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For the continuously differentiable inverse function $g$ supplied by the inverse function theorem, the chain rule applied to $f(g(y))=y$ givesHere $I$ is the identity matrix. The matrix inverse is taken at the recovered input $g(y)$, not at the output $y$. In one dimension this is the inverse function rule $g^{\prime }(y)=1/f^{\prime }(g(y))$.

The formula also describes sensitivity. For a small output perturbation $h$, differentiability gives $g(y+h)-g(y)=Dg(y)h+o(\Vert h\Vert )$. A large operator norm of $Dg(y)$ permits large amplification in some directions. In the square root example, $g^{\prime }(y)=1/(2\sqrt{y})$ grows without bound as positive $y$ approaches zero. Existence of a local inverse function therefore does not imply that recovery is insensitive to measurement error.

An inverse function reverses a bijection. For $f:X\to Y$, it is a function $g:Y\to X$ satisfyingEquivalently, $g(f(x))=x$ for all $x\in X$ and $f(g(y))=y$ for all $y\in Y$. We write $g=f^{-1}$. The superscript denotes reversal through function composition; it does not mean the pointwise multiplicative inverse $1/f(x)$. The preimage notation $f^{-1}(B)$ for a set $B$ is also distinct: a preimage exists even when no inverse function exists.

For a computed example, let $f:\mathbb{R}\to \mathbb{R}$ be $f(x)=2x+1$. Solving $y=2x+1$ for $x$ gives $g(y)=(y-1)/2$. Thus $f(1)=3$ and $g(3)=1$; direct substitution gives $g(f(x))=x$ and $f(g(y))=y$ throughout the real line. In contrast, $1/f(3)=1/7$, which does not undo $f(1)=3$.

Why must $f$ be a bijection? If $f(x_1)=f(x_2)$, applying $g$ gives $x_1=x_2$, proving injectivity. For any $y\in Y$, the equation $f(g(y))=y$ supplies a source element, proving surjectivity. Conversely, a bijection supplies exactly one predecessor for each $y$, defining a unique inverse function. A left inverse or right inverse requires only one of the two composition identities.

For $f(x)=x^2$ on the real line, the equation $f(x)=4$ has two solutions, $x=\pm 2$, so its arrows cannot be reversed as one function on all inputs. The restriction to $[0,\infty )$ removes that ambiguity. As a bijection from $[0,\infty )$ onto $[0,\infty )$, it has inverse function $g(y)=\sqrt{y}$. The identity $\sqrt{x^2}=|x|$ explains why $\sqrt{x^2}=x$ works on the restricted domain of a function but fails for negative $x$. ^[ref].

More generally, $(a,b)$ lies on the graph of a function $f$ precisely when $(b,a)$ lies on the graph of a function of its inverse function. This coordinate swap explains the geometric reflection and the horizontal line test. It also explains why inverse sensitivity is reciprocal locally: the inverse function rule divides by the original derivative at the corresponding input, when that derivative is nonzero.

The inverse function theorem answers a practical question: can nearby outputs be used to recover nearby inputs uniquely and smoothly? Its useful surprise is that a nonlinear function can be tested using its linear derivative. If the first-order changes preserve all input directions, then, on a sufficiently small region, the nonlinear map can also be reversed. This is the basis for treating a nonlinear map as a coordinate transformation.

Precisely, let $f:U\to {\mathbb{R}}^n$ be continuously differentiable on an open subset $U$ of Euclidean space, let $a\in U$, and put $b=f(a)$. If $Df(a)$ is an invertible linear map, there are open neighbourhoods $U_0$ of $a$ and $V_0$ of $b$ such that $f:U_0\to V_0$ is a bijection with a continuously differentiable inverse function $g$. Thus this restriction is a diffeomorphism. In coordinates the test is $\det Df(a)\ne 0$, using the Jacobian determinant. The derivative of a local inverse map is obtained by inverting the Jacobian matrix, at the corresponding input. ^[ref]

For a one-dimensional example, $f(x)=x^2$ has $f^{\prime }(1)=2$. Near $x=1$, the inverse function is $g(y)=\sqrt{y}$, and the inverse function rule gives $g^{\prime }(1)=1/2$. But $x^2$ has no inverse function on the whole real line, since $x$ and $-x$ give the same output. The theorem supplies a local restriction, not global injectivity. At zero, the derivative test fails; the nonnegative branch still has a square root inverse, but that inverse has no finite derivative at zero. ^[ref]

Here is a two-dimensional example with an invertible derivative everywhere and infinitely repeated outputs:The exponential function turns $s$ into a positive radius, while $\theta$ is the angle in polar coordinates. Restricting, for example, to $|s|<0.35$ and $|\theta |<0.6$ produces the reversible patch pictured below. On its image, with Cartesian coordinates $(u,v)$ and $u>0$, the inverse function isThe arctangent selects the angle near zero. Without an angle restriction, $f(s,\theta +2\pi k)=f(s,\theta )$ for every integer $k$, so there is no single global inverse function. This map is a local diffeomorphism but is not injective.

The short proof idea is to correct a guessed input using the inverse linear response. With $A=Df(a)$, solving $f(x)=y$ is equivalent to finding a fixed point ofAt $a$, the derivative $D{T}_y(a)=I-A^{-1}Df(a)$ vanishes. Continuity makes $T_y$ a contraction mapping on a sufficiently small closed ball, which it maps into itself when $y$ is sufficiently close to $b$. The contraction mapping theorem then gives existence and uniqueness there. Establishing the smooth dependence on $y$ completes the inverse theorem. This explains why the nonlinearity can be tolerated: its correction has a small local slope.^[ref]