Path Components as Equivalence Classes

Path Reachability

A path is a continuous way of moving from one point of a space to another. The word “moving” suggests a notion of reachability: two points are related when some path begins at one and ends at the other. To use this relation systematically, we must check that it is reflexive, symmetric and transitive. These checks depend on three constructions: a constant path, a reversed path and a concatenated path.

Throughout, let \((X,\mathcal{T})\) be a topological space and let \(I=[0,1]\) have its usual subspace topology from \(\mathbb{R}\). A path in \(X\) from \(x\) to \(y\) is a continuous map \(\gamma:I\to X\) satisfying \(\gamma(0)=x\) and \(\gamma(1)=y\). We write \(x\sim y\) when such a path exists. The relation \(\sim\) is defined on the points of \(X\); it does not require any algebraic structure on \(X\).

The Three Path Constructions

For any \(x\in X\), define \(c_x:I\to X\) by \(c_x(t)=x\). This is the constant path at \(x\). It is continuous: for every open set \(U\in\mathcal{T}\), the preimage \(c_x^{-1}(U)\) is either \(I\) (if \(x\in U\)) or \(\varnothing\) (if \(x\notin U\)), and both are open in \(I\).

If \(\gamma\) is a path from \(x\) to \(y\), define its reverse by \(\overline{\gamma}(t)=\gamma(1-t)\). The map \(t\mapsto1-t\) is continuous from \(I\) to \(I\), so \(\overline{\gamma}\), a composition of continuous maps, is continuous. Its endpoints are \(\overline{\gamma}(0)=\gamma(1)=y\) and \(\overline{\gamma}(1)=\gamma(0)=x\). Thus it is a path from \(y\) to \(x\).

For transitivity we need to join two paths at their shared endpoint. If \(\gamma\) is a path from \(x\) to \(y\) and \(\eta\) is a path from \(y\) to \(z\), define

At \(t=\frac12\), the first formula gives \(\gamma(1)=y\), and the second gives \(\eta(0)=y\); hence the formulas agree. The factor \(2\) uses the first half of the parameter interval to trace all of \(\gamma\), and the expression \(2t-1\) uses the second half to trace all of \(\eta\). The endpoint values are \(\lambda(0)=\gamma(0)=x\) and \(\lambda(1)=\eta(1)=z\).

We also need continuity, including at the joining parameter. On \(0\le t\le\frac12\), the formula is a composition of continuous maps; on \(\frac12\le t\le1\), it is also a composition of continuous maps. At \(t=\frac12\), let \(V\) be any open neighborhood of \(y=\lambda(\frac12)\). Continuity of \(\gamma\) at \(1\) gives \(\delta_1>0\) such that \(\gamma(s)\in V\) for \(s\in I\) with \(|s-1|<\delta_1\). Continuity of \(\eta\) at \(0\) gives \(\delta_2>0\) such that \(\eta(s)\in V\) for \(s\in I\) with \(|s|<\delta_2\). Take \(\delta=\frac12\min(\delta_1,\delta_2)\). If \(t\in I\) and \(|t-\frac12|<\delta\), then on the first half \(|2t-1|<\delta_1\), and on the second half \(|2t-1|<\delta_2\). Therefore \(\lambda(t)\in V\) on either side of the join. This proves continuity at the join, so \(\lambda\) is a path from \(x\) to \(z\).

Path Reachability Is an Equivalence Relation

Proof. For reflexivity, given \(x\in X\), the constant path \(c_x(t)=x\) begins and ends at \(x\). Thus \(x\sim x\). For symmetry, if \(x\sim y\), there is a path \(\gamma\) from \(x\) to \(y\); the reversed path \(\overline{\gamma}\) constructed above goes from \(y\) to \(x\), so \(y\sim x\). For transitivity, suppose \(x\sim y\) and \(y\sim z\). Choose paths \(\gamma\) from \(x\) to \(y\) and \(\eta\) from \(y\) to \(z\). The concatenation \(\lambda\) above is continuous, begins at \(x\), and ends at \(z\). Hence \(x\sim z\). All three defining conditions hold. If \(X=\varnothing\), there are no points for which a condition must be checked, so the relation on \(X\) is an equivalence relation vacuously. \(\square\)

The equivalence classes group points according to which other points they can reach by paths. They cover \(X\), since each \(x\in X\) belongs to \([x]\) by reflexivity. They are either equal or disjoint: if two classes have a point in common, symmetry and transitivity imply that their defining points are related, and then each class contains exactly the points in the other class. Here is the argument in detail.

Proof. Since \(x\sim x\), we have \(x\in[x]\), so every point of \(X\) belongs to a class. Now suppose \([x]\cap[y]\ne\varnothing\), and choose \(w\) in this intersection. Then \(w\sim x\) and \(w\sim y\). By symmetry, \(x\sim w\); by transitivity with \(w\sim y\), we obtain \(x\sim y\). If \(v\in[x]\), then \(v\sim x\), and \(x\sim y\), so \(v\sim y\); hence \(v\in[y]\). Thus \([x]\subseteq[y]\). Reversing the roles of \(x\) and \(y\) gives \([y]\subseteq[x]\), so the classes are equal. If they have no point in common, they are disjoint by definition. \(\square\)

Worked Examples

A Common Concatenation Error

A concatenation must use the whole first path on the first half of the parameter interval and the whole second path on the second half. A tempting but incorrect choice is to use \(\gamma(t/2)\) first and \(\eta((t+1)/2)\) second. For \(\gamma(t)=t\), a path from \(0\) to \(1\), and \(\eta(t)=1+t\), a path from \(1\) to \(2\), this proposed rule gives the left-hand value at the join \(t=\frac12\) as \(\gamma(\frac14)=\frac14\), while its right-hand value is \(\eta(\frac34)=\frac74\). The values do not match, so the piecewise map is not continuous at the join.

The correct rule gives \(\gamma(2t)=2t\) on the first half and \(\eta(2t-1)=1+(2t-1)=2t\) on the second half. At the join both formulas give \(1\); at the endpoints they give \(0\) and \(2\). The resulting map is the continuous path \(t\mapsto2t\) from \(0\) to \(2\). Matching the values at the join is necessary, and the parameter rescaling is what makes the concatenation trace both paths in order.

Path equivalence is useful because it turns the question “which points can be joined by paths?” into a relation whose classes can be studied as sets. The examples show that the answer depends on the topology: the standard real line has one class, a finite discrete space has one class per point, and the two-interval subspace has one class per interval.