Let $f:R^n\to R^n$ be a smooth, proper map where $n\geq 2$.
(a). Prove that $\lim_{|x|\to \infty} |f(x)|=\infty$ (i.e. for all $R>0$, there is $r>0$ such that if $|x|>r$ then $|f(x)|>R$).
(b). From (a) it follows that there is $r_0>0$ such that for all $x\in R^n$ that satisfy $|x|>r_0$: $f(x)\neq 0$.
Let $r_0$ be such a number.
For all $r\geq r_0$ define a map:
$\phi_r:S^{n-1}\to S^{n-1}$
by
$$\phi_r(x)=\frac{f(rx)}{\|f(rx)\|}$$
Prove that deg $\phi_r$ is independent of the choice of r.
(c). In part (b) symbols, assume that deg $\phi_r\neq 0$.
Prove that $f$ is surjective.
(Hint: with loss of generality assume that $0\in R^n$ is not in $Im(f)$ ).
Definitions:
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A proper map: a continuos map $f:X \to Y$ ($X,Y$ are topological spaces) is called proper if for all $K \subset Y$ compact, $f^{-1}(K)$ is compact.
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Let $f: M \to N$ be a smooth map between smooth manifolds.
We say that $y\in N$ is a singular value of $f$ is there is $x\in f^{-1}(y)$ such that $d_xf$ is not surjective, othetwise $y$ is called a regular value of $f$ (i.e. for all $x\in f^{-1}(y)$ , $d_xf$ is surjective). -
Homotopy:
Let $M,N$ be smooth manifolds, $\partial M=\emptyset$.
If $f_0, f_1: M\to N$ are smooth, then we say that a smooth map $F:M\times [0,1] \to N$ is a homotopy if $F(x,i)=f_i(x)$ for $i=0,1$ and $x\in M$.
For (a):
Trial 1: by the contrapositive.
Assume that $|f(x)|$ is bounded so $f$ is bounded too. And so the image of $f$ is contained in a closed ball (in $R^n$) that is a compact set, however the preimage of this ball would be $R^n$ which is not compact (because it is not bounded)?
Trial 2: consider a sequence (in $R^n$) that goes to $\infty$, if its image is bounded inside $U\subset R^n$, we take a compact set $K$ that contains a limit point so tge preimage of $K$ in not compact since it contains a subsequence of the initial sequence that goes to infinity.
How to write this formally?
For (b):
Let $r_2>r_1>r_0$ , define a homotopy:
$H:S^{n-1}\times [0,1]\to S^{n-1}$ by
$H(x,t)=\phi_{(1-t)r_1+tr_2}\phi_{r_1}$.
So if $t=0$, $H(x,0)=\phi_{ r_1}$ and if $t=1$, $H(x,1)=\phi_{r_2}$
Therefore $H $ is a homotopy (from $\phi_{r_1}$ to $\phi_{r_2}$).
Now using a theorem that states: if $M,N$ are smooth manifolds without boundary and have same dimension and have orientation. If $f_0, f_1$ are smooth, proper and there exists a proper homotopy between them that is a smooth, proper map $F:M\times [0,1] \to N$ which is a homotopy (from $f_0$ to $f_1$), then $\forall$
$y\in N$ a regular value of $f_0, f_1$ ,$deg_y f_0=deg_y f_1$.
In tgis case, $f_0=\phi_{r_1}, f_1=\phi_{r_2}$
Both are maps $S^{n-1}\to S^{n-1}$, they are smooth and proper since $f$ is so.
For (c), I tried to use the theorem:
If $f:M\to N$ is a smooth, proper map where $M, N$ are smooth manifolds without boundary. $M, N$ with same dimensions and have orientations, $N$ is connected.
So if $f$ is not surjective then $deg(f)=0$.
(If $f:M\to N$ is a smooth, proper, non surjective map then $deg (f)=0$).
I also thought about using homotopy here but did not manage to apply it.
Edited:
Part a: let $x_n$ be a sequence with $\|x_n\|\to \infty$ such that {$f(x_n): n$} is bounded. Choose a compact set $K\subset R^n$ to be $\overline {f(x_n):n}$ so it is a bounded and closed set in $R^n$ thus compact. However $f^{-1}(K)=\overline {x_n:n}$ is not compact since it is not bounded.
Part b: Let $r_2>r_1>r_0$ , define a homotopy:
$H:S^{n-1}\times [0,1]\to S^{n-1}$ by
$H(x,t)=\phi_{(1-t)r_1+tr_2}\phi_{r_1}$.
So if $t=0$, $H(x,0)=\phi_{ r_1}$ and if $t=1$, $H(x,1)=\phi_{r_2}$
Therefore $H $ is a homotopy, i.e. $\phi_{r_1}$ and $\phi_{r_2}$ are homotopic. Thus $deg (\phi_r)$ is independent of $r>r_0$. Why can I say that $\phi_r$ is smooth? (And so $H$ is smooth).
Part c: Assume that $f$ is non-surjective with $0$ not in $Im(f)$, then $\phi_r$ is defined for all $r\geq 0$.
By part b, since $\phi_r$ is indepedendent of the choice of $r$ so $deg(\phi_r)=deg(\phi_0)=$deg(constant) $=0$, a contradiction to the assumption.
Best Answer
For (a): The opposite of $\lim_{\|x\|\to\infty}\|f(x)\|=\infty$ is not “$f$ is bounded”. Rather it is: There is a sequence $x_n$ with $\|x_n\|\to\infty$ such that $\{f(x_n):n\}$ is bounded. Now you can get a contradiction to the compactness similarly as you attempted in Trial 2.
Your proof of (b) is the correct approach, but haven't you learnt a more general theorem about the homotopy invariance of the degree?
The approach to (c) is correct, and it is essentially the only possible approach: You need to use the so-called existence property of the degree (which is also sometimes formulated as part of the so-called additivity property); the homotopy invariance alone is not sufficient. One way to prove that $0$ lies in the image of $f$ is to combine it with the homotopy invariance: Assume by contradiction that $0$ does not lie in the image of $f$. What can you say about $\phi_r$ for $r<r_0$, in particular for $r=0$?