When Are Two Spaces Topologically the Same?
Two spaces can have different points and still have the same topological structure. To make this precise, we look for a bijection that translates the open-set structure in both directions: the map must be continuous, and its inverse must also be continuous. This two-way condition is what distinguishes topological equivalence from a mere correspondence between points.
Unless stated otherwise, \(\mathbb{R}\) has its standard topology, and intervals and subsets of \(\mathbb{R}\) have the subspace topology. Recall that a map \(f:X\to Y\) between topological spaces is continuous if \(f^{-1}(V)\) is open in \(X\) for every open set \(V\) in \(Y\). A bijection is a map that is both one-to-one and onto; for a bijection \(f\), its inverse \(f^{-1}:Y\to X\) sends each \(y\in Y\) to the unique \(x\in X\) with \(f(x)=y\).
The inverse condition is essential: continuity of \(f\) ensures that the map carries the topology in one direction without discontinuous jumps, while continuity of \(f^{-1}\) ensures the reverse correspondence works as well. In practical examples, the first step is usually to propose a formula for \(f\) and then find a formula for its inverse.
A Useful Test: Verify Both Inverse Identities
Proof. To prove that \(f\) is one-to-one, suppose \(f(x_1)=f(x_2)\). Applying \(g\) to both sides gives \(g(f(x_1))=g(f(x_2))\). The identity \(g\circ f=\operatorname{id}_X\) turns this into \(x_1=x_2\). To prove that \(f\) is onto, take any \(y\in Y\). The identity \(f\circ g=\operatorname{id}_Y\) gives \(f(g(y))=y\), so \(y\) is the image of \(g(y)\in X\). Thus \(f\) is bijective. For each \(y\in Y\), the equality \(f(g(y))=y\) says that \(g(y)\) is the unique preimage of \(y\), so \(g=f^{-1}\). If both maps are continuous, the definition of homeomorphism now applies. \(\square\)
Both identities matter: \(g\circ f=\operatorname{id}_X\) alone does not establish that \(f\) is onto \(Y\), and \(f\circ g=\operatorname{id}_Y\) alone does not establish that \(f\) is one-to-one. Checking both also makes the continuity check precise: once the inverse identities have been established, continuity of the two proposed maps is exactly what is required.
Worked Example: Rescaling a Closed Interval
Show that \([0,1]\) and \([2,5]\), each with the subspace topology from \(\mathbb{R}\), are homeomorphic. Define \(f:[0,1]\to[2,5]\) by \(f(x)=3x+2\).
First check the range. If \(x\in[0,1]\), then \(0\leq x\leq1\). Multiplying by \(3\) and adding \(2\) gives \(2\leq3x+2\leq5\), so \(f(x)\in[2,5]\). Conversely, for any \(y\in[2,5]\), the value \(x=(y-2)/3\) satisfies \(0\leq(y-2)/3\leq1\). This suggests the inverse \(g:[2,5]\to[0,1]\), \(g(y)=(y-2)/3\).
Check the inverse identities. For \(x\in[0,1]\), \(g(f(x))=((3x+2)-2)/3=3x/3=x\). For \(y\in[2,5]\), \(f(g(y))=3((y-2)/3)+2=(y-2)+2=y\). Thus the two maps are mutual inverses and \(f\) is a bijection.
Check continuity. For \(x_1,x_2\in[0,1]\), \(|f(x_1)-f(x_2)|=|(3x_1+2)-(3x_2+2)|=3|x_1-x_2|\). Given any \(\varepsilon>0\), choosing \(\delta=\varepsilon/3\) ensures that \(|x_1-x_2|<\delta\) implies \(|f(x_1)-f(x_2)|<\varepsilon\). For \(y_1,y_2\in[2,5]\), \(|g(y_1)-g(y_2)|=|y_1-y_2|/3\leq|y_1-y_2|\), so choosing \(\delta=\varepsilon\) proves continuity of \(g\). These estimates prove continuity in the subspace metrics. Therefore \(f\) is a homeomorphism.
Check: \(f(0)=3(0)+2=2\) and \(f(1)=3(1)+2=5\); the inverse sends these endpoints back to \(g(2)=(2-2)/3=0\) and \(g(5)=(5-2)/3=1\).
Homeomorphisms Between Different Kinds of Sets
A homeomorphism need not be an affine rescaling. The next examples use familiar functions to match spaces whose points and shapes are presented differently. We use the standard calculus facts that the exponential and logarithm functions are continuous on their respective domains, and that the displayed rational functions are continuous wherever their denominators are nonzero.
Worked Example: The Real Line and the Positive Real Numbers
Let \((0,\infty)=\{y\in\mathbb{R}:y>0\}\) have its subspace topology. Define \(f:\mathbb{R}\to(0,\infty)\) by \(f(x)=e^x\), and define \(g:(0,\infty)\to\mathbb{R}\) by \(g(y)=\ln y\).
For every \(x\in\mathbb{R}\), \(e^x>0\), so \(f\) takes values in \((0,\infty)\). For every \(y>0\), \(\ln y\) is defined and is real. The exponential and logarithm identities give, for all \(x\in\mathbb{R}\) and \(y>0\), \(g(f(x))=\ln(e^x)=x\) and \(f(g(y))=e^{\ln y}=y\). The maps are therefore mutual inverses, so \(f\) is a bijection and \(g=f^{-1}\).
The exponential function is continuous on \(\mathbb{R}\). The logarithm is continuous on the open interval \((0,\infty)\); consequently it is continuous as a map into \(\mathbb{R}\) with the stated subspace domain topology. Thus both maps are continuous, and they define a homeomorphism \(\mathbb{R}\cong(0,\infty)\).
Check: \(f(0)=e^0=1\) and \(g(1)=\ln 1=0\), consistent with \(g(f(0))=0\).
Worked Example: The Open Unit Interval and the Real Line
Give an explicit homeomorphism from \((0,1)\) to \(\mathbb{R}\). Both spaces have their standard or subspace topologies. Define \(f:(0,1)\to\mathbb{R}\) by \(f(x)=\ln\!\left(\frac{x}{1-x}\right)\). For \(0<x<1\), both \(x\) and \(1-x\) are positive, so the logarithm has a positive argument and \(f(x)\) is defined.
Propose \(g:\mathbb{R}\to(0,1)\) by \(g(y)=\frac{e^y}{1+e^y}\). Since \(e^y>0\), its denominator \(1+e^y\) is positive. Also \(e^y<1+e^y\), so \(0<g(y)<1\); hence \(g\) takes values in the required interval.
Check the first composition. If \(0<x<1\), then \(e^{f(x)}=x/(1-x)\). Substituting this into \(g\) gives \(\displaystyle g(f(x))=\frac{x/(1-x)}{1+x/(1-x)}=\frac{x/(1-x)}{((1-x)+x)/(1-x)}=x.\) The denominators are nonzero because \(1-x>0\).
Check the second composition. For \(y\in\mathbb{R}\), \(g(y)/(1-g(y))=e^y\): indeed, \(1-g(y)=1/(1+e^y)\), so the ratio is \(\bigl(e^y/(1+e^y)\bigr)/\bigl(1/(1+e^y)\bigr)=e^y\). Therefore \(f(g(y))=\ln(e^y)=y\). The two maps are mutual inverses.
The formula for \(f\) is continuous on \((0,1)\), since \(1-x\ne0\) there and its logarithm has positive argument. The formula for \(g\) is continuous on \(\mathbb{R}\), since \(1+e^y>0\) for every real \(y\). Thus \(f\) is a homeomorphism. As a numerical check, \(f(1/2)=\ln(1)=0\) and \(g(0)=1/(1+1)=1/2\).
Homeomorphism Is an Equivalence Relation
Topological equivalence behaves like equality for the purpose of grouping spaces: every space is equivalent to itself, equivalence works in either direction, and equivalences can be composed.
Proof. For any topological space \(X\), the identity map \(\operatorname{id}_X:X\to X\) is bijective and is its own inverse. For every open \(U\subseteq X\), \(\operatorname{id}_X^{-1}(U)=U\), so it and its inverse are continuous. Thus \(X\cong X\).
Next, suppose \(f:X\to Y\) is a homeomorphism. By definition, \(f^{-1}:Y\to X\) is a continuous bijection, and its inverse is \(f\), which is continuous. Thus \(f^{-1}\) is a homeomorphism, giving \(Y\cong X\).
Finally, suppose \(f:X\to Y\) and \(h:Y\to Z\) are homeomorphisms. Their composite \(h\circ f:X\to Z\) is a bijection, and its inverse is \(f^{-1}\circ h^{-1}\). The composition of continuous maps is continuous, so \(h\circ f\) is continuous and \(f^{-1}\circ h^{-1}\) is continuous. Therefore \(h\circ f\) is a homeomorphism, and \(X\cong Z\). These arguments also apply when one or more spaces are empty: the unique map from the empty space to itself is an identity, and a homeomorphism to or from an empty space can exist only when both spaces are empty. \(\square\)
A Common Pitfall: A Continuous Bijection May Not Be a Homeomorphism
Checking that a map is continuous and bijective is not enough. Its inverse must be continuous as well. For a concrete example, let \(X=\mathbb{R}\) with the standard topology and let \(Y=\mathbb{R}\) with the indiscrete topology \(\{\varnothing,\mathbb{R}\}\). Consider the identity function \(f:X\to Y\).
This map is bijective. It is continuous because the only open sets in \(Y\) are \(\varnothing\) and \(\mathbb{R}\), whose preimages under \(f\) are respectively \(\varnothing\) and \(\mathbb{R}\), both open in \(X\). But its inverse is not continuous: the set \((0,1)\) is open in \(X\), and its preimage under \(f^{-1}:Y\to X\) is \((0,1)\), which is not open in the indiscrete topology on \(Y\). Therefore \(f\) is not a homeomorphism.
When checking a proposed homeomorphism, keep the tasks separate: verify that each formula takes values in its stated codomain, verify both inverse identities, and verify continuity of both maps with respect to the topologies actually assigned to the spaces.
Check Your Understanding
For each question, identify the spaces and topologies before testing the maps.
- For \(f:[-1,1]\to[0,4]\) defined by \(f(x)=2x+2\), write an explicit inverse and verify both inverse identities.
- Why does continuity and bijectivity of a map not by itself establish that it is a homeomorphism?
- For the map from \((0,1)\) to \(\mathbb{R}\) in the worked example, verify directly that \(g(y)\in(0,1)\) for every \(y\in\mathbb{R}\).
- Suppose \(f:X\to Y\) has a continuous inverse and both inverse identities hold. What further continuity condition is needed for \(f\) to be a homeomorphism?
- Use the equivalence-relation theorem to explain why a chain of homeomorphisms \(X\cong Y\) and \(Y\cong Z\) implies \(X\cong Z\).