Proof of Inverse of a Continuous Strictly Increasing Function on an Interval: Continuity and Derivative

Preliminaries. The range of $f$ is its image $f(I)=J$, by Basic Properties of Functions: Equality, Composition, Identity, Inverse and Restriction (v), so $f$ is a map $I\to J$ (Functions, Values of a Function, and Functions from One Class to Another (iii)). The order $\le$ of $\mathbb{R}$ is total, so Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders (vi) applies to any two reals. For $u,v\in I$ with $u\le v$ we have $f(u)\le f(v)$: by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders (v), $u<v$ or $u=v$, and then $f(u)<f(v)$ or $f(u)=f(v)$.

Proof of (i). By The Real Line: Standing Notation and Background for Calculus on Intervals (vi), continuity of $f$ on $I$ is continuity on $I$ for the map from the subset $I$ of $(\mathbb{R},d_{\mathbb{R}})$ into $(\mathbb{R},d_{\mathbb{R}})$, so $J$ is an interval by Intermediate Value Theorem on a Closed Real Interval, and Continuous Images of Intervals (ii). Let $s,t\in I$ be distinct; by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders (vi) we may name them so that $s<t$. Then $f(s)<f(t)$, so $f(s)$ and $f(t)$ are distinct points of $J$, by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders (ii).

Proof of (ii). $f:I\to J$ is injective: let $u,u'\in I$ with $f(u)=f(u')$. If $u<u'$, then $f(u)<f(u')$, so $f(u)\neq f(u')$ by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders (ii), a contradiction; likewise $u'<u$ is impossible. Hence $u=u'$, by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders (vi). $f:I\to J$ is surjective: every $y\in J=f(I)$ is $f(u)$ for some $u\in I$, by Sets and Maps: Ordinary Notation (vii). So $f:I\to J$ is bijective, by Injective, Surjective and Bijective Functions between Classes (iii).

Values of the inverse. By (ii) and Basic Properties of Functions: Equality, Composition, Identity, Inverse and Restriction (vi), $g=f^{-1}$ is a map $J\to I$ with $g\circ f=\mathrm{id}_{I}$ and $f\circ g=\mathrm{id}_{J}$; by Basic Properties of Functions: Equality, Composition, Identity, Inverse and Restriction (ii) and Basic Properties of Functions: Equality, Composition, Identity, Inverse and Restriction (iii),

$$(\mathrm{V})\qquad g(f(u))=u\quad\text{for every }u\in I,\qquad f(g(y))=y\quad\text{for every }y\in J.$$

Since $I\subseteq\mathbb{R}$, $g$ is also a map $J\to\mathbb{R}$ (Functions, Values of a Function, and Functions from One Class to Another (iii)).

Proof of (iii). Let $y,z\in J$ with $y<z$, and let $s=g(y)$ and $t=g(z)$, so $f(s)=y$ and $f(t)=z$ by $(\mathrm{V})$. By Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders (vi), exactly one of $s<t$, $s=t$ and $t<s$ holds. If $s=t$, then $y=f(s)=f(t)=z$, and $y<z$ becomes $y<y$, contrary to Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders (iii). If $t<s$, then $z=f(t)<f(s)=y$, and with $y<z$ this gives $y<y$ by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders (iv), again a contradiction. Hence $s<t$, that is, $g(y)<g(z)$.

Proof of (iv). Fix $y_{0}\in J$ and put $x_{0}=g(y_{0})\in I$, so $f(x_{0})=y_{0}$ by $(\mathrm{V})$. By The Real Line: Standing Notation and Background for Calculus on Intervals (vi) it suffices to show: for every real $\varepsilon>0$ there is a real $\delta>0$ such that every $y\in J$ with $|y-y_{0}|<\delta$ satisfies $|g(y)-x_{0}|<\varepsilon$. Let $\varepsilon>0$ be real. The order arithmetic below is read in the type $\mathrm{Real}$, with the order of The Real Numbers as a Type (v) and the laws of The Real Numbers as a Type (vi). The choices are made in the following order: first the point $x_{+}$ and the number $\delta_{+}$, then $x_{-}$ and $\delta_{-}$, then $\delta$. By Closure Facts in an Ordered Field: Negatives, Differences, Quotients, Absolute Values, Powers, Natural Numbers, Halves and Minima (vii), $\varepsilon/2\in\mathbb{R}$ and $\varepsilon/2+\varepsilon/2=\varepsilon$, hence $0<\varepsilon/2<\varepsilon$.

The point to the right. Suppose there is $t\in I$ with $x_{0}<t$, and fix one such $t$. Then $0<t-x_{0}$ by Rules of Order in an Ordered Ring: Differences, Sums, Negatives, Products and Squares (i), and Closure Facts in an Ordered Field: Negatives, Differences, Quotients, Absolute Values, Powers, Natural Numbers, Halves and Minima (viii), applied to $t-x_{0}$ and $\varepsilon/2$, gives a real $c$ with $0<c\le t-x_{0}$ and $c\le\varepsilon/2$. Put $x_{+}=x_{0}+c$. Then $x_{0}<x_{+}\le t$, so $x_{+}\in[x_{0},t]\subseteq I$ by Basic Facts about Intervals of the Real Line (ii), and $x_{+}-x_{0}=c<\varepsilon$. Put $\delta_{+}=f(x_{+})-y_{0}$; since $f(x_{0})<f(x_{+})$, we have $0<\delta_{+}$ by Rules of Order in an Ordered Ring: Differences, Sums, Negatives, Products and Squares (i).

The point to the left. Suppose there is $t\in I$ with $t<x_{0}$, and fix one such $t$. As before there is a real $c'$ with $0<c'\le x_{0}-t$ and $c'\le\varepsilon/2$; put $x_{-}=x_{0}-c'$. Then $t\le x_{-}<x_{0}$, so $x_{-}\in[t,x_{0}]\subseteq I$ by Basic Facts about Intervals of the Real Line (ii), and $x_{0}-x_{-}=c'<\varepsilon$. Put $\delta_{-}=y_{0}-f(x_{-})$; then $0<\delta_{-}$.

At least one of them exists. $I$ contains two distinct points, so at least one of them, say $u$, differs from $x_{0}$; by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders (vi), $x_{0}<u$ or $u<x_{0}$.

The choice of $\delta$. If both $x_{+}$ and $x_{-}$ are defined, Closure Facts in an Ordered Field: Negatives, Differences, Quotients, Absolute Values, Powers, Natural Numbers, Halves and Minima (viii) gives a real $\delta$ with $0<\delta\le\delta_{+}$ and $\delta\le\delta_{-}$; if only $x_{+}$ is defined, put $\delta=\delta_{+}$; if only $x_{-}$ is defined, put $\delta=\delta_{-}$. In every case $0<\delta$, $\delta\le\delta_{+}$ when $x_{+}$ is defined, and $\delta\le\delta_{-}$ when $x_{-}$ is defined.

Verification. Let $y\in J$ with $|y-y_{0}|<\delta$, and put $x=g(y)$, so $f(x)=y$ by $(\mathrm{V})$. By Rules of Arithmetic and Order in an Ordered Field (x), $-|y-y_{0}|\le y-y_{0}\le|y-y_{0}|$. By Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders (vi) there are three cases.

If $x=x_{0}$, then $|g(y)-x_{0}|=|0|=0<\varepsilon$, by Rules of Arithmetic and Order in an Ordered Field (x).

If $x_{0}<x$, then $x$ witnesses the hypothesis of the right-hand construction, so $x_{+}$ is defined and $\delta\le\delta_{+}$. Suppose $x_{+}\le x$. Then $f(x_{+})\le f(x)=y$ by the preliminaries, so $\delta\le\delta_{+}=f(x_{+})-y_{0}\le y-y_{0}\le|y-y_{0}|<\delta$, and $\delta<\delta$ contradicts Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders (iii). Hence $x_{+}\le x$ fails, so $x<x_{+}$ by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders (vii). Thus $0<x-x_{0}<x_{+}-x_{0}<\varepsilon$, and $|g(y)-x_{0}|=|x-x_{0}|=x-x_{0}<\varepsilon$ (Differences, Quotients and Absolute Values of Real Numbers, Read in the Type Real (iii)).

If $x<x_{0}$, then $x_{-}$ is defined and $\delta\le\delta_{-}$. Suppose $x\le x_{-}$. Then $y=f(x)\le f(x_{-})$, so $\delta\le\delta_{-}=y_{0}-f(x_{-})\le y_{0}-y\le|y-y_{0}|<\delta$, a contradiction as before. Hence $x_{-}<x$ by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders (vii), so $0<x_{0}-x<x_{0}-x_{-}<\varepsilon$, and $|g(y)-x_{0}|=|x-x_{0}|=-(x-x_{0})=x_{0}-x<\varepsilon$ (Differences, Quotients and Absolute Values of Real Numbers, Read in the Type Real (iii)).

So $g$ is continuous at $y_{0}$ relative to $J$; as $y_{0}\in J$ was arbitrary, $g$ is continuous on $J$, by The Real Line: Standing Notation and Background for Calculus on Intervals (vi).

Proof of (v). Let $x\in I$, let $f$ be differentiable at $x$ with $f'(x)\neq0$, and put $y_{0}=f(x)\in J$; then $g(y_{0})=x$ by $(\mathrm{V})$. By Derivative of a Real Function on an Interval (ii), $f'(x)$ is a derivative of $f$ at $x$, so The Increment Form of the Derivative, and Differentiability Implies Continuity (i) gives a map $\varphi:I\to\mathbb{R}$, continuous at $x$ relative to $I$, with $\varphi(x)=f'(x)$ and

$$f(t)-f(x)=\varphi(t)\,(t-x)\qquad\text{for every }t\in I.$$

$\varphi$ has no zeros. $\varphi(x)=f'(x)\neq0$. Let $t\in I$ with $t\neq x$. Then $f(t)\neq f(x)$, as $f$ is injective by (ii), so $f(t)-f(x)\neq0$. If $\varphi(t)=0$, then $f(t)-f(x)=0\cdot(t-x)=0$ by Rules of Arithmetic and Order in an Ordered Field (i), a contradiction; so $\varphi(t)\neq0$.

The quotient map. Let $\chi=\varphi\circ g:J\to\mathbb{R}$, a map with $\chi(y)=\varphi(g(y))$ for $y\in J$, by Basic Properties of Functions: Equality, Composition, Identity, Inverse and Restriction (ii). The map $g:J\to\mathbb{R}$ takes its values in $I$, is continuous at $y_{0}$ relative to $J$ by (iv), and $\varphi$ is continuous at $g(y_{0})=x$ relative to $I$; so $\chi$ is continuous at $y_{0}$ relative to $J$, by the composition rule of The Real Line: Standing Notation and Background for Calculus on Intervals (vi) with $A=J$ and $B=I$. By the previous paragraph $\chi(y)\neq0$ for every $y\in J$, so the map $\psi=1/\chi:J\to\mathbb{R}$, $y\mapsto1/\chi(y)$, is defined (The Reciprocal and the Quotient of Real-Valued Maps with a Nonvanishing Denominator Are Maps (i)), and it is continuous at $y_{0}$ relative to $J$, by the reciprocal rule of The Real Line: Standing Notation and Background for Calculus on Intervals (vi).

The increment form for $g$. Let $y\in J$ and $t=g(y)\in I$, so $f(t)=y$ by $(\mathrm{V})$. Then

$$y-y_{0}=f(t)-f(x)=\varphi(t)\,(t-x)=\chi(y)\,\bigl(g(y)-g(y_{0})\bigr).$$

Multiplying by $\psi(y)$ and using $\psi(y)\,\chi(y)=1$ (Negatives, Reciprocals and the Strict Order in an Ordered Field, as Values (ii)) gives

$$g(y)-g(y_{0})=\psi(y)\,(y-y_{0})\qquad\text{for every }y\in J.$$

Conclusion. $J$ is an interval containing two distinct points, by (i). So The Increment Form of the Derivative, and Differentiability Implies Continuity (i), applied to the interval $J$, the map $g:J\to\mathbb{R}$, the point $y_{0}\in J$, the number $L=\psi(y_{0})$ and the map $\psi$, shows that $\psi(y_{0})$ is a derivative of $g$ at $y_{0}$. Hence $g$ is differentiable at $y_{0}=f(x)$, by Derivative of a Real Function on an Interval (ii), and since $g'(f(x))$ and $\psi(y_{0})$ are both derivatives of $g$ at $f(x)$, Uniqueness of the Derivative of a Real Function on an Interval (i) gives

$$g'(f(x))=\psi(y_{0})=\frac{1}{\varphi(g(y_{0}))}=\frac{1}{\varphi(x)}=\frac{1}{f'(x)}.$$