Building a Homeomorphism from a Formula
A formula can suggest a map between two familiar spaces, but a homeomorphism requires more than a well-defined continuous function. The map must be bijective, and its inverse must also be continuous. For subsets of the real line, useful constructions often come from solving the formula for the input, checking the exact range, and then checking continuity in the subspace topologies.
Unless stated otherwise, \(\mathbb{R}\) has its standard topology, and every interval or other subset of \(\mathbb{R}\) has the subspace topology. Recall that a map \(f:X\to Y\) is a homeomorphism if it is a bijection, \(f\) is continuous, and its inverse \(f^{-1}:Y\to X\) is continuous. The notation \(f^{-1}\) here means the inverse function, not the preimage of a set.
For a formula defined on an interval, the usual continuity rules from calculus establish continuity as a real-valued function wherever the formula is defined. If its image lies in a subset \(Y\subseteq\mathbb{R}\), it is also continuous as a map into \(Y\) with the subspace topology: for every open set \(V=Y\cap W\) in \(Y\), where \(W\) is open in \(\mathbb{R}\), its preimage is \(f^{-1}(V)=f^{-1}(W)\), since all values of \(f\) lie in \(Y\). The same reasoning applies to an inverse formula.
A Criterion for Monotone Formulas
Proof. Since \(f\) is a bijection, its inverse \(g:K\to J\) exists. We prove that \(g\) is continuous. First suppose \(f\) is strictly increasing. Fix \(y_0\in K\), and write \(x_0=g(y_0)\). Let \(U\) be any open neighborhood of \(x_0\) in \(J\). By the subspace topology, there is an \(\varepsilon>0\) such that \(J\cap(x_0-\varepsilon,x_0+\varepsilon)\subseteq U\).
If \(J\) contains a point below \(x_0\), choose \(a\in J\) with \(x_0-\varepsilon<a<x_0\). If \(J\) contains a point above \(x_0\), choose \(b\in J\) with \(x_0<b<x_0+\varepsilon\). Each choice is possible on its stated side because \(J\) is an interval. For an available lower-side point, strict increase gives \(f(a)<y_0\); for an available upper-side point, it gives \(y_0<f(b)\). Choose \(\delta>0\) smaller than each of the positive gaps \(y_0-f(a)\) and \(f(b)-y_0\) that occur. If neither side occurs, then \(J=\{x_0\}\), and \(g\) is continuous at \(y_0\) because its only possible value is \(x_0\).
For the other cases, suppose \(y\in K\) and \(|y-y_0|<\delta\). If \(y<y_0\), the lower side must occur, and \(y>f(a)\); strict increase implies \(a<g(y)<x_0\). If \(y>y_0\), the upper side must occur, and \(y<f(b)\); hence \(x_0<g(y)<b\). If \(y=y_0\), then \(g(y)=x_0\). In all cases \(g(y)\in J\cap(x_0-\varepsilon,x_0+\varepsilon)\subseteq U\). This proves continuity of \(g\) at \(y_0\). For a strictly decreasing \(f\), the same argument applies with the order of the inequalities reversed: values on the lower side of \(x_0\) map above \(y_0\), and values on the upper side map below \(y_0\). Thus the inverse is continuous in either case, and \(f\) is a homeomorphism. \(\square\)
This criterion is useful when a formula has a simple inverse but its continuity is not immediately visible from that inverse. For the examples below, we will also check the inverse formulas directly, so that the range and continuity requirements remain explicit.
Worked Example: Matching Closed Intervals by an Affine Formula
Construct a homeomorphism from \([1,4]\) to \([2,8]\), and verify its inverse.
The affine formula that sends \(1\) to \(2\) and \(4\) to \(8\) has slope \((8-2)/(4-1)=6/3=2\). Thus define \(f:[1,4]\to[2,8]\) by \(f(x)=2x\). If \(1\le x\le4\), then \(2\le2x\le8\), so the formula does take values in the stated target. It is injective because \(2x_1=2x_2\) implies \(x_1=x_2\). It is onto because for every \(y\in[2,8]\), \(x=y/2\) lies in \([1,4]\) and \(f(x)=y\). Therefore the inverse is \(g(y)=y/2\).
Both formulas are continuous as real-valued functions, being scalar multiples of the input. Since their values lie in the stated intervals, the subspace-topology observation above gives continuity as maps \([1,4]\to[2,8]\) and \([2,8]\to[1,4]\). Hence \(f\) is a homeomorphism.
Check: \(f(1)=2\) and \(f(4)=8\), while \(g(2)=1\) and \(g(8)=4\). For every \(x\in[1,4]\), \(g(f(x))=(2x)/2=x\); for every \(y\in[2,8]\), \(f(g(y))=2(y/2)=y\).
The same affine method works for intervals with matching endpoint types. Given real numbers \(a<b\) and \(c<d\), the formula \(f(x)=c+\frac{d-c}{b-a}(x-a)\) sends \((a,b)\) onto \((c,d)\), and also sends \([a,b]\) onto \([c,d]\). Its inverse is \(f^{-1}(y)=a+\frac{b-a}{d-c}(y-c)\). The positive slopes preserve the endpoint order; open or included endpoints are carried to open or included endpoints, respectively.
Sending a Bounded Open Interval to the Real Line
An affine function cannot send a bounded interval onto all of \(\mathbb{R}\), but a formula that grows without bound toward the endpoints can. A logarithm of a ratio provides such a construction. The interval endpoints must be excluded because the ratio in the formula is defined only in the interior.
Worked Example: A Homeomorphism from \((2,5)\) to \(\mathbb{R}\)
Define \(f:(2,5)\to\mathbb{R}\) by \(f(x)=\ln\!\left(\frac{x-2}{5-x}\right)\). For \(2<x<5\), both \(x-2\) and \(5-x\) are positive, so the logarithm is defined. Solve \(t=f(x)\) for \(x\): exponentiating gives \(e^t=(x-2)/(5-x)\), so \(e^t(5-x)=x-2\), hence \(5e^t+2=x(1+e^t)\). Therefore the candidate inverse is \(g(t)=\frac{2+5e^t}{1+e^t}\).
For every \(t\in\mathbb{R}\), \(e^t>0\), so \(g(t)-2=\frac{3e^t}{1+e^t}>0\) and \(5-g(t)=\frac{3}{1+e^t}>0\). Thus \(g(t)\in(2,5)\). Substitution gives \(\frac{g(t)-2}{5-g(t)}=\frac{3e^t/(1+e^t)}{3/(1+e^t)}=e^t\), so \(f(g(t))=\ln(e^t)=t\). For \(x\in(2,5)\), the equation used to solve for \(x\) gives \(g(f(x))=x\). These two identities show that the maps are inverses, and in particular that \(f\) is a bijection onto \(\mathbb{R}\).
The numerator and denominator of the ratio defining \(f\) are continuous, and the denominator is nonzero on \((2,5)\). The ratio is positive there, so composing it with the continuous logarithm gives a continuous \(f\). The formula for \(g\) is continuous on \(\mathbb{R}\), since \(1+e^t>0\) for every real \(t\). Thus both maps are continuous with the indicated subspace topologies, and \(f\) is a homeomorphism.
Check: At \(x=7/2\), the ratio is \((7/2-2)/(5-7/2)=(3/2)/(3/2)=1\), so \(f(7/2)=\ln1=0\). The inverse gives \(g(0)=(2+5)/(1+1)=7/2\). At \(x=17/4\), the ratio is \((9/4)/(3/4)=3\), so \(f(17/4)=\ln3\); substituting \(t=\ln3\) into \(g\) gives \((2+15)/(1+3)=17/4\).
The same calculation works on any bounded open interval \((a,b)\), where \(a<b\). The formula \(x\mapsto\ln\!\left(\frac{x-a}{b-x}\right)\) has inverse \(t\mapsto\frac{a+be^t}{1+e^t}\). The positivity of \(x-a\), \(b-x\), and \(e^t\) ensures that each formula is defined on its entire stated domain and that the inverse lies in \((a,b)\).
A Half-Open Interval and a Common Range Error
A simpler rational formula gives a homeomorphism from a half-open bounded interval to a half-line. As with the logarithmic construction, checking the image before claiming a target is essential.
Worked Example: From \([0,1)\) to \([0,\infty)\)
Define \(f:[0,1)\to[0,\infty)\) by \(f(x)=\frac{x}{1-x}\). For \(0\le x<1\), the denominator is positive and the numerator is nonnegative, so \(f(x)\ge0\). To solve \(y=x/(1-x)\), multiply by \(1-x\) to obtain \(y-yx=x\), and then \(y=x(1+y)\). Thus the candidate inverse is \(g(y)=\frac{y}{1+y}\).
If \(y\ge0\), then \(0\le y/(1+y)<1\): the first inequality follows from \(y\ge0\), and the strict upper bound follows from \(y<1+y\). Also, \(1-g(y)=1/(1+y)>0\), and therefore \(f(g(y))=\bigl(y/(1+y)\bigr)/\bigl(1/(1+y)\bigr)=y\). For \(x\in[0,1)\), \(1+f(x)=1/(1-x)\), so \(g(f(x))=(x/(1-x))/(1/(1-x))=x\). These identities prove that the maps are inverse bijections.
The denominator \(1-x\) is nonzero on \([0,1)\), and \(1+y\) is nonzero on \([0,\infty)\); the quotient rules therefore show that \(f\) and \(g\) are continuous on their domains. By the subspace topology, they are continuous as maps between the displayed spaces. Hence \(f\) is a homeomorphism.
Check: \(f(0)=0\), and \(f(1/3)=(1/3)/(2/3)=1/2\). Applying the inverse gives \(g(1/2)=(1/2)/(3/2)=1/3\), as required.
Common pitfall. A range error is to claim that \(x/(1-x)\) maps \([0,1)\) onto \(\mathbb{R}\) merely because its denominator is nonzero. For every \(x\in[0,1)\), \(x\ge0\) and \(1-x>0\), so \(x/(1-x)\ge0\). In particular, the value \(-1\) is not in its image. The correct target is \([0,\infty)\), and the inverse calculation confirms that every value in this target is attained.
For each construction, the interval endpoints and the sign of the formula determine the appropriate range. Solving for the inverse then checks that no target values are missed, while continuity must be verified in both directions. These checks are what turn a plausible formula into a homeomorphism.
Check Your Understanding
For each question, specify the domain and target topologies as subspace topologies from \(\mathbb{R}\), and verify the range before concluding that a formula is a homeomorphism.
- Find an affine homeomorphism from \([0,2]\) to \([-3,3]\), and write its inverse.
- For the map \(x\mapsto \ln\!\left(\frac{x}{1-x}\right)\) on \((0,1)\), solve for the inverse and verify its range.
- Use the monotonicity theorem to explain why a continuous, strictly decreasing bijection between two intervals is a homeomorphism.
- Find the inverse of \(x\mapsto x/(1-x)\) on \([0,1)\), and state its exact range.
- Why does the formula \(x\mapsto x/(1-x)\) not define a map from \([0,1)\) onto \(\mathbb{R}\)?