Definition of submanifold
Let $M$ be an n-manifold and $S$ a subspace of $M$. $S$ is said a submanifold of dimension $d$ of $M$ is for every point $p$ of $M$, there is a chart $(U,x)$ of $M$ such that $x(U\cap S)=x(U)\cap (\mathbb{R}^d\times \{0\}^{n-d})$. In other words, if the points $q$ that fall in $U$ are characterized of the $n-d$ equations $x^{d+1},...,x^{n}(q)=0$. Such a chart is called a privileged chart(with respect to S) and determins a chart $(U \cap S,x_{U \cap S})$, called submanifold chart (if we identify $\mathbb{R}^d$ with$\mathbb{R}^d \times \{0\}^{n-d}$)
If $(U,x), (V,y)$ are privileged charts, we have that $x(U\cap V \cap S)$ is an open set in $x(U\cap S)$ and so in $\mathbb{R}^d$ and $y_{V\cap S} \circ (x_{U\cap S})^{-1}=(y\circ x^{-1})_{x(U \cap V \cap S)}: x(U \cap V \cap S) \to y(U \cap V \cap S)$ is differentiable, so that the submanifold charts form an atlas and determine a d-dimensional differential structure over $S$. With this structure, $S$ is a d-manifold. A d-submanifold is always thought of with this canonical or natural submanifold structure
Proposition: Each chart of the canonical structure of a submanifold S is locally a submanifold chart.
Proof. Let $p$ be a point in the domain of the chart $(V,y)$ of $S$. By definition,...(1)  there exists a privileged chart $(U,x)$ of $M$ at $p$ that we can supposed such that $U \cap S \subset V$ and $ x(U)=A \times B$ with $A$ an open set at $0$ in $\mathbb{R}^d$ and B an open set at $0$ in  $\mathbb{R}^{n-d}$...(2)
By means of the chart $x$, the canonical projection $\sigma : A\times B \to A\times \{0\}$ induces a sort of projection $\pi = x^{-1} \circ \sigma \circ x :U \to U \cap S$ which is differentiable ...(3)
Thus we have that $(U,z)$ with $z=(y^1\circ \pi, ...,y^d\circ \pi,x^{d+1},...,x^n)$ with values in $y(U\cap S)\times B$ is a privileged chart and $z|_{U\cap S}=y|_{U\cap S}$...(4)
My questions are:
1 They start using a definition of what?. If it is the definition of a d-submanifold,as I would expect, why aren't they using the definition above, that is $x(U\cap S)=x(U)\cap (\mathbb{R}^d\times \{0\}^{n-d})$)?
2 Why can we suppose $U \cap S \subset V$ and why do we need that? and why do we take opens sets at $0$ necessarily ?)
3 Why is $\pi = x^{-1} \circ \sigma \circ x$ differentiable? My guess is that $\sigma $is differentiable by definition of product topology (if that is applicable here), but still i would need to know that $x $is differentiable for  the composition to be differentiable as well, but all I know is that $x$ is a homeomorphism, being $(U,x)$ a chart
4 Why is  $z|_{U\cap S}=y|_{U\cap S}$ and why do we need to mention it ?
I'm sorry for all these questions, but I have been around this for days and they are all related to this short proposition,so I don't think it made sense to make independent questions.... can someone please shed some light?
 A: *

*Here they are indeed using the definition of a $d$-manifold, but now requiring as a new assumption the open subset $x(U)$ to be an elementary one of the product space $\mathbb{R}^d\times\mathbb{R}^{n-d}$. We can always assume this point since elementary subsets form a topological basis of the topology of the product space.


*The result we want to show is the existence of a privileged chart $(U,x)$ at $p$ with $U\cap S\subset V$ such that $x|_{U\cap S}=y|_{U\cap S}$, so this inclusion is required for the last equality to make sense. Since $V$ is an open set in the subspace topology, then it is by definition of the form $W\cap S$ with $W$ an open set of $M$: take $\widetilde{U}=U\cap W$ if needed. The fact that we choose the chart for the open set $B$ to be at $0$ is important to check the definition you gave of a submanifold, but there are equivalent definitions of being a submanifold where you replace $\mathbb{R}^{d}\times\{0\}^{n-d}$ by $\mathbb{R}^d\times\{q\}$ with $q\in\mathbb{R}^{n-d}$ an arbitrary point (the "difference" is that in this second definition we ask $S$ to locally look like an affine subspace of $\mathbb{R}^n$, meanwhile in the first one we ask it to locally look like a linear subspace of $\mathbb{R}^n$: of course they are equivalent by the mean of translations $x\mapsto x- q$).


*When you choose an atlas on a manifold, you are actually making the charts diffeomorphisms between open sets of your manifolds and open sets of $\mathbb{R}^n$ with the usual differentiable structure (just remark that if $x:U\to x(U)\subset\mathbb{R}^n$ is a chart, then $\mathrm{id}_{\mathbb{R^n}}\circ x\circ x^{-1}=\mathrm{id}_{\mathbb{R}^n}$ is a differentiable map between open sets of $\mathbb{R}^n$). Since $\sigma$ is a linear map between open subsets of $\mathbb{R}^n$, then differentiable, and that both $x$ and $x^{-1}$ are diffeomorphisms then differentiable by the above argument, the composition is differentiable.


*This is the result we wanted to show, namely the chart of the canonical structure locally being the restriction of a privileged chart. You can first check that it is a chart by finding an explicit inverse, then show that it is a privileged one using the definition, then show that it coincides with $y$ when plotted at points in $U\cap S$ using that $\pi|_{U\cap S}=\mathrm{id}_{U\cap S}$.
