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.


2 Answers 2


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.

  • $\begingroup$ I'm having trouble understanding your definition of G. There are multiple functions $f_1,\ldots , f_m$ because of the values $y_1,\ldots,y_m$ to be equal to those, and if that is the case, wouldn't each of the coordinates of the range be equal to 0? $\endgroup$ Commented Nov 26, 2012 at 19:55
  • $\begingroup$ The function $f$ goes from $\mathbb{R}^n$ to $\mathbb{R}^m$, i.e. you give it $n$ numbers and it gives you $m$ numbers back. The numbers you give it are $x_1,\ldots,x_n$ while the numbers it gives you are $f_1,\ldots,f_m.$ In longhand: $$f(x) = f(x_1,\ldots,x_n) = (f_1(x_1,\ldots,x_n), \ldots, f_m(x_1,\ldots,x_n)).$$ Does this help? $\endgroup$ Commented Nov 26, 2012 at 20:12
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    $\begingroup$ The the graph of $f$ is given by the equation $G(x,y)=0$. As you said: the coordinates of the range of the graph will all be zero. This is what the IFT does: It takes a system of equations and tells you if you get a smooth, parametrisable manifold as the solution space. The trick is to find a system of equations whose solution gives you what you're interested in. $\endgroup$ Commented Nov 26, 2012 at 20:23
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    $\begingroup$ But applying the IFT surely is an overkill... You have a global chart $\mathbb{R}^n \to \operatorname{graph}(f) \subseteq \mathbb{R}^{n+m}$ given by $x \mapsto (x,f(x))$ which is a smooth bijection onto its image and its inverse is given by the restriction of the projection $(x,y) \mapsto x$ to the graph. $\endgroup$
    – Martin
    Commented Apr 23, 2013 at 1:01
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    $\begingroup$ Actually, applying the IFT seems to be circular in this case. The proof of the fact that the pre-image of a regular value is a submanifold proceeds by showing that you get charts by writing the preimage locally as graph of a function. $\endgroup$
    – Martin
    Commented Apr 23, 2013 at 1:10

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.

  • $\begingroup$ Let me try to reiterate. This matrix comes from the derivative of my graph function where $I_n$ is the derivative of $t$ and $f'(t)$ is the derivative of $f(t)$. That I believe I understand. I'm not sure about how rank $n$ means that the range is a manifold or even exactly what it is to be a submanifold. Could you please elaborate a little on that? $\endgroup$ Commented Nov 26, 2012 at 19:38
  • $\begingroup$ It's been long since I had Calculus, so I'm not sure I'll give the exact explanation, but I'll try. $\endgroup$
    – yo'
    Commented Nov 26, 2012 at 19:40
  • $\begingroup$ Your further explanation as well as the other answer were very helpful. +1 Thank you. $\endgroup$ Commented Nov 26, 2012 at 20:05

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