Which part(s) of the definition of smooth manifold below excludes the possibility of a manifold containing it's boundary? This question is about why a manifold can't contain a boundary. I am reading about differential forms and manifolds in Adams and Essex Calculus 2.
As far as I read, and as stated in the book

A manifold $M$ in $\mathbb{R}^n$ does not itself contain any boundary points ....

However, the definition of a smooth manifold in the book (which is probably not the most general as this is an introductory text) reads (slightly abbreviated)

A subset $M$ of $\mathbb{R}^n$ is a k-manifold of dimension $k\leq n$ if for every $\mathbf{x} \in M$ there exists an open set $U \subset \mathbb{R}^n$, containing $\mathbf{x}$ and a smooth function $\mathbf{f}:U\rightarrow \mathbb{R}^{n-k}$: such that the two following conditions hold: i) the part of M in U is specified by the equation $\mathbf{\mathbf{f}(\mathbf{x})} = \mathbf{0}$, ii) the linear transformation from $\mathbb{R}^{n}$ into $\mathbb{R}^{n-k}$ given by the Jacobian of $\mathbf{f}$ is onto $\mathbb{R}^{n-k}$.

What i don't understand is what part of this definition excludes the possibility of a manifold containing it's boundary.
As an example, say a manifold $M$ could be defined by $M = \{(x,y)\in \mathbb{R}^2 | y=x^2 \wedge 0 \leq x \leq 1\} $. Then we can let $f = y-x^2$. Then for any $(x,y)$ in $M$ we can have a small disk ($U$) around any $(x,y)$ which would be mapped to $\mathbb{R}$ by $f$, the part of $M$ inside the disk would be specified by  $\mathbf{f}=0$, and the Jacobian $\begin{bmatrix}-2x & 1\end{bmatrix}$ is onto $\mathbb{R}$.
All seems good to me, but the points $(0,0)$, $(1,0)$ would seem to be boundary points. Where am I going wrong?
 A: In the sentence

the part of $M$ in $U$ is specified by the equation $f(x)=0$,

It has to be understood as
$$\tag{1} M\cap U = \{(x, y) \in U | f(x, y) =0\},$$ instead of just
$$\tag{2} M\cap U \subset \{(x, y) \in U | f(x, y) =0\}.$$
In your example of $M, f$. Let $\mathbf x = (1,1)$ and $U$ be the open set
$$U = \{ (x, y)\in \mathbb R^2 : x\in (0.5, 1.5), |x^2-y|<1\}.$$
Then
$$ \{(x, y) \in U : f(x, y) = 0\} = \{ (x, y)\in U: x^2= y\}$$
strictly contains
$$U\cap M = \{ (x, y) \in U | x^2=y,\ \text{ and } x\in (0.5, 1]\}.$$
So your example is not a manifold, since the sentence should be understood as equality.
To see clearly that this should be the correct interpretation: if you merely require inclusion (2) instead of equality (1), then every subset $X$ which lies inside $\mathbb R\times \{0\} $ would be a $1$-manifold: let $X \subset \mathbb R\times \{0\} \subset \mathbb R^2$ be any subset. Then with $f(x, y) = x$, (2) is satisfied and thus $X$ would be a "manifold" if you assume just (2).
A: You need homeomorphic charts from open subsets of $\mathbb R$ to open subsets of $M$. But there is no such chart to any open subset of $M$ containing its boundary points, because the open line element is not homeomorphic to the half-closed (or closed) line element. This already excludes $M$ from being a topological manifold, let alone a smooth one.
