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.
 A: 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.
A: This is an application of the Implicit Function Theorem. You have a function $f : \mathbb{R}^n \to \mathbb{R}^m$ and you construct the graph given by: $\{ (x,y) \in \mathbb{R}^n \times \mathbb{R}^m : y = f(x) \}.$ Let me define a new map, say, $G : \mathbb{R}^{n+m} \to \mathbb{R}^m$ given by $G : (x,y) \mapsto y-f(x).$ I have defined this the way I have so that the graph of $f$ is the zero-level set of $G$, i.e. the graph of $f$ is the set of $(x,y) \in \mathbb{R}^n \times \mathbb{R}^m$ such that $G(x,y) = 0.$ 
In brutal detail this map is really:
$$G : (x_1,\ldots,x_n,y_1,\ldots,y_m) \mapsto (y_1-f_1(x_1,\ldots,x_n),\ldots,y_m-f_m(x_1,\ldots,x_n)) \, .$$
We need to calculate the Jacobian Matrix of $G$. A quick calculation will show you that:
$$ J_G = \left[\begin{array}{c|c} -J_f & I_m \end{array}\right] ,$$
where $J_f$ is the $m \times n$ Jacobian matrix of $f$ and $I_m$ is the $m \times m$ identity matrix. The matrix $J_G$ is an $m \times (m+n)$ matrix.
To be able to apply the IFT, we need to show that $0$ is a regular value of $G$. (After all, the graph of $f$ is $G^{-1}(0).$) We can do this by showing that none of the critical points get sent to 0 by $G$. Notice that $G$ has no critical points because $J_G$ always has maximal rank, i.e. $m$. This is clearly true since the identity matrix $I_m$ has rank $m$.
It follows that the graph of $f$ is a smooth, parametrisable $(n+m)-m=n$ dimensional manifold in a neighbourhood of each of its points.
