Dgdrxrt

Inverse Function Theorem. Let $f: mathbbR^n to mathbbR^n$ be a $C^1$ function. If $det Df_a neq 0$, there is open sets $U, V$ such that $f: U to V$ is a diffeomorphism $C^1$ ($a in U$ and $f(a) in V$).

Why is the continuously differentiability necessary for this version? I'm trying to find a example with $f$ just differentiable, $det Df_a neq 0$ but $f$ is not a local diffeomorphism (at the point $a$)

Thanks for the advance!

Not a particularly flashy example, but consider the function $f: mathbbRto mathbbR$ given by
$$
f(x) = begincases
x+2x^2sin(frac1x)&textif xneq 0\
0&textif x=0
endcases.
$$
You can show that $f$ is differentiable on all of $mathbbR$ with $f'(0) = 1$, but $f'$ is discontinuous at 0. To see that $f$ is not a local diffeomorphism at the origin, you can find a sequence of intervals that approach the origin on which $f$ is decreasing, but since $f'(0)>0$ we can find a point $a$ with $0<a$ with $f(0)<f(a)$, so $f$ can't be a one-to-one mapping on any neighborhood of zero.

In detail, we have
$$
f'(x) = begincases
1 + 4xsin(frac1x)-2cos(frac1x)&textif xneq 0\
1&textif x=0
endcases.
$$
As $x$ approaches 0 the $4xsin(frac1x)$ term can be made arbitrarily small while the $2cos(frac1x)$ term oscillates between $-2$ and $2$. Consequently, we can find a sequence of points $x_nto 0$ with $0<x_n+1<x_n$ such that $f'(x_n) = -1$ say. Since $f'$ is continuous away from the origin, $f$ is decreasing on a small neighborhood of each $x_n$ by the mean value theorem.

By the definition of the derivative we have
$$
f'(0) = lim_xto 0fracf(x)-f(0)x-0 = lim_xto 0fracf(x)x = 1>0.
$$
In particular, for $a$ sufficiently close to zero and positive we have $f(a)>0$. Now if we take one of our $x_n$'s such that $0<x_n<a$ then $f$ is decreasing on some interval contained in $(0, a)$. But if $f(0)=0$, $f(a)>0$, and $f$ is decreasing on some interval between $0$ and $a$ then by continuity, $f$ can't be one-to-one on $[0,a]$.

More intuition than rigor: First verify that if $xle f(x) le x+x^2$ on $(-1,1),$ then $f'(0)=1.$

Now consider this situation: $1>a_1>1/2 >a_2>1/3 >a_3>1/4cdots.$ Choose the $a_n$ so close to $1/n$ that the line segments $[a_n,1/n]times 1/n$ lie between the graphs of $x$ and $x+x^2.$ We can then connect these line segments together, smoothly, so that the result is the graph of a smooth function on $(0,1)$ that stays between the graphs of $x,x+x^2.$

Call this function $f$ and then extend it to $(-1,0]$ by setting $f(x)=x,xle 0.$ Then $f$ is differentiable on $(-1,1)$ and $f'(0)=1.$ Yet $f$ is constant on each of the intervals $[a_n,1/n].$ This shows $f$ is not $1-1$ on any interval containing $0.$