# Whitney Embedding Theorem (Proposition 6.15 Lee Smooth Manifolds)

In the Compact case of the proof for Whitney's Embedding Theorem presented in John M. Lee's Introduction to Smooth Manifolds, why do we define the function $$F$$ this way?

I don't quite get why we need to last $$m$$ components which are just the bump functions i.e. why do we define the map particularly into $$\mathbb{R}^{nm + m}$$ It does seem like the function takes each ball $$B_i$$ in the finite cover of $$M$$ and maps it into $$B\mathbb{R}^{n+1}$$, giving us a map into $$(\mathbb{R}^{n+1})^m$$. Is this related in any way?

Let's first look at the map $$p \mapsto \rho_1(p) \phi_1(p),$$ defined for $$p \in B_1$$ for the time being.

1. Near the center of $$B_1$$, we'll have $$\phi_1(p)$$ is approximately $$1$$, so this will look like an embedding of $$B_1$$ into $$R^n$$.

2. Near the edge of $$B_1$$, $$\phi_1(p)$$ will be almost zero, so this just collapses the outer regions of that ball down towards the origin.

Let's take a concrete example. Suppose that you have a 1-manifold --- the real line! --- and a coordinate chart $$\phi$$ that maps points of some interval in $$M$$ to $$[-1, 1]$$. We'll also build a partition-of-unity function $$\rho$$ by defining $$\rho(p) = (1 - \phi(p))^2.$$ ---basically it's "1" at the center of the 'interval' and fades off to 0 by the ends of the interval. It's not $$C^\infty$$, but it'll be at least $$C^1$$ when extended by $$0$$ outside the chosen interval.

What does the image of the interval look like? Under $$\phi$$ alone, it'd just be the open interval from $$-1$$ to $$1$$ -- great! But when you multiply by $$\rho$$, the "ends" fold back, and the image looks more like a staple in a piece of paper: a piece of wire with its ends folded back towards the middle.

Let's make this even more concrete: Let's suppose our interval is $$[2, 4]$$. Then $$\phi(t) = t-3 \\ \rho(t) = 1 - (t-3)^2$$ and $$h(t) = \phi(t)\rho(t) = (t-3) - (t-3)^3.$$ The derivative of $$h$$ is $$h'(t) = 1 - 3(t-3)^2$$ which is $$0$$ when $$(t-3)^2 = 1/3$$, i.e. $$t = 3 + \frac{1}{\sqrt{3}}$$. So $$h$$ isn't all that nice a map. If you stacked up a bunch of these (one for each interval of the form $$I_n = (n-1, n+1)$$) along various axes in $$R^K$$ for some large $$K$$, what would the derivative look like at some point $$t$$? Well $$t$$ lies $$I_n$$ for at most 2 values of $$n$$ , so most of your coordinates are zeroes. But for those two remaining coordinates, you've got something where projection on one axis has zero derivative somewhere in the overlap interval, and projection on the other has a zero derivative somewhere in that interval, etc. And if you're unlucky, those zeroes happen at the same point, and your map isn't a diffeomorphism.

Why does adding the $$\rho$$-coordinates fix this? I don't know offhand, and I don't have a copy of Lee's book, but I'll bet you find the derivative of the map $$F$$ computed within the next couple of paragraphs, and I'll bet further that the $$\rho$$-coordinates will play heavily in showing that the derivative has maximal rank everywhere.

But you can at least see from this concrete example that simply using only the $$\rho\phi$$ coordinates won't really work in general.

• Thank you for the detailed reply! Commented Mar 11 at 19:58