The graph of a smooth real function is a submanifold Given a function $f: \mathbb{R}^n \rightarrow \mathbb{R}^m $ which is smooth, show that
$$\operatorname{graph}(f) = \{(x,f(x)) \in \mathbb{R}^{n+m} : x \in \mathbb{R}^n\}$$
is a smooth submanifold of $\mathbb{R}^{n+m}$.
I'm honestly completely unsure of where or how to begin this problem. I am interested in definitions and perhaps hints that can lead me in the right direction.

The map $\mathbb R^n\mapsto \mathbb R^{n+m}$ given by $t\mapsto (t, f(t))$ has the Jacobi matrix $\begin{pmatrix}I_n\\f'(t)\end{pmatrix}$, which has a full rank $n$ for all $t$ (because of the identity submatrix). This means that its value range is a manifold. Is there anything unclear about it?
How is this a proof that it is a manifold?
A manifold of rank $n$ is such set $X$ that for each $x\in X$ there exists a neighborhood $H_x\subset X$ such that $H_x$ is isomorphic to an open subset of $\mathbb R^n$. In this case, the whole $X=graph(f)$ is isomophic to $\mathbb R^n$. The definition of a manifold differs, often it is required for the isomophism to be diffeomophism, which is true here as well.
Think of it this way: A manifold $X$ of rank $2$ is something, in which: wherever someone makes a dot there by a pen, I can cut a piece of $X$ and say to this person: "See, my piece is almost like a piece of paper, it's just a bit curvy.
The definition of manifold might seems strage here because here you can take the neighborhood as the whole $X$. This is not always the case: A sphere is a manifold as well, but a whole sphere is not isomorphic to $\mathbb R^2$, you have to take only some cut-out of it.
