From Based Homotopy to Equivalence Classes
A based loop is not classified only by its parametrized formula: two different loops may be continuously deformed into one another while keeping their base point fixed. Endpoint-preserving homotopy makes this comparison precise. We will prove that it is an equivalence relation, so it divides the based loops into classes, each collecting exactly the loops with the same based homotopy type.
Let \((X,\mathcal{T})\) be a topological space, let \(x_0\in X\), and give \(I=[0,1]\) its standard subspace topology from \(\mathbb{R}\). The square \(I\times I\) has the product topology.
We use \(t\) for position along a loop and \(s\) for the deformation parameter. This distinction matters: changing \(s\) moves between loops, while changing \(t\) changes the parametrization along a loop. In particular, keeping the endpoints fixed means the equations \(H(0,s)=H(1,s)=x_0\) must hold for every value of \(s\), not just at the start and end of the homotopy.
Why Based Homotopy Is an Equivalence Relation
Proof. Let \(\gamma,\eta,\theta:I\to X\) be loops based at \(x_0\).
Reflexivity. Define \(H:I\times I\to X\) by \(H(t,s)=\gamma(t)\). This map is continuous because it is the composition of the continuous projection \((t,s)\mapsto t\) with the continuous map \(\gamma\). It satisfies \(H(t,0)=H(t,1)=\gamma(t)\), and \(H(0,s)=H(1,s)=x_0\). Thus \(\gamma\simeq_{x_0}\gamma\).
Symmetry. Suppose \(H\) is an endpoint-preserving homotopy from \(\gamma\) to \(\eta\). Define \(K(t,s)=H(t,1-s)\). The map \((t,s)\mapsto(t,1-s)\) is continuous, so \(K\) is continuous. Substitution gives \(K(t,0)=H(t,1)=\eta(t)\) and \(K(t,1)=H(t,0)=\gamma(t)\). Also, \(K(0,s)=K(1,s)=x_0\) because \(H\) fixes both endpoints for every parameter. Therefore \(\eta\simeq_{x_0}\gamma\).
Transitivity. Suppose \(H\) is an endpoint-preserving homotopy from \(\gamma\) to \(\eta\), and \(K\) is one from \(\eta\) to \(\theta\). Define \(L:I\times I\to X\) by $$ L(t,s)= \begin{cases} H(t,2s),&0\le s\le \tfrac12,\\ K(t,2s-1),&\tfrac12\le s\le 1. \end{cases} $$ At \(s=\tfrac12\), the first formula gives \(H(t,1)=\eta(t)\), and the second gives \(K(t,0)=\eta(t)\). Thus the formulas agree where their domains overlap.
For completeness, the pasting fact used here is that two continuous maps on closed subsets covering a space combine to give a continuous map if they agree on the overlap. Indeed, for any closed set \(F\subseteq X\), the preimage of \(F\) under the combined map is the union of its preimages in the two closed pieces. Each such preimage is closed in its piece and hence closed in the whole space; their finite union is closed. The closed-set test for continuity then proves the fact. Here the pieces \(I\times[0,\tfrac12]\) and \(I\times[\tfrac12,1]\) are closed and cover \(I\times I\), so \(L\) is continuous. Finally, \(L(t,0)=\gamma(t)\), \(L(t,1)=\theta(t)\), and each formula gives \(L(0,s)=L(1,s)=x_0\). Hence \(\gamma\simeq_{x_0}\theta\). All three properties hold, proving the theorem. \(\square\)
What an Equivalence Class Records
Proof. Reflexivity gives \(\gamma\simeq_{x_0}\gamma\), so \(\gamma\in[\gamma]\). If \(\gamma\simeq_{x_0}\eta\), symmetry and transitivity imply that a based loop is homotopic to \(\gamma\) exactly when it is homotopic to \(\eta\); hence \([\gamma]=[\eta]\). Conversely, if \([\gamma]=[\eta]\), then \(\gamma\in[\gamma]=[\eta]\), so \(\gamma\simeq_{x_0}\eta\). Now suppose two classes have a common member \(\lambda\). Then \(\lambda\simeq_{x_0}\gamma\) and \(\lambda\simeq_{x_0}\eta\) for representatives \(\gamma,\eta\) of those classes. Symmetry and transitivity give \(\gamma\simeq_{x_0}\eta\), so the classes are equal. Thus distinct classes are disjoint, and the classes partition the set of all loops based at \(x_0\). \(\square\)
The class depends on the space \(X\), its topology, and the chosen base point. It is not simply the set of loops with the same image: the defining condition is the existence of a continuous deformation that fixes the base point at both ends throughout.
Worked Example: Two Loops in the Real Line
Give \(\mathbb{R}\) its standard topology and choose \(x_0=0\). Consider the loops $$ \gamma(t)=4t(1-t),\qquad \eta(t)=2t(1-t). $$
At \(t=0\), both formulas give \(0\), and at \(t=1\), both give \(0\). They are continuous polynomial maps into \(\mathbb{R}\), so both are loops based at \(0\). Define $$ H(t,s)=(1-s)\gamma(t)+s\eta(t)=(4-2s)t(1-t). $$
This is continuous as a polynomial in \(t\) and \(s\). Substituting \(s=0\) gives \(H(t,0)=4t(1-t)=\gamma(t)\), and substituting \(s=1\) gives \(H(t,1)=2t(1-t)=\eta(t)\). At \(t=0\) and \(t=1\), the factor \(t(1-t)\) is zero, so \(H(0,s)=H(1,s)=0\) for all \(s\in I\). Thus \(\gamma\simeq_0\eta\), and the two loops belong to the same class.
Check: at \(t=\tfrac12\), \(\gamma(\tfrac12)=4(\tfrac12)(\tfrac12)=1\), while \(\eta(\tfrac12)=2(\tfrac12)(\tfrac12)=\tfrac12\). The homotopy gives \(H(\tfrac12,s)=(4-2s)(\tfrac14)=1-\tfrac{s}{2}\), which starts at \(1\) and ends at \(\tfrac12\). Its endpoints remain \(0\), as required.
Worked Example: All Based Loops in the Unit Interval
Give \(X=I=[0,1]\) the subspace topology from \(\mathbb{R}\), and take \(x_0=0\). The loops \(\gamma(t)=t(1-t)\) and \(\eta(t)=2t(1-t)\) are continuous and have value \(0\) at \(t=0\) and \(t=1\). For \(t\in I\), \(0\le t(1-t)\le\tfrac14\), so \(0\le\gamma(t)\le\tfrac14\) and \(0\le\eta(t)\le\tfrac12\); both maps take values in \(I\).
More generally, let \(\lambda:I\to I\) be any loop based at \(0\). Define $$ H(t,s)=(1-s)\lambda(t). $$ This map is continuous. Since \(0\le\lambda(t)\le1\) and \(0\le1-s\le1\), it satisfies \(0\le H(t,s)\le1\), so its values stay in \(I\). It starts at \(H(t,0)=\lambda(t)\), ends at \(H(t,1)=0\), and has \(H(0,s)=H(1,s)=0\). Thus every based loop in \(I\) is endpoint-preserving homotopic to the constant loop at \(0\). By transitivity, any two such loops are homotopic, so there is exactly one based-loop class.
Check: for the particular loop \(\eta(t)=2t(1-t)\), the homotopy gives \(H(\tfrac12,s)=(1-s)2(\tfrac12)(\tfrac12)=\tfrac{1-s}{2}\), which lies between \(0\) and \(\tfrac12\) for \(s\in I\). At \(s=0\) it is \(\tfrac12\), and at \(s=1\) it is \(0\).
Worked Example: Based Loops in a Discrete Space
Let \(X=\{a,b\}\) have the discrete topology, in which every subset is open, and choose \(x_0=a\). A based loop \(\gamma:I\to X\) must be constant. To verify this, define \(f:X\to\mathbb{R}\) by \(f(a)=0\) and \(f(b)=1\). This map is continuous because the preimage of every open set in \(\mathbb{R}\) is a subset of the discrete space \(X\), and every such subset is open. If a continuous loop \(\gamma\) took the value \(b\) at some \(t\), then the continuous map \(f\circ\gamma:I\to\mathbb{R}\) would take both the value \(0\) at \(t=0\) and the value \(1\) at that \(t\). The intermediate value theorem would force it to take the value \(\tfrac12\), which is impossible because \(f\circ\gamma\) only takes values \(0\) and \(1\). Therefore \(\gamma(t)=a\) for every \(t\).
There is just one based loop, the constant loop at \(a\). It is homotopic to itself by the constant homotopy, so there is exactly one class, consisting of that loop. This example differs from the interval example: in the interval there are many loops but only one class, whereas here there is only one loop to begin with.
Check: the constant map \(\gamma(t)=a\) satisfies \(\gamma(0)=\gamma(1)=a\), and \(H(t,s)=a\) is continuous into the discrete space and satisfies all three endpoint-preserving homotopy equations.
A Common Mistake: Reversing the Wrong Parameter
To prove symmetry, reverse the homotopy parameter \(s\), not the loop parameter \(t\). For a concrete distinction, let \(X=\mathbb{R}\) with its standard topology, base point \(0\), and let \(\gamma(t)=t(1-t)(2t-1)\). This is a based loop because its values at \(t=0\) and \(t=1\) are both \(0\). Reversing the loop parameter gives \(\gamma(1-t)=-\gamma(t)\), which is a different loop: \(\gamma(\tfrac14)=\tfrac14\cdot\tfrac34\cdot(-\tfrac12)=-\tfrac{3}{32}\), while \(\gamma(1-\tfrac14)=\gamma(\tfrac34)=\tfrac{3}{32}\).
A mistaken symmetry argument might set \(K(t,s)=\gamma(1-t)\) and claim that this reverses a homotopy from \(\gamma\) to itself. But \(K(t,0)=\gamma(1-t)\), which is not in general \(\gamma(t)\); this formula changes the loop parameter and does not establish symmetry. The correct construction, for a homotopy \(H\) from \(\gamma\) to \(\eta\), is \(K(t,s)=H(t,1-s)\). It starts at \(\eta\), ends at \(\gamma\), and preserves both endpoints for every \(s\).
Equivalence classes therefore provide a precise way to say when based loops represent the same deformation type. Equality of classes means there is an endpoint-preserving homotopy between their representatives; it does not mean the two parametrized maps are identical.
Check Your Understanding
For each question, keep the base point fixed throughout any homotopy you propose.
- Write down the three equations that a homotopy from a loop \(\gamma\) to a loop \(\eta\) based at \(x_0\) must satisfy.
- Given a homotopy \(H\) from \(\gamma\) to \(\eta\), what formula gives a homotopy from \(\eta\) to \(\gamma\)?
- Why do two endpoint-preserving homotopy classes that have a common member have to be equal?
- In the discrete space \(\{a,b\}\) based at \(a\), can a continuous based loop take the value \(b\)? Explain using the intermediate value theorem.
- In \([0,1]\) based at \(0\), give a homotopy from an arbitrary based loop \(\lambda\) to the constant loop at \(0\), and check that its values stay in \([0,1]\).