# Question about proof of the Rank Theorem from Lee's Smooth Manifolds

The Rank Theorem: Given a map, $$F: M \rightarrow N$$ of constant rank, r, there exist smooth charts $$(U,\phi)$$ and $$(V, \psi)$$ such that

$$\psi \circ F \circ \phi^{-1} (x^1,...,x^r,x^{r+1},...,x^m) = (x^1,...,x^r,0,...,0)$$

In the book he defines, $$\phi: U_0 \rightarrow \tilde{U}_0$$, where $$\tilde{U}_0$$ is an open cube, with $$\phi(x,y) = (Q(x,y), y)$$ where $$x$$ is short hand for $$x^1,...,x^r$$ and $$y$$ is short hand for $$x^{r+1},...,x^m$$ which he relabels as $$y^{r+1},...,y^m$$, with $$F(x,y) = (Q(x, y), R(x, y))$$.

Then he arrives at $$D(F \circ \phi^{-1}) = \begin{pmatrix} I & 0 \\ \frac{\partial \tilde{R^i}}{\partial x^j} & \frac{\partial \tilde{R^i}}{\partial y^j} \end{pmatrix}$$

He mentions that it is necessary for $$\tilde{U}_0$$ to be an open cube so that $$\tilde{R}$$ is independent of $$(y^1,...,y^{m-r})$$, based on $$D(F \circ \phi^{-1})$$ above, but I thought the same held based on rank arguments alone.

He also uses the fact that $$\tilde{U}_0$$ later on, where he says,

Because $$\tilde{U}_0$$ is a cube and $$F \circ \phi^{-1}$$ has the form (4.6), it follows that $$F \circ \phi^{-1}(\tilde{U}_0) \subseteq V_0$$

My question is, why is $$\tilde{U}_0$$ being a cube needed for the above statements?

I'm struggling with the same result too.

I agree one does not need $$\tilde{U}_0$$ to be a cube in order to the derivatives vanish.

For the second argument, I think the important fact about being a cube is to be connected, so you can be sure you stay in the domain of the chart $$(V,\psi)$$ despite the compositions.

Note that $$\psi(V_0)$$ is a connected subset of $$\mathbb{R}^n$$, which I think (haven't proof) is a cube.

• I actually asked some people offline. I need to write up the answer. For the first part, the derivatives vanishing isn't sufficient to say that $\tilde{R}(x,y) = \tilde{R}(x,0)$.
– Jeff
Dec 15 '18 at 2:46
• For one, if you have something like a T-shaped region $\tilde{R}$ might not even be defined at $\tilde{R}(x,0)$. A second problem would be if $\tilde{U_0}$ is disconnected. In this case, even though the derivative with respect to $y$ could be 0 in each region $\tilde{R}(x,y)$ could be at different constant values with respect to $y$ in each region. The cube avoids these problems.
– Jeff
Dec 15 '18 at 2:54
• You’re right, I found this interested fact to figure it out the importance of connected sets in the proof, and it’s related with the corollary following this theorem math.stackexchange.com/questions/252130/… Dec 27 '18 at 2:04