Transporting Loops Between Base Points
The fundamental group \(\pi_1(X,x_0)\) records homotopy classes of loops that start and end at \(x_0\). If \(x_0\) and \(x_1\) are different points, loops based at \(x_1\) cannot be placed directly into this group: they have the wrong base point. A path from \(x_0\) to \(x_1\) gives a way to modify each such loop. Travel from \(x_0\) to \(x_1\), traverse the loop, and return to \(x_0\) along the reverse path. This construction produces a loop at \(x_0\).
Let \(X\) be a topological space. A path from \(x\) to \(y\) is a continuous map \(\gamma:I\to X\), where \(I=[0,1]\), with \(\gamma(0)=x\) and \(\gamma(1)=y\). A loop based at \(x\) is a path \(\alpha\) with \(\alpha(0)=\alpha(1)=x\). Two loops based at the same point are based-homotopic if there is a continuous homotopy between them that keeps both endpoints fixed at that point throughout.
The two formulae for a concatenation agree at the joining time: both give the common endpoint \(\alpha(1)=\beta(0)\). The pieces are continuous on closed subintervals and agree at their shared endpoint, so the finite pasting lemma makes the concatenation continuous. The same construction, with thirds of the parameter interval, defines a concatenation of three paths. Changing the subdivision of a path into finitely many pieces does not change its based-homotopy class; equivalently, the associativity of concatenation holds for classes, as established in the earlier material on concatenation.
The Change-of-Base-Point Map
Here \([\alpha]\) denotes the based-homotopy class of \(\alpha\). The endpoint check explains the order in this definition: \(\gamma\) ends at \(x_1\), where \(\alpha\) starts and ends, and \(\overline{\gamma}\) then returns to \(x_0\). Thus the concatenation starts and ends at \(x_0\). For the expression to define a function on classes, it must also give the same output class whenever \(\alpha\) is replaced by a based-homotopic loop. That is the key well-definedness check.
Proof. Suppose \(\alpha\) and \(\beta\) are loops based at \(x_1\) representing the same class. By the definition of based-homotopy, there is a continuous map \(H:I\times I\to X\) such that \(H(0,t)=\alpha(t)\), \(H(1,t)=\beta(t)\), and \(H(s,0)=H(s,1)=x_1\) for every \(s,t\in I\). Define $$ K(s,t)= \begin{cases} \gamma(3t),&0\le t\le \frac13,\\ H(s,3t-1),&\frac13\le t\le \frac23,\\ \gamma(3-3t),&\frac23\le t\le 1. \end{cases} $$ At \(t=\frac13\), the first and second pieces both equal \(x_1\). At \(t=\frac23\), the second and third pieces both equal \(x_1\). Each piece is continuous and the pieces agree on their overlaps, so the finite pasting lemma shows that \(K\) is continuous. For every \(s\), \(K(s,0)=\gamma(0)=x_0\) and \(K(s,1)=\gamma(0)=x_0\). Also, \(K(0,\cdot)\) traverses \(\gamma,\alpha,\overline{\gamma}\) and \(K(1,\cdot)\) traverses \(\gamma,\beta,\overline{\gamma}\), each with the indicated thirds parametrization. Thus \(K\) is a homotopy that fixes the loop endpoints at \(x_0\), and the two output loops represent the same class. The assignment is therefore independent of the representative \(\alpha\), as required. \(\square\)
The theorem establishes a function on loop classes; it does not, by itself, establish further properties of that function. In particular, constructing a loop at \(x_0\) is not the same as proving that the resulting function has an inverse. The examples below focus on the construction and on checking its representatives directly.
Worked Example: A Loop in the Real Line
Use the standard topology on \(X=\mathbb{R}\). Let \(x_0=0\), \(x_1=2\), and let \(\gamma(t)=2t\), a path from \(0\) to \(2\). Take the loop at \(2\) given by \(\alpha(t)=2+\sin(2\pi t)\). Its endpoints are both \(2\), since \(\sin(0)=\sin(2\pi)=0\).
Compute \(\gamma*\alpha*\overline{\gamma}\) using three equal parameter intervals. On the first third, the path is \(\gamma(3t)=6t\). On the middle third, it is \(\alpha(3t-1)=2+\sin(2\pi(3t-1))=2+\sin(6\pi t-2\pi)=2+\sin(6\pi t)\). On the final third, the reverse path is \(\overline{\gamma}(3t-2)=\gamma(1-(3t-2))=6-6t\). Hence a representative of the output class is $$ \beta(t)= \begin{cases} 6t,&0\le t\le \frac13,\\ 2+\sin(6\pi t),&\frac13\le t\le \frac23,\\ 6-6t,&\frac23\le t\le 1. \end{cases} $$
Check: At \(t=\frac13\), the first piece is \(6(\frac13)=2\), and the middle piece is \(2+\sin(2\pi)=2\). At \(t=\frac23\), the middle piece is \(2+\sin(4\pi)=2\), and the last piece is \(6-6(\frac23)=2\). The initial point is \(\beta(0)=0\) and the final point is \(\beta(1)=0\), so the computed path is indeed a loop at \(x_0\). For a further value, \(t=\frac{5}{12}\) lies in the middle third and gives \(2+\sin(6\pi\cdot\frac{5}{12})=2+\sin(\frac{5\pi}{2})=3\).
Worked Example: Transporting a Loop on the Circle
Give \(S^1=\{z\in\mathbb{C}:|z|=1\}\) the subspace topology from \(\mathbb{C}\). Let \(x_0=1\), \(x_1=-1\), and take \(\gamma(t)=e^{i\pi t}\). This path starts at \(e^0=1\) and ends at \(e^{i\pi}=-1\). Use the loop \(\alpha(t)=e^{i(\pi+2\pi t)}\) at \(-1\): its initial value is \(e^{i\pi}=-1\) and its final value is \(e^{i3\pi}=-1\).
On the first third of the concatenation, the formula is \(e^{i3\pi t}\). On the middle third, substituting \(3t-1\) into \(\alpha\) gives \(e^{i(\pi+2\pi(3t-1))}=e^{i(6\pi t-\pi)}\). On the last third, continuing the angle from \(3\pi\) along the reverse of the first half-circle gives \(e^{i(5\pi-3\pi t)}\). This is the reverse path on that segment because \(5\pi-3\pi t\) and \(3\pi-3\pi t\) differ by \(2\pi\), so their complex exponentials agree. Therefore the output loop based at \(1\) is represented by $$ \beta(t)= \begin{cases} e^{i3\pi t},&0\le t\le \frac13,\\ e^{i(6\pi t-\pi)},&\frac13\le t\le \frac23,\\ e^{i(5\pi-3\pi t)},&\frac23\le t\le 1. \end{cases} $$
Check: At \(t=\frac13\), the first expression is \(e^{i\pi}=-1\), and the second is \(e^{i\pi}=-1\). At \(t=\frac23\), the second is \(e^{i3\pi}=-1\), and the third is \(e^{i3\pi}=-1\). At \(t=0\) the value is \(1\), and at \(t=1\) it is \(e^{i2\pi}=1\). For \(t=\frac{5}{12}\), the middle expression gives \(e^{i(6\pi\cdot\frac{5}{12}-\pi)}=e^{i3\pi/2}=-i\). These endpoint and sample-value checks agree with the three pieces of the construction.
Worked Example: A Constant Loop and an Explicit Contraction
Give \(X=\mathbb{R}^2\) its Euclidean topology. Let \(x_0=(0,0)\), \(x_1=(1,0)\), and \(\gamma(t)=(t,0)\). Let \(c\) be the constant loop at \(x_1\), so \(c(t)=(1,0)\) for every \(t\). The constructed loop \(\beta=\gamma*c*\overline{\gamma}\), parametrized by thirds, is $$ \beta(t)= \begin{cases} (3t,0),&0\le t\le \frac13,\\ (1,0),&\frac13\le t\le \frac23,\\ (3-3t,0),&\frac23\le t\le 1. \end{cases} $$
Its endpoints are \(\beta(0)=(0,0)\) and \(\beta(1)=(0,0)\). To check its class directly, define \(H(s,t)=(1-s)\beta(t)\) for \(s,t\in I\), with scalar multiplication in \(\mathbb{R}^2\). This map is continuous, \(H(0,t)=\beta(t)\), and \(H(1,t)=(0,0)\) for every \(t\). Since \(\beta(0)=\beta(1)=(0,0)\), we also have \(H(s,0)=H(s,1)=(0,0)\) for every \(s\). Thus this particular transported loop is based-homotopic to the constant loop at \(x_0\). The calculation illustrates the construction and its output without assuming a general statement about all loops in \(\mathbb{R}^2\).
A Common Ordering Error
Common pitfall. Concatenation is order-sensitive: \(\alpha*\beta\) means travel along \(\alpha\) first, then along \(\beta\). A tempting but incorrect expression is \(\overline{\gamma}*\alpha*\gamma\). For a concrete endpoint check, take \(X=\mathbb{R}\), \(x_0=0\), \(x_1=2\), and \(\gamma(t)=2t\). The reverse path \(\overline{\gamma}\) starts at \(2\), while \(\alpha\), a loop at \(2\), also starts at \(2\). After traversing \(\overline{\gamma}\), the path ends at \(0\), so it cannot be followed by \(\alpha\), which starts at \(2\). That concatenation is not defined. Likewise, \(\gamma*\alpha\) starts at \(0\) but ends at \(2\), so it is not a loop at \(0\). The correct order is \(\gamma*\alpha*\overline{\gamma}\): its successive endpoints are \(0\to2\), \(2\to2\), and \(2\to0\).
The base-point path is part of the data in the construction. Reversing its direction changes which fundamental groups are the source and target. The well-definedness proof addresses a different issue: once \(\gamma\) is fixed, changing the representative loop within its based-homotopy class does not change the output class.
Check Your Understanding
For each question, keep track of the initial and terminal points as well as the homotopy class.
- If \(\gamma\) runs from \(x_0\) to \(x_1\) and \(\alpha\) is a loop at \(x_1\), state the starting and ending point of each of \(\gamma\), \(\alpha\), and \(\overline{\gamma}\), then verify the endpoints of their concatenation in the prescribed order.
- In the standard topology on \(\mathbb{R}\), take \(\gamma(t)=3t\) from \(0\) to \(3\), and take the constant loop at \(3\). Write the three pieces of the transported loop using thirds of the parameter interval.
- In the circle example, evaluate the middle formula for the transported loop at \(t=\frac12\), and check that the result lies in \(S^1\).
- Explain why a based homotopy between two loops at \(x_1\) gives a based homotopy between their transported loops at \(x_0\). Which endpoint conditions on the homotopy are essential?
- For a path \(\gamma\) from \(x_0\) to \(x_1\), explain why \(\overline{\gamma}*\alpha*\gamma\) cannot in general be used to form a loop at \(x_0\) when \(\alpha\) is based at \(x_1\).