Conjugation and Dependence on the Base-Point Path

Why the Path Can Matter

Changing the base point of a fundamental group requires a path connecting the old and new base points. The resulting groups are isomorphic, but the isomorphism uses the chosen path. If a different path is chosen, the two isomorphisms need not be equal: their difference is measured by conjugation in the group at the new base point. This tutorial derives that formula and determines exactly when the choice of path makes no difference.

Let $(X,\mathcal{T})$ be a topological space. A path from $x_0$ to $x_1$ is a continuous map $\alpha:[0,1]\to X$ with $\alpha(0)=x_0$ and $\alpha(1)=x_1$. Write $\bar\alpha(t)=\alpha(1-t)$ for the reversed path. We use $\rho*\sigma$ to mean first traverse $\rho$, then $\sigma$; the endpoint of $\rho$ must be the starting point of $\sigma$. The fundamental group $\pi_1(X,x)$ consists of based homotopy classes of loops at $x$, with multiplication induced by concatenation. A based homotopy keeps the loop's base point fixed.

Definition: Given a path $\alpha$ from $x_0$ to $x_1$, the base-point-change map is $$ \Phi_\alpha:\pi_1(X,x_0)\longrightarrow\pi_1(X,x_1),\qquad \Phi_\alpha([\gamma])=[\bar\alpha*\gamma*\alpha]. $$ The path first carries the new base point back to the old one, the loop $\gamma$ is traversed, and the path then returns to $x_1$. The earlier base-point-change result establishes that this map is an isomorphism.

All path expressions below are interpreted up to homotopy fixing their endpoints. We use the standard path facts that concatenation is associative up to such homotopy, and that $\rho*\bar\rho$ and $\bar\rho*\rho$ are homotopic, relative to their endpoints, to the appropriate constant paths. These are the path cancellation and associativity results established in the earlier material on fundamental groups.

The Conjugation Formula

Suppose $\alpha$ and $\beta$ are two paths from $x_0$ to $x_1$. Their difference gives a loop at $x_1$: travel from $x_1$ back to $x_0$ along $\beta$, then from $x_0$ to $x_1$ along $\alpha$. Denote its class by $\delta$.

Theorem (Path-dependence formula): For paths $\alpha,\beta:x_0\to x_1$, put $$ \delta=[\bar\beta*\alpha]\in\pi_1(X,x_1). $$ Then for every $g\in\pi_1(X,x_0)$, $$ \Phi_\beta(g)=\delta\,\Phi_\alpha(g)\,\delta^{-1}. $$ Thus $\Phi_\beta$ is obtained from $\Phi_\alpha$ by conjugation by $\delta$.

Proof. Let $g=[\gamma]$. The inverse of $\delta=[\bar\beta*\alpha]$ is represented by the reversed loop $\bar\alpha*\beta$, since reversing the concatenation reverses the order and reverses each path. Using the definition of $\Phi_\alpha$, the product on the right is represented by $$ (\bar\beta*\alpha)*(\bar\alpha*\gamma*\alpha)*(\bar\alpha*\beta). $$ By associativity up to endpoint-fixed homotopy, we may group adjacent paths as $$ \bar\beta*(\alpha*\bar\alpha)*\gamma*(\alpha*\bar\alpha)*\beta. $$ Here $\alpha*\bar\alpha$ is a loop at $x_0$ that contracts, relative to its endpoints, to the constant path at $x_0$. Cancelling both such backtracks leaves $\bar\beta*\gamma*\beta$. Its class is $\Phi_\beta(g)$, which proves the formula. $\square$

The location and direction of the loop $\delta$ matter. It is a loop at $x_1$, so it conjugates elements of $\pi_1(X,x_1)$, not elements of $\pi_1(X,x_0)$. The formula also specifies the order: with our convention that paths are traversed from left to right, the conjugator is $[\bar\beta*\alpha]$.

Exactly When Do the Maps Agree?

The center of a group $G$, written $Z(G)$, is the set of elements that commute with every element of $G$. The conjugation formula turns equality of the two base-point-change maps into a question about the center of the target group.

Theorem (Equality criterion): For paths $\alpha,\beta:x_0\to x_1$, let $\delta=[\bar\beta*\alpha]$. Then $$ \Phi_\alpha=\Phi_\beta \quad\Longleftrightarrow\quad \delta\in Z\bigl(\pi_1(X,x_1)\bigr). $$

Proof. Suppose first that $\Phi_\alpha=\Phi_\beta$. The path-dependence formula gives $\delta\,\Phi_\alpha(g)\,\delta^{-1}=\Phi_\alpha(g)$ for every $g\in\pi_1(X,x_0)$. Since $\Phi_\alpha$ is an isomorphism, it is onto: every $h\in\pi_1(X,x_1)$ equals $\Phi_\alpha(g)$ for some $g$. Therefore $\delta h\delta^{-1}=h$ for every such $h$, which is equivalent to $\delta h=h\delta$ for every $h$. Thus $\delta$ lies in the center.

Conversely, suppose $\delta\in Z(\pi_1(X,x_1))$. Then $\delta h\delta^{-1}=h$ for every $h\in\pi_1(X,x_1)$. In particular, take $h=\Phi_\alpha(g)$ for any $g\in\pi_1(X,x_0)$. The path-dependence formula gives $\Phi_\beta(g)=\Phi_\alpha(g)$ for every $g$, so the maps are equal. $\square$

Worked Example: Homotopic Choices of Path

Suppose $\alpha$ and $\beta$ are homotopic while their endpoints remain fixed. Show that their base-point-change maps agree.

Since $\beta$ is homotopic to $\alpha$ relative to endpoints, $\bar\beta$ is homotopic to $\bar\alpha$ relative to endpoints. Consequently, the loop $\bar\beta*\alpha$ is homotopic to $\bar\alpha*\alpha$. The latter traverses $\alpha$ backwards and then forwards, so it contracts to the constant loop at $x_1$. Thus $\delta$ is the identity element of $\pi_1(X,x_1)$.

For any $g\in\pi_1(X,x_0)$, the formula now gives $$ \Phi_\beta(g)=\delta\,\Phi_\alpha(g)\,\delta^{-1} =e\,\Phi_\alpha(g)\,e^{-1} =\Phi_\alpha(g), $$ where $e$ is the identity element. Therefore $\Phi_\alpha=\Phi_\beta$.

Check: the identity loop conjugates every loop class to itself, as the displayed computation verifies. The relevant homotopy must keep both endpoints fixed; a homotopy that moves an endpoint does not establish that $\bar\beta*\alpha$ is the identity class at this fixed base point.

In fact, the same path-cancellation facts show that $[\bar\beta*\alpha]$ is the identity precisely when $\alpha$ and $\beta$ are homotopic relative to endpoints. One direction was just proved. For the converse, if $\bar\beta*\alpha$ contracts, prepend $\beta$ to this loop. The path $\beta*\bar\beta$ then cancels, leaving a path homotopic to $\alpha$; hence $\beta$ and $\alpha$ are homotopic relative to endpoints. This gives a useful sufficient condition for agreement, while the center criterion is more general: a nonidentity difference loop can still produce equal maps if its class is central.

Worked Example: An Abelian Target Group

Consider a base-point-change situation in which $\pi_1(X,x_1)$ is isomorphic to the additive group $\mathbb{Z}$. In additive notation, conjugation has the form $\delta+h-\delta$. Take the difference class to correspond to $4$, and let $h$ correspond to $-3$. Then $$ 4+(-3)-4=1-4=-3. $$ The result is the original element $h=-3$. More generally, for any integers $d$ and $h$, $d+h-d=h$, because integer addition is commutative. Thus every element is central, and the equality criterion says that any two base-point-change maps with this target group agree.

Check: substituting the chosen values into the conjugation expression gives $4+(-3)+(-4)=1-4=-3$, exactly the original class. This example uses only the stated group structure; it does not require a computation of which spaces have this fundamental group.

Worked Example: A Nonabelian Target Group

To see how maps can differ, work in the permutation group $S_3$, with products composed from right to left. Suppose the target fundamental group has elements corresponding to $\delta=(123)$ and $h=(12)$. Conjugating a transposition $(a\,b)$ by a permutation $\sigma$ gives $(\sigma(a)\,\sigma(b))$. Since $(123)$ sends $1$ to $2$ and $2$ to $3$, we obtain $$ (123)(12)(132)=(23). $$ The result $(23)$ is not $(12)$, so this $\delta$ does not commute with $h$.

Because $\Phi_\alpha$ is onto, there is an element $g\in\pi_1(X,x_0)$ with $\Phi_\alpha(g)=h$. For that element, the path-dependence formula gives $\Phi_\beta(g)=(23)$, while $\Phi_\alpha(g)=(12)$. Hence the two maps differ. This calculation is conditional on the indicated group elements occurring in the target; it illustrates the group-theoretic test without asserting a particular space has this fundamental group.

Check: $(23)\ne(12)$, since the first permutation exchanges $2$ and $3$, while the second exchanges $1$ and $2$. Therefore the conjugation is genuinely nontrivial.

A Common Order Error

Common pitfall. Reversing the two paths in the difference loop changes the conjugator. For the convention used here, the correct loop is $\bar\beta*\alpha$, and the formula is $\Phi_\beta(g)=\delta\Phi_\alpha(g)\delta^{-1}$. A mistaken calculation might instead use $\delta^{-1}\Phi_\alpha(g)\delta$.

The distinction can be checked in $S_3$. Let $\delta=(123)$ and $h=(12)$, again composing right to left. The correct conjugation gives $$ (123)(12)(132)=(23), $$ whereas the reversed conjugation gives $$ (132)(12)(123)=(13). $$ Indeed, conjugation by $(132)$ sends $1$ to $3$ and $2$ to $1$, so it sends $(12)$ to $(31)=(13)$. These answers are different. The correction is not to choose whichever order gives a convenient result: trace the paths from their endpoints. The loop $\bar\beta*\alpha$ starts and ends at $x_1$, and the theorem's proof fixes its position on the left of $\Phi_\alpha(g)$.

What the Criterion Tells Us

The isomorphism given by a path is not automatically independent of that path. The precise issue is whether the difference loop acts trivially by conjugation on the target fundamental group. If the target group is abelian, every such action is trivial. If the difference class is central, the maps agree even when that class is not the identity. If it fails to be central, some element of the target group is changed by conjugation, and surjectivity of $\Phi_\alpha$ ensures that this change is detected on an element of the source group.

This distinction is useful whenever base-point change is used to compare loop classes: an isomorphism between groups does not by itself specify a unique identification. The path supplies that identification, and the conjugation formula records exactly how another choice modifies it.

Check Your Understanding

Use the path-dependence formula and the equality criterion to answer these questions.

For paths $\alpha,\beta:x_0\to x_1$, at which base point is the loop $\bar\beta*\alpha$ based?
If $[\bar\beta*\alpha]$ is central but not the identity, must the base-point-change maps agree? Explain.
Why does surjectivity of $\Phi_\alpha$ matter in proving that equality of the maps forces $[\bar\beta*\alpha]$ to be central?
In an abelian target group written additively, simplify $d+h-d$ and explain what this says about the choice of path.
With right-to-left permutation composition, compute $(123)(23)(132)$. Does $(123)$ commute with $(23)$?