@iwriteonbananas Here's a little something. If $f : \Bbb R^n \to \Bbb R^n$ is a $C^1$ map such that $Df(a)$ is invertible, then $f$ has a local inverse near $a$, correct?
Here's a global version of the theorem: $f : \Bbb R^n \to \Bbb R^n$ be $C^1$ and $Df(x)$ invertible at all $x \in \Bbb R^n$. If moreover for all sequences $\{x_k\}$, $\lim_{\|x_k\| \to \infty} f(\|x_k\|) = \infty$, $f$ has a global inverse.
@Mambo But it doesn't satisfy the second condition. Take the sequence $\{x_k\}$ where $x_k = [0, k]$. $\|f(x_k)\| = e^0 = 1$. Yet $\|x_k\| = k \to \infty$.
Well, $f(x) = \sin(x)$. Derivative is uniformly bounded above and below by $1$ and $-1$ respectively, but clearly $f$ is not proper, because it is also bounded.
(AKA: Show your condition does not imply Mambo's, that the derivative is uniformly bounded.) You have already seen an example that shows that smooth maps with uniformly bounded invertible differentials still need not be proper.
@BalarkaSen An example where the function is $C^1$ map and derivative is invertible at every point, and the inverse of derivative at every point is uniformly bounded, still the function does not have global inverse
Then, question: Suppose the differential is everywhere invertible and uniformly bounded away from 0 and infty; that is, $c < \|Df\| < C$ everywhere, $0 < c < C < \infty$.
I am trying to find an example where the function is $C^1$ map and derivative is invertible at every point, and the inverse of derivative at every point is uniformly bounded, still the function does not have global inverse
I guess even in this case function will have global inverse.
@Balarka: There were two further problems for you up above. Find an example of a diffeomorphism whose derivative is unbounded; and investigate what happens when you can uniformly bound the derivative away from 0 and infty.
@MikeMiller Completely silly. $f : \Bbb R^2 \to \Bbb R$, $f(x, y) = x$ works. $\|Df\| = 1$ which is obviously bounded away from both $0$ and $\infty$, and given the sequence $x_k = [0, k]$, $\|f(x_k)\| = 0$ even though $\|x_k\| \to \infty$. So $f$ is not proper.
$Df$ is supposed to be invertible everywhere. I thought that was clear from context. My apologies for lack of clarity.
In everything we're talking about $Df$ should be invertible. Our goal is to take the local inverse function theorem and see whether or not we can make it global.
$\exp(- 1/x^2)$ on $\mathbb R$ is a counterexample, since $C||x||^{-\alpha} \exp(-1/x^2)\to 0$ as $x \to 0$ (for all $\alpha$) (And thus must eventually by smaller than $1$). You can extend this onto the complex plane by taking $f(x+iy)=\exp(-1/x^2)$ and it is smooth
Is there a same extension of perpendicularity to less than (or more) 90 degrees as in there is a surface of a Riemann Sphere where the definition of parallel lines is interpreted differently?
Well I don't know but there are certainly ways to generalize the notion of perpendicularity. Also there is Non-Euclidean geometry where the parallel postulate can be different from the Euclidean parallel postulate.
@ropstah because 90 is half of 180? it don't mean that sarcastically: it guarantees that rotating a line by 90 degrees twice gives back the original line
"Starting from $x\equiv 3\pmod{7}$, this means that $x$ is of the form $x=7k+3$ for some integer $k$. Substituting into the second congruence, we get $$7k+3 \equiv 2\pmod{5}.$$ Since $7\equiv 2\pmod{5}$, and $3\equiv -2\pmod{5}$, this is equivalent to $\color{blue}{2k-2\equiv 2\pmod{5}}$. Dividing through by $2$ (which we can do since $\gcd(2,5)=1$) we get $k-1\equiv 1\pmod{5}$, or $k\equiv 2\pmod{5}$. That is, $k$ is of the form $k=5r+2$ for some integer $r$." - Could someone please explain how they got the blue bit?
@Ropstah For example, two vectors in an inner product space are orthogonal if their inner product is zero, where orthogonality means the same thing as perpendicularity for non-zero Euclidean vectors in 2 or 3 dimensions. (For the record I think a better terminology for this is is that it "extends" the notion rather than generalizes it.)
But take what I say with a grain of salt because most of the time I'm only 50% sure that what I am saying is true :P
