# A “Manifold with Boundary” Question

I'm given the following scenario: Letting $U$ be an open subset of $\mathbb{R}^n$ and $f,g:U\rightarrow \mathbb{R}$, two smooth functions such that $f(\vec x)\lt g(\vec x)$ for all $\vec x\in U$, we'll consider the set $M$ of all pairs $(\vec x, y)$ such that $x\in U$ and $y\in [f(\vec x),g(\vec x)]$. I want to know that $M$ is a manifold with boundary under these conditions, in particular the dimension and what the boundary is. In the event that the definitions of manifold with boundary differ, I'm using the following:

An n-dim manifold with boundary in $\mathbb{R}^N$ is a subset M such that for all $\vec x \in M$ there exists and open subset $V\subseteq \mathbb{R}^N$ containing $\vec x$, an open subset $U\subseteq \mathbb{R}^n$, and an embedding $\psi:U\rightarrow \mathbb{R}^N$ such that $\psi(U\cap H^n)=V\cap M$ (where $H^n$ is the usual upper-half space).

So I'm thinking I can cover the manifold with two embeddings namely $\psi_1(\vec x, z)=(\vec x, f(\vec x)+z)$ and $\psi_2(\vec x, z)=(\vec x, g(\vec x)-z)$. Am I on the right track here? Certainly the graphs of f and g are part of the boundary but is there more? And more generally, what can go wrong in the event the $f\lt g$ condition isn't satisfied? As always, if I've poorly articulated anything, lemme know!

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The definition of a manifold with boundary you give there is incomplete; according to your definition any chart necessarily contains boundary points (the images under $\psi$ of the boundary points of $U \cap H$). You also have to allow the same statement without intersecting $U$ with $H$.
Anyways, in your case you can cover $M$ with two charts like that precisely as you do. The boundary will precisely consist of the graphs of $f$ and $g$ restricted to $U$, as you say (notice it is important that $U$ is open). So you are definitely on the right track.
To your other question: The set $M(U,f,g) = \{(x,y) | x \in U, y \in [f(x),g(x)]\}$ will be a manifold precisely if $f(x) < g(x)$ for all $x \in U$. This is not hard to prove, but i will just give a moral example:
Consider $U = \mathbb R$, $f(x) = -x^2, g(x) = x^2$. If $M(U,f,g)$ would be a manifold it must be $2$-dimensional. But if you remove the point $0$ from it you will disconnect it, but you can not disconnect a (connected) manifold of dimension $\geq$ 2 by removing a point.