Loops, Maps, and Basepoints
A fundamental group records loops based at one specified point. That specification matters: a loop is a map \(\alpha:I\to X\) with \(\alpha(0)=\alpha(1)=x_0\), and the group \(\pi_1(X,x_0)\) consists of homotopy classes of such loops, where homotopies keep the endpoints at \(x_0\). We use the convention that \(\alpha*\beta\) traverses \(\alpha\) first and then \(\beta\). Thus the product in \(\pi_1(X,x_0)\) is \([\alpha][\beta]=[\alpha*\beta]\).
If \(f:X\to Y\) is continuous and \(f(x_0)=y_0\), composing a loop in \(X\) with \(f\) gives a loop in \(Y\) at \(y_0\). The resulting group homomorphism is the basic way maps enter fundamental-group calculations.
Proof. If \(\alpha\) and \(\beta\) represent the same element of \(\pi_1(X,x_0)\), there is a homotopy \(H:I\times I\to X\) from \(\alpha\) to \(\beta\) that fixes both endpoints at \(x_0\). The map \(f\circ H\) is a homotopy from \(f\circ\alpha\) to \(f\circ\beta\), and it fixes the endpoints at \(y_0\). Therefore \(f_*([\alpha])=f_*([\beta])\), so the definition does not depend on the representative. Also \(f\circ(\alpha*\beta)=(f\circ\alpha)*(f\circ\beta)\), with the same first-then-second convention. Hence \(f_*([\alpha][\beta])=f_*([\alpha])f_*([\beta])\). For composition, \((g\circ f)\circ\alpha=g\circ(f\circ\alpha)\), so the two induced maps agree on every loop class. Composing with \(\operatorname{id}_X\) leaves each loop unchanged, giving the identity homomorphism. \(\square\)
For an inclusion, the formula has a direct interpretation: \(i_*([\alpha])\) is the class of the very same path, now regarded as a loop in the larger space. Its class may change because the larger space permits homotopies that were unavailable in the subspace.
Worked Example: The Boundary Circle Included in the Disk
Give \(D^2=\{z\in\mathbb{C}:|z|\leq1\}\) and \(S^1=\{z\in\mathbb{C}:|z|=1\}\) their subspace topologies from \(\mathbb{C}\). Let \(i:S^1\to D^2\) be inclusion and use the basepoint \(1\). Consider the loop \(\alpha(t)=e^{2\pi i t}\), \(0\leq t\leq1\). Its image under inclusion is \(i\circ\alpha(t)=e^{2\pi i t}\), now viewed as a loop in the disk.
In the disk this loop is homotopic, with its basepoint fixed, to the constant loop \(c_1(t)=1\). An explicit homotopy is \(H(s,t)=(1-s)e^{2\pi i t}+s\). For each \(s,t\), the triangle inequality gives \(|H(s,t)|\leq(1-s)|e^{2\pi i t}|+s|1|=(1-s)+s=1\), so \(H(s,t)\in D^2\). Also \(H(0,t)=\alpha(t)\), \(H(1,t)=1\), and \(H(s,0)=H(s,1)=1\). Thus \(i_*([\alpha])=[c_1]\), the identity element of \(\pi_1(D^2,1)\). The calculation concerns this loop class; the inclusion formula itself does not require that every loop in the circle be represented by this particular loop.
Check: At \(s=0\), the formula gives \(H(0,t)=e^{2\pi i t}\); at \(s=1\), it gives \(H(1,t)=1\). The endpoint values remain \(1\) at every \(s\), and the displayed modulus bound verifies that the whole homotopy stays in the disk.
Changing the Basepoint Along a Path
Let \(\gamma:I\to X\) be a path from \(x_0\) to \(x_1\), so \(\gamma(0)=x_0\) and \(\gamma(1)=x_1\). Write \(\bar\gamma(t)=\gamma(1-t)\) for the reverse path. A loop at \(x_0\) can be turned into one at \(x_1\): travel from \(x_1\) back to \(x_0\) along \(\bar\gamma\), traverse the original loop, and return to \(x_1\) along \(\gamma\).
Proof. The concatenated path starts at \(x_1\), travels to \(x_0\), follows the loop \(\alpha\), and ends at \(x_1\), so it is a loop at the required basepoint. A homotopy of \(\alpha\) that fixes its endpoints can be placed between the fixed paths \(\bar\gamma\) and \(\gamma\); this gives a homotopy of the concatenated loops fixing \(x_1\). Hence \(T_\gamma\) is well-defined. Concatenation of loops and cancellation of a path followed by its reverse give \(T_\gamma([\alpha][\beta])=[\bar\gamma*\alpha*\beta*\gamma]=[\bar\gamma*\alpha*\gamma][\bar\gamma*\beta*\gamma]\). The equality is an equality of homotopy classes: the middle \(\gamma*\bar\gamma\) on the right contracts, with endpoints fixed, to the constant path at \(x_0\). The same path-cancellation homotopy gives \(T_{\bar\gamma}(T_\gamma([\alpha]))=[\alpha]\) and \(T_\gamma(T_{\bar\gamma}([\delta]))=[\delta]\) for loops \(\alpha\) at \(x_0\) and \(\delta\) at \(x_1\). Thus \(T_\gamma\) is a homomorphism with inverse \(T_{\bar\gamma}\). \(\square\)
The order of these paths is forced by their endpoints. With our convention, \(\bar\gamma*\alpha*\gamma\) starts and ends at \(x_1\). Writing \(\gamma*\alpha*\bar\gamma\) instead generally does not even give a path with matching endpoints, so it cannot be the basepoint-change formula for this direction of travel.
Worked Example: Moving a Loop on the Circle
Give \(S^1\) its usual subspace topology, take \(x_0=1\), \(x_1=-1\), and let \(\gamma(t)=e^{\pi i t}\), the upper semicircle from \(1\) to \(-1\). Use the standard winding-number description of the fundamental group of the circle. Suppose \(\alpha\) is a loop at \(1\) with winding number \(n\in\mathbb{Z}\). The transported loop is \(\bar\gamma*\alpha*\gamma\), based at \(-1\).
Track a continuous choice of angle along its three pieces. Start with angle \(\pi\) at \(-1\). Along \(\bar\gamma\), the angle goes from \(\pi\) to \(0\), a change of \(-\pi\). Along \(\alpha\), it changes by \(2\pi n\), ending at angle \(2\pi n\). Along \(\gamma\), it changes by \(+\pi\). The total change is \(-\pi+2\pi n+\pi=2\pi n\). Dividing by \(2\pi\), the transported loop has winding number \(n\). In this example the basepoint changes, but the integer label is unchanged. The path formula remains important: it specifies the loop at the new basepoint even when this particular group’s winding-number calculation makes the result look unchanged.
Check: If \(n=2\), the three angle changes are \(-\pi\), \(4\pi\), and \(+\pi\); their sum is \(4\pi\), giving winding number \(4\pi/(2\pi)=2\), as required.
Paths, Inclusions, and Covering Spaces
Suppose \(A\subseteq X\), \(a_0,a_1\in A\), and \(\gamma\) is a path in \(A\) from \(a_0\) to \(a_1\). The same path can be viewed in \(X\). Changing basepoint before including a loop, or including it before changing basepoint, must give the same class. This is the compatibility that lets computations made in different pieces of a cover be compared.
Proof. Represent \([\alpha]\) by a loop \(\alpha\) at \(a_0\). Applying basepoint change in \(A\) and then inclusion gives the loop class represented in \(X\) by \(i\circ(\bar\gamma*\alpha*\gamma)\). Since \(i\) is inclusion, this is the path \(\bar\gamma*\alpha*\gamma\) viewed in \(X\). Including first gives the same loop \(\alpha\) in \(X\) and then applies the same concatenation with \(\bar\gamma\) and \(\gamma\). The resulting representative is identical, so the classes agree. \(\square\)
In a cover \(X=U\cup V\), the cleanest setup is to choose \(p\in U\cap V\) and use \(p\) as the basepoint in \(U\), \(V\), and \(U\cap V\). The inclusions then induce maps with matching basepoints. If loops in \(U\) or \(V\) have instead been described at other points, choose paths from those points to \(p\) within the relevant set and transport the loop classes before comparing them. For a loop in \(U\cap V\), the chosen path should lie in \(U\cap V\); then the proposition guarantees that its transport agrees whether the loop is regarded in the intersection, in \(U\), or in \(V\).
Worked Example: A Common Basepoint for Two Convex Pieces
Give \(D^2\) its usual topology and define \(U=\{(x,y)\in D^2:x<1/2\}\) and \(V=\{(x,y)\in D^2:x>-1/2\}\). These are open in the subspace topology on \(D^2\), and they cover \(D^2\): if a point were in neither, it would satisfy \(x\geq1/2\) and \(x\leq-1/2\), which is impossible. Their intersection is the strip \(\{(x,y)\in D^2:-1/2<x<1/2\}\).
Choose \(p=(0,0)\) in the intersection, \(u=(-1/4,0)\in U\), and \(v=(1/4,0)\in V\). The straight paths \(\gamma_U(t)=(-(1-t)/4,0)\) from \(u\) to \(p\), and \(\gamma_V(t)=((1-t)/4,0)\) from \(v\) to \(p\), stay in \(U\) and \(V\), respectively. For example, the first path has \(x\)-coordinate between \(-1/4\) and \(0\), hence always less than \(1/2\); the second has \(x\)-coordinate between \(0\) and \(1/4\), hence always greater than \(-1/2\). Both paths also lie in the intersection.
Thus any loop class initially based at \(u\) can be transported to \(p\) using \(\bar\gamma_U*\alpha*\gamma_U\); a loop class based at \(v\) can be transported using \(\bar\gamma_V*\beta*\gamma_V\). For a loop based at \(u\) that lies in \(U\cap V\), the first transported loop lies in the intersection as well, and its class maps to the same loop class when included into either \(U\) or \(V\). This example illustrates the basepoint bookkeeping; it does not require computing the groups of these sets.
Check: At \(t=0\), \(\gamma_U(0)=u\) and \(\gamma_V(0)=v\); at \(t=1\), both paths equal \(p\). The inequalities \(-1/4\leq -(1-t)/4\leq0<1/2\) and \(-1/2<0\leq(1-t)/4\leq1/4\) verify their stated membership for every \(t\in I\).
When the Chosen Path Matters
There may be more than one path from \(x_0\) to \(x_1\). Their basepoint-change isomorphisms need not agree. If \(\gamma\) and \(\delta\) are two such paths, then \(\lambda=\bar\delta*\gamma\) is a loop at \(x_1\), and for \(a\in\pi_1(X,x_0)\), \(T_\delta(a)=[\lambda]\,T_\gamma(a)\,[\lambda]^{-1}\). Indeed, writing \(a=[\alpha]\), the expression on the right is represented by \(\bar\delta*\gamma*\bar\gamma*\alpha*\gamma*\bar\gamma*\delta\). The adjacent path-reverse pairs contract, leaving \(\bar\delta*\alpha*\delta\), which represents \(T_\delta(a)\). Thus changing the chosen path changes the identification by conjugation. If the group is not commutative, that conjugation can change elements.
Common pitfall. A basepoint is not merely a label that can be dropped when writing an inclusion-induced map. For instance, if a loop in \(U\) is based at \(u\) while the intersection group is based at \(p\), the expression “include the loop into \(U\cap V\)” is not a based homomorphism with the stated basepoints: its image loop is still based at \(u\), not \(p\). First transport using a path in the intersection from \(u\) to \(p\), if one is available. Then the inclusion and transport compatibility proposition applies. If no such path exists, this particular basepoint comparison cannot be made by that route.
In practice, record each space together with its basepoint, record the direction and location of each path, and write the conjugating loop in the order dictated by those endpoints. This keeps the inclusion homomorphisms comparable when a cover is assembled.
Check Your Understanding
Use the convention that a product of paths traverses the first path and then the second.
- For a based continuous map \(f:(X,x_0)\to(Y,y_0)\), write the formula for \(f_*\) and explain why it respects loop multiplication.
- If \(\gamma\) runs from \(x_0\) to \(x_1\), write the loop at \(x_1\) obtained from a loop \(\alpha\) at \(x_0\).
- What is the inverse of the basepoint-change isomorphism \(T_\gamma\), and why?
- State the compatibility between basepoint transport and inclusion when the transport path lies in the subspace.
- In a cover \(X=U\cup V\), why is it useful to choose a common basepoint in \(U\cap V\)?