# Thread: Milnor - Topology from the Differentiable Viewpoint ... 2 questions

1. ## Milnor - Topology from the Differentiable Viewpoint ... 2 questions

Hello,

I have 2 questions concerning the proof of the "Homogeneity Lemma" at p. 22-24.

First:
On p. 24 he writes: "But clearly, with suitable choice of c and t, the diffeomorphism F_t will carry the origin to any desired point in the open unit ball."

How does this work? How do I have to choose c and t?

Second:
A few lines later:
"hence the above argument shows that every point sufficiently close to y is "isotopic" to y." [isotopic means: It exists a diffeomorphism f from N to itself, f(x)=y and f is smoothly isotopic to the identity map of N]

I want to prove that.
Let y be an interior point of N, U a open neighborhood of y that is diffeomorphic to R^n via f, i.e. $f:U \rightarrow \mathbb{R}^n$. You can choose f, so that f(y)=0.
I want to show, that every x in $f^{-1}(B(0,1))$ is isotopic to y (B(0,1) is the open unit ball).
For $x \in f^{-1}(B(0,1))$ you can choose a diffeomorphism F_t [construced in the first part of the proof] with F_t(0)=f(x), since f(x) is in the open unit ball. Then $A:=f^{-1} \circ F_t \circ f$ is a diffeomorphism from U to U, that is smoothly isotopic to the identity map of U and A(y)=x. But that doesn't do, since we need a diffeomorphism from N to N.

The only thing I'm not able to show is, that i can extend A to a Diffeomorphism B from N to N. I know that A(x)=x for every x that is not in $f^{-1}(B(0,1))$.
Define:
$B: N \rightarrow N$ B(x)=A(x) for x in U and B(x)=x for x not in U.
I can't show that B is smooth. I use the definition from page 1 of the book. If x is within U or if x is an exterior point of U, it is clear that B is smooth at x. How does it work, if x is a boundary point of U, i.e. if every neighborhood of x cointains a point of U and a point of N\U.

I would be really thankful if anyone could help me.

engmaths.

edit: I could post the pages of the book, but not sure if I'm allowed to (copyright).

2. ## Re: Milnor - Topology from the Differentiable Viewpoint ... 2 questions

I took a break, and then thought everything over. This is what I got:

let x be a boundary point of U, i.e. in every neighborhood of x in N is a point of U and a point of N\U.

I want to show the following: (*)
You can choose a neighborhood of A of x in N so that there is not point of $f^{-1}(B(0,1))$ in A.

If I have shown this, the defined map B is the identity on A and therefore smooth in x.

Proof of (*):
.... that was wrong ....

engmaths

3. ## Re: Milnor - Topology from the Differentiable Viewpoint ... 2 questions

1. Suppose we want to move 0 to a point p inside the unit ball. Let c=p/|p|, then the differential equation system is $\frac{dx}{dt}=\frac{\phi}{|p|} p$, since p is different from 0, we can choose a non-zero component, say $p^i$, we have $\frac{dt}{dx^i}=\frac{|p|}{\phi p^i}$, now choose $t = \int_0^{p^i} \frac{|p|}{\phi p^i} dx^i$, we'll have $F_t(0)=p$

4. ## Re: Milnor - Topology from the Differentiable Viewpoint ... 2 questions

So actually the first question is, a particle starts from 0 running towarding p on a straight line, with a velocity $x'=\phi c$, when is it going to reach p?

5. ## Re: Milnor - Topology from the Differentiable Viewpoint ... 2 questions

2. You don't need to prove the statement "every point sufficiently close to y is "isotopic" to y", since the whole previous paragraph was doing that, by constructing a system of differential equations and solving it.