Two Different Kinds of Equivalence
A homeomorphism identifies two spaces exactly: it is a bijection that is continuous and whose inverse is continuous. A homotopy equivalence allows a weaker comparison. It asks for continuous maps in both directions whose composites can be continuously deformed to the identity maps. The maps themselves need not be bijections, and the composites need not equal the identities at every point.
We will make this distinction explicit using a one-point space and the unit interval. The example is especially useful because there is no bijection at all between these spaces, yet there are maps in both directions that satisfy the homotopy-equivalence conditions.
The domains matter in these definitions. If \(f:X\to Y\) and \(g:Y\to X\), then \(g\circ f\) is a map from \(X\) to \(X\), while \(f\circ g\) is a map from \(Y\) to \(Y\). Each composite is compared with the identity map on its own domain.
A Point and the Unit Interval
Worked Example: Homotopy Inverse Maps Between a Point and an Interval
Let \(P=\{p\}\) have its unique topology \(\{\varnothing,P\}\), and let \(I=[0,1]\) have the standard subspace topology inherited from \(\mathbb{R}\). Define \(f:P\to I\) and \(g:I\to P\) by \(f(p)=0\) and \(g(x)=p\) for every \(x\in I\). Both maps are continuous: every map from \(P\) is continuous because its only open subsets are \(\varnothing\) and \(P\); for \(g\), the preimage of either open subset of \(P\) is \(\varnothing\) or \(I\), both open in \(I\).
First compute the composite on \(P\). Its only input is \(p\), and \((g\circ f)(p)=g(f(p))=g(0)=p\). Thus \(g\circ f=\operatorname{id}_P\).
Next compute the composite on \(I\). For every \(x\in I\), \((f\circ g)(x)=f(g(x))=f(p)=0\). So \(f\circ g\) is the constant map from \(I\) to \(I\) with value \(0\); it is not the identity map, since \((f\circ g)(1)=0\) whereas \(\operatorname{id}_I(1)=1\).
Define \(H:I\times I\to I\) by \(H(x,t)=(1-t)x\). This formula is continuous: the coordinate functions \(x\) and \(t\) are continuous on the product, and sums and products of real-valued continuous functions are continuous. Also \(0\leq (1-t)x\leq1\) whenever \(x,t\in I\), so its values lie in \(I\). At the two ends of the parameter interval, \(H(x,0)=(1-0)x=x\) and \(H(x,1)=(1-1)x=0\). Therefore \(H\) is a homotopy from \(\operatorname{id}_I\) to \(f\circ g\). Together with \(g\circ f=\operatorname{id}_P\), this verifies both required conditions, so \(P\) and \(I\) are homotopy equivalent.
Check: at \(x=\tfrac{2}{3}\) and \(t=\tfrac{1}{4}\), the homotopy gives \(H(\tfrac{2}{3},\tfrac{1}{4})=(1-\tfrac{1}{4})\tfrac{2}{3}=\tfrac{3}{4}\cdot\tfrac{2}{3}=\tfrac{1}{2}\), which lies in \(I\). At \(t=0\) it returns the input \(x\), and at \(t=1\) it returns \(0\), as required.
This pair is not homeomorphic. The set \(P\) has one element, whereas \(I\) contains at least the two distinct elements \(0\) and \(1\). No bijection from \(I\) to \(P\) exists: any function \(I\to P\) takes both \(0\) and \(1\) to \(p\), so it is not injective. Since a homeomorphism must be a bijection, there is no homeomorphism between \(P\) and \(I\). In this example, homotopy equivalence holds even though there is no bijection, continuous or otherwise.
Proof. Suppose \(h:X\to Y\) is a homeomorphism, and let \(k:Y\to X\) be its continuous inverse. Set \(f=h\) and \(g=k\). By the inverse identities, \(g\circ f=\operatorname{id}_X\) and \(f\circ g=\operatorname{id}_Y\). A map \(u:X\to X\) is homotopic to itself: the map \(H:X\times I\to X\) defined by \(H(x,t)=u(x)\) is continuous, since it is the composite of the continuous projection onto \(X\) and \(u\), and its values at \(t=0\) and \(t=1\) both equal \(u(x)\). Apply this to each identity map. The two composites are therefore homotopic to the required identities, so \(X\) and \(Y\) are homotopy equivalent. \(\square\)
The implication goes in this direction because a homeomorphism supplies exact inverse maps, which automatically satisfy the weaker homotopy conditions. The point-and-interval example shows that the converse fails: homotopy-equivalent spaces need not be homeomorphic, and need not even admit a bijection.
Changing the Chosen Point in the Interval
Worked Example: Choosing the Other Endpoint
Keep \(P=\{p\}\) with its unique topology and \(I=[0,1]\) with its standard topology. This time define \(f:P\to I\) by \(f(p)=1\), and define \(g:I\to P\) by \(g(x)=p\). As before, both maps are continuous: every map from \(P\) is continuous, and the preimages under \(g\) of open subsets of \(P\) are \(\varnothing\) or \(I\).
For the composite on \(P\), \((g\circ f)(p)=g(1)=p\), so \(g\circ f=\operatorname{id}_P\). For the composite on \(I\), every \(x\in I\) satisfies \((f\circ g)(x)=f(p)=1\). Thus \(f\circ g\) is constant with value \(1\).
Define \(H:I\times I\to I\) by \(H(x,t)=(1-t)x+t\). This is continuous because it is built from continuous coordinate functions using sums and products. For \(x,t\in I\), it is a convex combination of \(x\) and \(1\): \(H(x,t)=(1-t)x+t\cdot1\), with nonnegative coefficients \(1-t\) and \(t\) whose sum is \(1\). Since both \(x\) and \(1\) lie in \(I\), \(H(x,t)\in I\). Its endpoint values are \(H(x,0)=x\) and \(H(x,1)=1\), so it joins \(\operatorname{id}_I\) to \(f\circ g\).
Check: with \(x=\tfrac{1}{4}\) and \(t=\tfrac{1}{2}\), the value is \(H(\tfrac{1}{4},\tfrac{1}{2})=(1-\tfrac{1}{2})\tfrac{1}{4}+\tfrac{1}{2}=\tfrac{1}{8}+\tfrac{4}{8}=\tfrac{5}{8}\), which is in \(I\). At \(t=0\) the formula gives \(x\), and at \(t=1\) it gives \(1\). These maps also make \(P\) and \(I\) homotopy equivalent.
The two examples use different maps \(f\): one selects \(0\), the other selects \(1\). In each case \(g\) is the map from the interval to the point, and \(f\circ g\) is the constant map on the interval at the selected value. The homotopy verifies that this composite can be joined to the identity. It does not say that \(f\) and \(g\) are inverse functions; they are not.
A Homeomorphism Gives a Stronger Comparison
Worked Example: Two Homeomorphic Intervals
Let \(I=[0,1]\) and \(J=[-1,1]\), both with their standard subspace topologies from \(\mathbb{R}\). Define \(h:I\to J\) by \(h(x)=2x-1\), and \(k:J\to I\) by \(k(y)=(y+1)/2\). For \(x\in I\), \(-1\leq 2x-1\leq1\), so \(h\) takes values in \(J\). For \(y\in J\), \(0\leq (y+1)/2\leq1\), so \(k\) takes values in \(I\). Both maps are continuous, since affine real-valued functions are continuous.
Compute both composites for arbitrary inputs: \((k\circ h)(x)=((2x-1)+1)/2=2x/2=x\), and \((h\circ k)(y)=2((y+1)/2)-1=y+1-1=y\). Thus \(k\) is the continuous inverse of \(h\), and \(h\) is a homeomorphism. In particular, these maps are also homotopy inverses: each composite is already the identity, so the constant-in-time homotopies \(H_I(x,t)=x\) and \(H_J(y,t)=y\) verify the homotopy conditions.
Check: \(h(0)=2\cdot0-1=-1\), and \(k(-1)=(-1+1)/2=0\). Also \(h(1)=2\cdot1-1=1\), and \(k(1)=(1+1)/2=1\). These endpoint computations agree with the composite formulas.
This example illustrates the difference in what must be checked. For a homeomorphism, the maps must be exact inverses and both must be continuous. For a homotopy equivalence, the maps must be continuous, but the composites only need to be homotopic to the identities. Exact equality of composites is sufficient, but it is not required.
A Common Composition Pitfall
Common pitfall. With \(f:P\to I\) and \(g:I\to P\) from the first example, it is incorrect to say that \(f\) collapses the interval to the point. Its domain is \(P\), so \(f\) sends the point \(p\) to \(0\); it is \(g\) that sends every point of \(I\) to \(p\). A related wrong calculation is to claim that \(f\circ g\) is the identity on \(I\). For the input \(x=1\), the actual calculation is \((f\circ g)(1)=f(g(1))=f(p)=0\), while \(\operatorname{id}_I(1)=1\). The correction is that \(f\circ g\) is constant at \(0\), and it is this constant map that is homotopic to the identity. The other composite has the different domain \(P\): \((g\circ f)(p)=p\).
Keeping the domains visible prevents both errors. Write \(f:X\to Y\) and \(g:Y\to X\) before composing; then check \(g\circ f:X\to X\) against \(\operatorname{id}_X\), and \(f\circ g:Y\to Y\) against \(\operatorname{id}_Y\). This simple check is essential because the two composites usually have different domains and codomains.
Homeomorphism records exact topological sameness, while homotopy equivalence permits a broader comparison through maps whose composites return to the identities up to homotopy. The point and interval make the difference sharp: they are homotopy equivalent, but have different cardinalities, so no bijection between them can exist. This is why homotopy equivalence should not be treated as a weakened form of homeomorphism that still requires a one-to-one correspondence; it does not require a bijection at all.
Check Your Understanding
For each question, keep the domain and codomain of every map explicit.
- For the maps \(f:P\to I\), \(f(p)=0\), and \(g:I\to P\), \(g(x)=p\), write the domains and codomains of \(g\circ f\) and \(f\circ g\).
- For the homotopy \(H(x,t)=(1-t)x\), verify the endpoint values and explain why \(H(x,t)\) stays in \(I\) for all \(x,t\in I\).
- Why does the existence of homotopy inverse maps between \(P\) and \(I\) not imply that a bijection between them exists?
- Prove that if \(f:X\to Y\) and \(g:Y\to X\) are continuous exact inverses, then they are homotopy inverses.
- In the example with \(h(x)=2x-1\), compute \(k(h(\tfrac{3}{4}))\) directly and compare it with \(\tfrac{3}{4}\).