Homotopy Equivalence Is an Equivalence Relation

Homotopy Equivalence as a Relation

A homotopy equivalence does not require two spaces to have inverse maps whose composites are exactly identity maps. Instead, those composites need only be homotopic to the identities. This distinction matters when proving that homotopy equivalence is transitive: the inverse maps must be composed in the right order, and the homotopies must then be joined with matching endpoints.

Throughout, all products with \(I=[0,1]\) have the product topology. A homotopy between continuous maps \(a,b:X\to Y\) is a continuous map \(H:X\times I\to Y\) such that \(H(x,0)=a(x)\) and \(H(x,1)=b(x)\) for every \(x\in X\). We write \(a\simeq b\). The identity map on \(X\) is \(\operatorname{id}_X\).

Each direction in the definition is required: one composite is a map from \(X\) to \(X\), and the other is a map from \(Y\) to \(Y\). The maps \(f\) and \(g\) need not be inverse maps in the strict sense that both composites equal identity maps.

Two Facts About Homotopies

We will use two elementary constructions. First, a homotopy can be composed with continuous maps on either side. More precisely, if \(H:a\simeq b\), \(\alpha:X'\to X\) is continuous, and \(\beta:Y\to Y'\) is continuous, then \(\beta\circ a\circ\alpha\simeq\beta\circ b\circ\alpha\). A homotopy is given by \((x',t)\mapsto\beta(H(\alpha(x'),t))\). This map is continuous because it is a composition of continuous maps, using the product topology on \(X'\times I\).

Second, homotopies can be concatenated when the ending map of the first is the starting map of the second. If \(H:a\simeq b\) and \(K:b\simeq c\), define

At \(t=\frac12\), the first formula gives \(H(x,1)=b(x)\), and the second gives \(K(x,0)=b(x)\), so the formulas agree. Each formula is continuous on its closed part of \(X\times I\). The finite closed pasting lemma says that functions continuous on finitely many closed subsets covering a space combine to a continuous function if they agree on overlaps. Thus \(J\) is continuous, with \(J(x,0)=a(x)\) and \(J(x,1)=c(x)\); hence \(a\simeq c\). This gives the precise way we will join the homotopies below.

The Equivalence-Relation Theorem

Proof. Reflexivity. For any space \(X\), choose \(f=g=\operatorname{id}_X\). Both composites are \(\operatorname{id}_X\). The map \(H:X\times I\to X\) defined by \(H(x,t)=x\) is continuous, since it is the projection onto \(X\), and satisfies \(H(x,0)=H(x,1)=x\). Thus both required homotopies hold. If \(X=\varnothing\), the same map is the unique map from \(\varnothing\times I\), so the argument also covers the empty space.

Symmetry. Suppose \(X\) is homotopy equivalent to \(Y\), witnessed by \(f:X\to Y\) and \(g:Y\to X\), with \(g\circ f\simeq\operatorname{id}_X\) and \(f\circ g\simeq\operatorname{id}_Y\). To witness that \(Y\) is homotopy equivalent to \(X\), use \(g\) as the map from \(Y\) to \(X\) and \(f\) as the map from \(X\) to \(Y\). The two required composites are \(f\circ g\) on \(Y\) and \(g\circ f\) on \(X\), and the given homotopies verify them. Therefore the relation is symmetric.

Transitivity. Suppose \(X\) is homotopy equivalent to \(Y\), with maps \(f:X\to Y\) and \(g:Y\to X\), and \(Y\) is homotopy equivalent to \(Z\), with maps \(u:Y\to Z\) and \(v:Z\to Y\). Choose homotopies \(H:g\circ f\simeq\operatorname{id}_X\), \(K:f\circ g\simeq\operatorname{id}_Y\), \(L:v\circ u\simeq\operatorname{id}_Y\), and \(M:u\circ v\simeq\operatorname{id}_Z\). We claim that \(u\circ f:X\to Z\) and \(g\circ v:Z\to X\) are homotopy inverses.

For the composite on \(X\), first use \(L\), with \(f\) before it and \(g\) after it. The resulting homotopy is \(P(x,t)=g(L(f(x),t))\), from \(g\circ v\circ u\circ f\) to \(g\circ f\). Follow it by \(H\), which goes from \(g\circ f\) to \(\operatorname{id}_X\). The endpoints match, so concatenation gives \((g\circ v)\circ(u\circ f)\simeq\operatorname{id}_X\).

For the composite on \(Z\), use \(K\) with \(v\) before it and \(u\) after it. This gives \(Q(z,t)=u(K(v(z),t))\), from \(u\circ f\circ g\circ v\) to \(u\circ v\). Follow it by \(M\), from \(u\circ v\) to \(\operatorname{id}_Z\). Concatenating gives \((u\circ f)\circ(g\circ v)\simeq\operatorname{id}_Z\). Both composites therefore satisfy the definition of homotopy equivalence, proving transitivity. \(\square\)

Worked Constructions

Composition Order and a Common Pitfall

The transitivity proof depends on types as well as formulas. Starting with \(x\in X\), the map \(f\) sends \(x\) to \(Y\), then \(u\) sends it to \(Z\). Thus the map from \(X\) to \(Z\) is \(u\circ f\), not \(f\circ u\). In the reverse direction, \(v\) first sends \(z\in Z\) to \(Y\), and \(g\) then sends that point to \(X\), so the map is \(g\circ v\).

Common pitfall. One might write \(v\circ g\) as the reverse composite. But \(g\) has output in \(X\), while \(v\) accepts inputs in \(Z\); the composition \(v\circ g\) is not defined for the interval maps in the preceding example, because \(X=[0,1]\) and \(Z=[0,3]\) are different domains for the next map. The valid composite is \(g\circ v\). It sends \(z\in[0,3]\) to \(v(z)=2z/3\in[0,2]\), then to \(g(2z/3)=z/3\in[0,1]\). A correct transitivity argument must verify both that each composite is defined and that it has the required domain and codomain.

There is a second, related error: replacing “homotopic to the identity” with “equal to the identity.” The constant-map example shows why this is not permitted. At \(x=3/5\), the composite has value \(0\), not \(3/5\), yet the displayed homotopy verifies the definition. In transitivity, the composite homotopies need not become equalities; they are joined because the ending map of one homotopy is exactly the starting map of the next.

The three properties proved here let us use homotopy equivalence consistently as a relation: every space belongs to its own class, an equivalence can be read in either direction, and equivalences can be chained. The transitive construction also gives a practical method: compose the forward maps in forward order, compose the inverse maps in reverse order, and check the two composite homotopies separately.