Homotopic Maps as an Equivalence Relation

Homotopy Relates Maps With the Same Domain and Codomain

Homotopy is a way to say that two continuous maps can be deformed into one another through continuous maps. The definition involves a map on a product with the unit interval, so continuity is a condition on both the domain point and the time parameter. Once a homotopy is given, three simple operations produce homotopies that establish reflexivity, symmetry, and transitivity.

Let \(X\) and \(Y\) be topological spaces, with topologies understood, and let \(I=[0,1]\) have the subspace topology inherited from the standard topology on \(\mathbb{R}\). Give \(X\times I\) the product topology. We consider continuous maps from \(X\) to \(Y\).

The maps \(f\) and \(g\) must have the same domain and codomain for this relation to apply. We will use the product topology in checking continuity. In particular, if \(r:I\to I\) is continuous, then \((x,t)\mapsto(x,r(t))\) is continuous: the preimage of a basic open rectangle \(U\times V\) is \(U\times r^{-1}(V)\), which is open in \(X\times I\). The same reasoning applies when the time interval is restricted to a subinterval of \(I\).

The Three Homotopy Operations

Proof. Fix continuous maps \(f,g,h:X\to Y\) as needed for each part.

Reflexivity. Define \(C:X\times I\to Y\) by \(C(x,t)=f(x)\). This is \(f\circ\pi_1\), where \(\pi_1:X\times I\to X\) is the first projection; the projection is continuous in the product topology, so \(C\) is continuous. Also \(C(x,0)=f(x)\) and \(C(x,1)=f(x)\) for every \(x\). Thus \(f\simeq f\).

Symmetry. Suppose \(H:X\times I\to Y\) is a homotopy from \(f\) to \(g\). Define \(H^{\mathrm{rev}}(x,t)=H(x,1-t)\). The function \(r(t)=1-t\) is continuous from \(I\) to \(I\), and the product map \((x,t)\mapsto(x,1-t)\) is continuous by the product-topology observation above. Hence \(H^{\mathrm{rev}}\) is continuous. Its endpoints are \(H^{\mathrm{rev}}(x,0)=H(x,1)=g(x)\) and \(H^{\mathrm{rev}}(x,1)=H(x,0)=f(x)\). Therefore \(g\simeq f\).

Transitivity. Suppose \(H\) is a homotopy from \(f\) to \(g\), and \(K\) is a homotopy from \(g\) to \(h\). Define \[ L(x,t)= \begin{cases} H(x,2t),&0\le t\le \tfrac12,\\ K(x,2t-1),&\tfrac12\le t\le 1. \end{cases} \] Both formulas are defined on their stated intervals: \(2t\in I\) when \(0\le t\le\tfrac12\), and \(2t-1\in I\) when \(\tfrac12\le t\le1\). At the joining time, the first formula gives \(H(x,1)=g(x)\), and the second gives \(K(x,0)=g(x)\). Thus the formulas agree where both apply.

For completeness, the continuity argument uses the following closed-set pasting fact. If a space is the union of finitely many closed subsets and continuous functions on those subsets agree on their overlaps, the function they define on the whole space is continuous. Indeed, for any closed \(F\subseteq Y\), its preimage on each closed subset is closed relative to that subset, hence closed in the whole space; the preimage on the union is the finite union of these closed sets and is closed. Apply this fact to the closed subsets \(X\times[0,\tfrac12]\) and \(X\times[\tfrac12,1]\) of \(X\times I\). The restrictions defining \(L\) are continuous because they are compositions of \(H\) or \(K\) with the continuous maps \((x,t)\mapsto(x,2t)\) or \((x,t)\mapsto(x,2t-1)\), respectively. The two restrictions agree at the overlap, as checked above, so \(L\) is continuous. Finally, \(L(x,0)=H(x,0)=f(x)\) and \(L(x,1)=K(x,1)=h(x)\). Thus \(f\simeq h\), proving transitivity. These constructions also work if \(X\) is empty, since the required functions and endpoint conditions are then defined on an empty domain. \(\square\)

The time rescaling is essential: each given homotopy occupies half of the new time interval. The first half runs through all of \(H\), and the second half runs through all of \(K\). This is what makes their common endpoint \(g\) the value at the join.

Worked Examples of the Formulas

A Common Mistake and Why the Relation Matters

Common pitfall. In the second half of the concatenation, using \(K(x,2t)\) instead of \(K(x,2t-1)\) fails to start the second homotopy at its initial value. In the preceding example, this incorrect formula would give \(K(*,2t)=1+2t\) for \(\tfrac12\le t\le1\). At the join \(t=\tfrac12\), the first piece gives \(H(*,1)=1\), but the incorrect second piece gives \(K(*,1)=2\). The formulas disagree, so they do not define the intended continuous concatenation. The corrected rescaling gives \(K(*,2(\tfrac12)-1)=K(*,0)=1\), matching the first piece. The subtraction of \(1\) maps the second half of the time interval onto the full interval \(I\), starting at time \(0\).

These three constructions show why homotopy is an equivalence relation: a map is homotopic to itself, a homotopy can be run backward, and two homotopies with a shared endpoint can be run in succession. Consequently, continuous maps with fixed domain \(X\) and codomain \(Y\) can be grouped into equivalence classes under homotopy. This grouping records which maps can be connected through a continuous family, while keeping the domain and codomain fixed.