The Inverse Function Theorem Completed: Examples and Counterexamples

By [L1] and [L3], $DP(r,\theta )=\left(\begin{array}{cc}\cos \theta & -r\sin \theta \\ \sin \theta & r\cos \theta \end{array}\right),\det DP(r,\theta )=r\mathrm{.}$ Thus [L4] and [L2] give the asserted local diffeomorphism and orientation wherever $r\mathrm{\ne }0$.

On the principal strip, equality of two images first gives equality of the positive radii by the Pythagorean identity. Translating each negative angle by $2\pi$ puts both angles into the half-open interval of the unit-circle parametrization without changing sine or cosine; its injectivity then gives equality of the original angles. The injective regular-map theorem makes this restriction a diffeomorphism onto its open image. On the full positive-radius domain, $(r,\theta )$ and $(r,\theta +2\pi )$ are distinct with the same image, so global injectivity fails there.

The coordinate functions are smooth, so [L2] makes $F(x,y)=(x,xy)$ smooth. For an increment $(h,k)$, $F(x+h,y+k)-F(x,y)-(h,yh+xk)=(0,hk)\mathrm{.}$ Since $2\mathrm{\mid }hk\mathrm{\mid }\le h^2+k^2$, the remainder divided by $\sqrt{h^2+k^2}$ tends to zero. Hence $DF(x,y)=\left(\begin{array}{cc}1& 0\\ y& x\end{array}\right),\det DF(x,y)=x\mathrm{.}$ On $U$, direct substitution gives $G(F(x,y))=(x,y)$ and $F(G(u,v))=(u,v)$.

Step 1.1 verifies the hypotheses and conclusion of [L1], so $F:U\to U$ is a smooth diffeomorphism. On the full plane, $F(0,y)=(0,0)$ for every $y$, and hence no point $(0,v)$ with $v\mathrm{\ne }0$ lies in its image.

Repeated use of [L3] and [L5] makes the polynomial terms smooth, while [L2] and [L5] make the exponential terms smooth; hence $F$ is smooth. Direct substitution gives $F(0,0,0)=0$, and [L2]–[L4] give $D_{(y,z)}F(0,0,0)=\left(\begin{array}{cc}0& -3\\ 1& 1\end{array}\right),$ whose determinant is $3$, while $D_{x}F(0,0,0)=(0,1{)}^T$.

Step 1.1 satisfies [L1], so unique smooth $y(x),z(x)$ solve the system near zero. Its derivative formula gives $\left(\genfrac{}{}{0px}{}{y^{\mathrm{\prime }}(0)}{z^{\mathrm{\prime }}(0)}\right)=-D_{(y,z)}F(0{)}^{-1}{D}_{x}F(0)=\left(\genfrac{}{}{0px}{}{-1}{0}\right)$.

By [L1], [L4], and [L5], the successive derivatives of the cube map are $3x^2$, $6x$, $6$, and then zero, so it is smooth. By [L7], the cube is strictly increasing on the nonnegative half-line; the identity $(-x{)}^3=-x^3$ then gives strict increase on the whole line. For $y\ge 0$, [L6] supplies $x\ge 0$ with $x^3=y$, while for $y<0$ the negative of the cube root of $-y$ maps to $y$. Thus the cube map is onto and hence bijective. By [L3] it and its inverse are continuous, so it is a homeomorphism and therefore open. Its derivative at zero is zero.

Suppose its inverse $g$ were differentiable at zero. Applying [L2] to $g(x^3)=x$ at zero would give $g^{\mathrm{\prime }}(0)f^{\mathrm{\prime }}(0)=1$, but step 1.1 makes the left side zero. Therefore the inverse is not differentiable at zero. The map $x\mapsto x^3$ is a smooth open bijection of $\mathbb{R}$ with derivative zero at the origin, but its inverse is not differentiable there.

Every point of $U$ has a neighbourhood contained in exactly one component, and $f$ is constant there. The zero linear map therefore leaves a zero remainder in [L1], so $Df(a)=0$ for every $a\in U$.

The values at $-3\mathrm{/}2$ and $3\mathrm{/}2$ are respectively $1$ and $-1$. Thus the total derivative of $f$ is zero at every point of $U$, but $f$ is not constant on $U$.

By [L1], $(f(h)-f(0))\mathrm{/}h=1+2h\sin (1\mathrm{/}h)\to 1$, so $f^{\mathrm{\prime }}(0)=1$. For $x\mathrm{\ne }0$, the algebra, chain, and power rules with [L2] give $f^{\mathrm{\prime }}(x)=1+4x\sin (1\mathrm{/}x)-2\cos (1\mathrm{/}x)\mathrm{.}$

Put $x_n=1\mathrm{/}(2\pi n)$ and $y_n=1\mathrm{/}((2n+1)\pi )$ for $n\ge 1$. By [L3] and [L4], step 1.1 gives $f^{\mathrm{\prime }}(x_n)=-1$ and $f^{\mathrm{\prime }}(y_n)=3$. Both sequences tend to zero, so derivatives of both signs occur in every neighbourhood of zero.

Suppose $f$ were injective on an interval about zero. It is continuous there, so [L5] would make it strictly increasing or strictly decreasing. Difference quotients show that the derivative of an increasing differentiable function is nonnegative and that of a decreasing one is nonpositive, contradicting step 2.1. Thus $f^{\mathrm{\prime }}(0)=1$ is invertible but no local inverse exists.

By [L3], $DS(x,y)=\left(\begin{array}{cc}2x& -2y\\ 2y& 2x\end{array}\right)\mathrm{.}$ At $(x,y)\mathrm{\ne }(0,0)$, the matrix $\frac{1}{2(x^2+y^2)}\left(\begin{array}{cc}x& y\\ -y& x\end{array}\right)$ is its two-sided inverse, so [L2] and [L1] make $S$ locally invertible at every point of $U$.

Nevertheless $S(1,0)=(1,0)=S(-1,0)$. Thus $S$ is not injective and has no global inverse, despite its everywhere-invertible derivative.

The map in [L1] satisfies the $C^1$ bijection hypothesis of the false claim.

Its inverse fails even to be differentiable at zero by [L1], and therefore cannot be $C^1$. Hence the claim is false.

The construction in [L1] satisfies the derivative hypothesis and violates the conclusion, so it refutes the claim.

The comparison with [L2] identifies connectedness as the missing hypothesis: adding it restores exactly the stated equivalence.

The cube map in [L1] is open and $C^1$, but its derivative at zero is the zero linear map, which is not invertible.

Therefore openness does not imply derivative invertibility. The valid result [L2] is only the forward sufficient implication from derivative invertibility to openness.