Spivak Calculus on Manifolds - Definition of a k-dimensional manifold

Question

Above is one of the two definitions of a k-dimensional manifold in Spivak Calculus on Manifolds.

Could someone explain to me why $f'(y)$ must have rank k?

Thank you.

Answer 1

With apologies, this is really more of a motivation than a definition, but the way I like to think about it is this: you want a smooth manifold to have enough of the properties of $\mathbb{R}^k$ so that you can generalize notions from calculus or analysis to them. One thing that smooth functions in $\mathbb{R}^k$ have that you might want is the Jacobian, which is telling you something about how your function is varying in different directions at a point.

Requiring $f'$ to have rank $k$ at all points in the manifold is exactly what will allow you to export this property to the more general setting of manifolds. We don't want rank less than $k$, because then at that point we lose some information about functions defined on our manifold, and we don't want rank greater than $k$ because you're hoping for something with only $k$ dimensions.

Ahhh, in response to your clarification, let's reason it out: $f$, in the proof, is a function $W \subset \mathbb{R}^k \to \mathbb{R}^n$, and $H$ is a function from $\mathbb{R}^n \to \mathbb{R}^k$. The idea is that because the composition of functions is the identity (at least on this subset of $\mathbb{R}^k$), its Jacobian must also be the identity $k\times k$ matrix. But when is this possible? Only when $f'$ is one-to-one as a linear transformation, i.e. when its matrix has rank $k$.

Answer 2

The really simple version is that you are going (forgetting about manifolds) $\mathbb{R}^k\to \mathbb{R}^n$, so that is a $n\times k$ matrix. But if the derivative (which says how much "like a $k$-plane" the function is) doesn't have the right rank, then you could possibly lose a dimension even though the function itself is 1-1.

For instance, take the function $f(t)=(t^2,t^3)$ from the reals to $\mathbb{R}^2$. An open set around zero is certainly going on a one-to-one basis to the curve $y^2=x^3$ here, and the derivative exists everywhere $(2t,3t^2)$. (I guess the inverse is continuous too, though I'm too lazy to check that now.)

But the derivative does NOT have rank $k=1$ at $t=0$, it's the zero matrix. And indeed the image isn't a manifold, it has a singularity/cusp.

See here or here for probably a better answer. Also, there are topological manifolds, which if I recall correctly only require continuity. Hope I got this all right.