The Base-Point Track and the Conjugation Formula
A homotopy between two maps need not keep a chosen base point fixed. Its value at that point traces a path in the target, so the induced homomorphisms have different target base points. The path tells us how to compare them: travel back along the track, traverse the first image loop, and then travel forward along the track. The order matters, because each path must start where the preceding path ends.
We use the convention that \(\alpha*\beta\) means first follow \(\alpha\), then follow \(\beta\). Paths are maps from \(I=[0,1]\), and a path homotopy relative to endpoints is a homotopy through paths that keeps both endpoints fixed. A loop based at \(y\) is a path whose initial and final points are both \(y\); its path-homotopy class is written \([\alpha]\in\pi_1(Y,y)\). The fundamental group operation is induced by concatenation.
Concatenations may be parenthesized in either order without changing their path-homotopy class: changing parentheses only changes the speeds on the three successive pieces, and a linear interpolation between the two time allocations gives a homotopy relative to endpoints. A constant path at the relevant endpoint acts as an identity up to such a reparameterization. We will also use cancellation: \(h*\bar h\) and \(\bar h*h\) are homotopic relative to endpoints to the corresponding constant paths. For example, \(h*\bar h\), a loop at \(y_0\), contracts by shortening the portion of \(h\) that is traversed: at stage \(u\in I\), use \(h(2su)\) for \(0\le s\le 1/2\), and \(h(2(1-s)u)\) for \(1/2\le s\le1\). At \(u=1\) this is \(h*\bar h\), and at \(u=0\) it is constant at \(y_0\).
Proof. Let \(\alpha\) and \(\delta\) be loops based at \(y_0\). By the definition of the group operation, \[ \beta_h([\alpha][\delta])=[\bar h*\alpha*\delta*h]. \] On the other hand, the product \(\beta_h([\alpha])\beta_h([\delta])\) is represented by \[ (\bar h*\alpha*h)*(\bar h*\delta*h). \] After reassociating, the middle part is \(h*\bar h\), which is homotopic relative to endpoints to the constant path at \(y_0\). Removing it leaves \(\bar h*\alpha*\delta*h\). Thus the two products give the same path-homotopy class, as required. \(\square\)
How an Unbased Homotopy Gives the Formula
Proof. Fix a loop \(\gamma:I\to X\) based at \(x_0\), and define \(K:I\times I\to Y\) by \(K(s,t)=H(\gamma(s),t)\). This is continuous as a composition of continuous maps. Its four edges are \(f\circ\gamma\) along the bottom, \(h\) along the right, \(g\circ\gamma\) along the top, and \(h\) along the left. The two routes across the square from \((0,0)\) to \((1,1)\)—bottom then right, or left then top—are homotopic relative to their endpoints. To see this directly, parametrize each route on the two halves of \(I\), calling the resulting paths \(P\) and \(Q\). Both lie in the square and have the same endpoints. The paths \((1-v)P(s)+vQ(s)\), for \(0\le v\le1\), stay in the square and fix those endpoints. Composing with \(K\) gives a path homotopy, so \[ [(f\circ\gamma)*h]=[h*(g\circ\gamma)]. \] Now prepend \(\bar h\) to both paths. This is composable: \(\bar h\) starts at \(g(x_0)\), ends at \(f(x_0)\), and each path on the right-hand side of the preceding equality starts at \(f(x_0)\). The resulting paths are loops based at \(g(x_0)\). The left side becomes \(\bar h*(f\circ\gamma)*h\); on the right, the consecutive paths \(\bar h*h\) cancel up to a homotopy relative to endpoints, leaving \(g\circ\gamma\). Therefore the stated loop classes agree. Since this holds for every \([\gamma]\), it proves the homomorphism formula. \(\square\)
The formula is a statement about homomorphisms with their base points specified: \(f_*\) has target \(\pi_1(Y,f(x_0))\), while \(g_*\) has target \(\pi_1(Y,g(x_0))\). If \(h\) is constant, these base points coincide and \(\beta_h\) is the identity map, recovering the based-homotopy case. If the track is not constant, omitting it can give a loop at the wrong base point or the wrong conjugate.
Worked Example: A Rotation of the Circle
Give \(S^1\) its subspace topology from \(\mathbb C\), and take \(X=Y=S^1\), \(x_0=1\). Define \[ H(z,t)=e^{2\pi i(t/4)}z,\qquad f(z)=z,\qquad g(z)=iz. \] At \(t=0\), \(H(z,0)=z=f(z)\); at \(t=1\), \(H(z,1)=e^{\pi i/2}z=iz=g(z)\). The base-point track is \(h(t)=e^{\pi it/2}\), from \(1\) to \(i\). Let \(\gamma(s)=e^{2\pi is}\), the once-around loop based at \(1\). Then \(g\circ\gamma(s)=i e^{2\pi is}\), a loop based at \(i\).
Check the conjugation formula by recording angles in turns, so that angle \(r\) represents \(e^{2\pi ir}\). The path \(\bar h*(\gamma)*h\), with its three pieces assigned equal thirds of the parameter interval, has angle \[ \theta_P(u)= \begin{cases} \frac14-\frac34u,&0\le u\le\frac13,\\ 3u-1,&\frac13\le u\le\frac23,\\ 1+\frac34(u-\frac23),&\frac23\le u\le1. \end{cases} \] Its initial angle is \(\theta_P(0)=1/4\), corresponding to \(i\), and its final angle is \(\theta_P(1)=5/4\), also corresponding to \(i\). The loop \(g\circ\gamma\) has angle \(\theta_Q(u)=1/4+u\), with the same endpoints. The formula \[ e^{2\pi i((1-v)\theta_P(u)+v\theta_Q(u))},\qquad 0\le u,v\le1, \] is a homotopy between these loops that fixes both endpoints: at \(u=0\) the angle is always \(1/4\), and at \(u=1\) it is always \(5/4\). This explicitly verifies the conjugation relation for this loop.
Check: the first piece of the concatenation begins at angle \(1/4\), travels back to \(0\), the middle piece goes from \(0\) to \(1\), and the last piece goes from \(0\) to \(1/4\). The final angle \(5/4\) therefore represents the same point \(i\) as the starting angle \(1/4\), as a loop based at \(g(x_0)\) must.
Worked Example: Cyclically Changing the Starting Point of a Loop
Let \(Y=S^1\vee S^1\), the quotient of two copies of \(S^1\) obtained by identifying their marked points, with the quotient topology. Denote the identified point by \(o\). Let \(X=S^1\) with base point \(1\). Define a loop \(L:S^1\to Y\) by traversing the first circle once and then the second circle once. More explicitly, for \(0\le s\le1\), use the first circle point \(e^{4\pi is}\) when \(0\le s\le1/2\), and the second circle point \(e^{4\pi i(s-1/2)}\) when \(1/2\le s\le1\). At \(s=0,1/2,1\) the values are \(o\), so the pieces join continuously and \(L\) defines a loop.
Write \(a\) and \(b\) for the first- and second-circle traversals, respectively. Set \(f(z)=L(z)\) and \(H(z,t)=L(e^{\pi it}z)\). This is a homotopy from \(f\) to \(g(z)=L(-z)\). Its base-point track is \(h(t)=L(e^{\pi it})\): as \(t\) runs from \(0\) to \(1\), this traverses the first circle once, so \(h=a\). The loop \(f\circ\gamma\), for \(\gamma(s)=e^{2\pi is}\), is \(a*b\). The loop \(g\circ\gamma\) starts at the wedge point halfway through the parameter of \(L\), so it traverses \(b\) and then \(a\), namely \(b*a\).
The right-hand side of the theorem is represented by \[ \bar h*(f\circ\gamma)*h=\bar a*(a*b)*a. \] Here \(\bar a*a\) contracts at the wedge point, leaving \(b*a\), exactly the loop \(g\circ\gamma\). Thus the formula accounts for the change in starting point even though the homotopy moves that point around the first circle.
Check: the concatenation starts at \(o=g(1)\): \(\bar a\) starts at \(o\), ends at \(o\), \(a*b\) starts and ends at \(o\), and the final \(a\) returns to \(o\). The cancellation gives a loop at \(o\), as required. No claim that the two induced homomorphisms are equal is needed; the formula specifies their relationship through the track.
Worked Example: A Moving Base Point in an Interval
Let \(X=S^1\) and let \(Y=I=[0,1]\) have their usual topologies. Take \(x_0=1\), and define \(f(z)=(1+\operatorname{Re}z)/2\), \(g(z)=0\), and \(H(z,t)=(1-t)(1+\operatorname{Re}z)/2\). These formulas give a continuous homotopy: \(H(z,0)=f(z)\), \(H(z,1)=g(z)\), and every value lies in \(I\). The base-point track is \(h(t)=1-t\), which runs from \(f(1)=1\) to \(g(1)=0\).
Every loop \(\alpha\) in \(I\) based at \(y\in I\) is homotopic to the constant loop at \(y\) by \[ C(s,u)=(1-u)\alpha(s)+uy. \] Each value lies in \(I\), and \(C(0,u)=C(1,u)=y\), so this is a homotopy relative to endpoints. Thus both target fundamental groups in the theorem are trivial. More explicitly, for any loop \(\gamma\) based at \(1\), the conjugated loop \(\bar h*(f\circ\gamma)*h\) starts at \(0\), travels to \(1\), follows \(f\circ\gamma\), and returns to \(0\); it is a loop in \(I\) and contracts by the displayed formula with \(y=0\). The loop \(g\circ\gamma\) is constant at \(0\). Hence both sides represent the identity in \(\pi_1(I,0)\), in agreement with the theorem.
A Common Orientation Error
Common pitfall. Reversing the track in the wrong place can make the proposed concatenation impossible. In the circle example, \(h\) starts at \(1\) and ends at \(i\), while \(f\circ\gamma\) is based at \(1\). Thus \(h*(f\circ\gamma)\) is not defined: the first path ends at \(i\), but the second starts at \(1\). The correct loop based at \(i\) is \(\bar h*(f\circ\gamma)*h\): \(\bar h\) starts at \(i\) and ends at \(1\), the loop \(f\circ\gamma\) starts and ends at \(1\), and \(h\) returns to \(i\). In general, always list the initial and final point of each path before concatenating. For the same reason, the constant path \(c_0\) at \(0\) can be followed by \(h\) when \(h\) starts at \(0\) and ends at \(1\); the reverse order would not match endpoints.
The conjugation formula is useful because it separates two effects: the homotopy compares the maps, while the base-point track specifies how their target fundamental groups are identified. It does not assert that the homomorphisms are literally equal when their target base points differ. If the track is constant, the change-of-base-point map reduces to the identity. If the track is not constant, the conjugation is part of the precise conclusion.
Check Your Understanding
For each question, write down the base points and check path endpoints before using concatenation.
- If \(h\) runs from \(y_0\) to \(y_1\), what are the initial and final points of \(\bar h*\alpha*h\) when \(\alpha\) is based at \(y_0\)?
- In the theorem, identify the initial and final points of \(f\circ\gamma\), \(g\circ\gamma\), and the base-point track \(h\).
- For a constant base-point track, explain why \(\beta_h\) is the identity on the relevant fundamental group.
- In the wedge-of-circles example, what are the track \(h\), the loop \(f\circ\gamma\), and the loop \(g\circ\gamma\), in terms of \(a\) and \(b\)?
- Why is \(h*(f\circ\gamma)\) generally not a valid concatenation when \(h\) runs from \(f(x_0)\) to \(g(x_0)\)?