algebraic-topologydifferential-geometrydifferential-topologygeometric-topologymorse-theory
The question is given below:
And here is exercise (16):
And here is the solution to exercise(17)
But I have difficulties in understanding the following parts of the solution:
1-Why the codomain of the defined $h$ in the second line is $\mathbb{R}$?
2-Why $h$ is clearly smooth as stated in the third line?
3-Why we are using U x {0}, what is the importance of using the singleton $0$?
4-Why K x {0} is compact? and why this leads to that $h > 2\delta$ for some $\delta > 0$?
5-And by which property of continuity of h, there exists an open set $U'$ such that $h > \delta$ on $U'$?
6-When usually Tube lemma is applied, when we need what, we apply it?
Thank you!
Related Solutions
A difficulty in understanding a part of solution of Q.1.3.10 in Allan Pollack and Guillemin.
OK, let's break the exercise into steps.First, the "then $f$ maps $Z$ diffeomorphically onto $f(Z)$" part.
Since you have that $f$ in one-one on $Z$, and immersion at every $x\in Z$, you only need to prove that the map $f|_Z: Z \rightarrow f(Z)$ is a proper map, because then it would be an embedding, and an embedding is a diffeomorphism onto its image (pages 17, 18). Just take a compact $K \subset f(Z)$, since $f(Z)$ is given the subspace topology, it is also compact in $Y$, and since $Y$ is Hausdorff, it is closed in $Y$. Then since $f|Z$ is continuous as a map with image in $Y$ (just the restriction of a continuous map), $f^{-1}(K)$ is closed in $Z$, but $Z$ compact implies it is compact in $Z$. So $f$ is also proper, hence a diffeomorphism.
Now, we need to prove that $f$ is a little more, is a diffeomorphism of an open neighborhood of $Z$ in $X$ onto an open neighborhood of $f(Z)$ in $Y$.
Why is it that it's enought to prove $f$ to be one-one is an open neighborhood of $Z$ in $X$ (Exercise 5)? Because Exercise 5 tells you that a local diffeomorphism that is one-one is actually a diffeomorphism. Suppose you find an open nighborhood $U$ of $Z$ in $Y$ with $f|_U$ one-one, we could (if necessary) intersect $U$ with a neighborhood $V$ of $Z$ such that $f|V$ is a local diffeomorphism (this $V$ you get it by just taking a finite union of neighborhoods in $X$ of points in $Z$ restricted to wich $f$ is a diffeomorphism, local condition). Then you can assume there's a $U$ neighborhood of $Z$ in $X$ such that $f|_U: U\rightarrow f(U)$ is a local diffeo and one-one, applying Exercise 5 we get that $f|_U$ is a diffeomorphism.
There are only two things left to prove: Exercise 5, and the injectivity in a neighborhood of $Z$ in $X$.
For the first one, just observe that if $f: A\rightarrow B$ is a one-one local diffeo, then exists $f^{-1}: f(A)\rightarrow A$ and both $f$, $f^{-1}$ are smooth by just using their local expression, the identity. Then $f$ is a diffeomorphism onto its image, that hence is going to be open in $B$.
For the second one, suppose $f$ is not one-one restricted to any neighborhood of $Z$ in $X$.
Define $U_n=$ {$x \in X| d(x,Z)<\frac{1}{n}$} (your manifold X is sitting inside some $\mathbb{R}^k$ by definition, and if you don't take Guillemin's convention, just use Whitney's Embedding Theorem). There are {$a_n$}, {$b_n$} such that $a_n, b_n \in U_n$ and $f(a_n)=f(b_n)$. Clearly $d(a_n, Z), d(b_n, Z)$ tend to zero. Since the distance is continuous and $Z$ is closed in $X$ (again $Z$ compact inside $X$ Hausdorff is closed), then there are $z_1, z_2\in Z$ with $z_1=\lim_{n\to\infty} a_n$, $z_2=\lim_{n\to\infty} b_n$. Since $f$ is continuous, $f(z_1)=f(z_2)$. This contradicts the injectivity of $f$ restricted to $Z$ if $z_1\neq z_2$. So that $z_1=z_2=z$.
Now that contradicts the fact that $df_z$ is an isomorphism, since this implies that there's a neighborhood of $z$ in $X$ restricted to which $f$ is a diffeomorphism, in particular one-one.
Stack of records theorem from Allan Pollack and Guillemin differential topology.(Q.1.4.7)
By the Regular Level Set Theorem, since $y$ is a regular value of $f:X\to Y$, we have that $f^{-1}(y)$ is smooth embedded submanifold of $X$ of codimension (in $X$) equal to the dimension of $Y$. Since $\dim{X}=\dim{Y}$, we have that $f^{-1}(y)$ is a $0$-dimensional submanifold. Since $f$ is continuous, the preimage of any closed set is closed. So $f^{-1}(y)$ is closed, and hence compact, since it's a closed subset of the compact set $X$.
Begin Edit-$f^{-1}(y)$ is discrete since it's a $0$-dimensional manifold. Since it's compact and discrete it must be finite. Indeed, if it was infinite, then it would contain a limit point. Hence every neighborhood of said point would contain other points in the set contradicting discreteness. Hence $f^{-1}(y)=\{x_1,...,x_n\}$.-End Edit
It's a local diffeomorphism about each $x_j\in f^{-1}(y)$ by the inverse function theorem (since $y$ is a regular value $df_{x_j}$ has full rank at each $x_j$).
The remainder of the proof is just chasing open sets until you have one small enough with all the desired properties.
