Proving Base-Point Change Is an Isomorphism

Transporting Loops Between Base Points

A loop is based not only at a point of a space but also at a choice of that point: its initial and final values must both equal the base point. A path between two points lets us transport loop classes from one fundamental group to the other. The key to proving this transport is reversible is to travel back along the same path in the reverse direction.

Throughout, $X$ is a topological space with topology $\mathcal{T}$. A path from $x$ to $y$ is a continuous map $\gamma:[0,1]\to X$ with $\gamma(0)=x$ and $\gamma(1)=y$. Its reversal is $\bar{\gamma}(t)=\gamma(1-t)$, a path from $y$ to $x$. We use the left-to-right concatenation convention: $\alpha*\beta$ means first traverse $\alpha$, then $\beta$, and it is defined when $\alpha(1)=\beta(0)$.

Definition: A based loop at $x$ is a path $\alpha$ with $\alpha(0)=\alpha(1)=x$. Two loops at $x$ are based homotopic if there is a continuous homotopy between them that keeps both endpoints equal to $x$ throughout. The fundamental group $\pi_1(X,x)$ consists of their based homotopy classes, with multiplication $[\alpha][\beta]=[\alpha*\beta]$.

For paths $\alpha$ and $\beta$, their concatenation uses the first half of the parameter interval for $\alpha$ and the second half for $\beta$. Concatenation of path classes is associative, and the constant loop $c_x(t)=x$ represents the identity. These are the previously established associativity and identity properties of the fundamental group.

Let $\gamma$ be a path from $x_0$ to $x_1$. If $\alpha$ is a loop at $x_0$, the correctly based loop at $x_1$ is $\bar{\gamma}*\alpha*\gamma$: it first travels from $x_1$ to $x_0$, traverses $\alpha$, and returns from $x_0$ to $x_1$.

Definition: The base-point-change map along $\gamma$ is $$ \Phi_\gamma:\pi_1(X,x_0)\longrightarrow\pi_1(X,x_1), \qquad \Phi_\gamma([\alpha])=[\bar{\gamma}*\alpha*\gamma]. $$

The Map Is Well-Defined and Preserves Multiplication

First check the types of the paths in the definition. The path $\bar{\gamma}$ starts at $x_1$ and ends at $x_0$, while $\alpha$ starts and ends at $x_0$, and $\gamma$ starts at $x_0$ and ends at $x_1$. Thus both concatenations are defined, and the full path is a loop at $x_1$.

If $\alpha$ and $\alpha'$ are based homotopic at $x_0$, put that homotopy between the fixed paths $\bar{\gamma}$ and $\gamma$. This gives a homotopy between $\bar{\gamma}*\alpha*\gamma$ and $\bar{\gamma}*\alpha'*\gamma$ that keeps the endpoints at $x_1$. The homotopy is continuous because its three pieces agree at their joining points. Therefore $\Phi_\gamma$ does not depend on the representative of $[\alpha]$.

Proposition: The map $\Phi_\gamma$ is a group homomorphism.

Proof. Let $\alpha$ and $\beta$ be loops at $x_0$. By the definition of multiplication and associativity of concatenation on path classes, $[\Phi_\gamma([\alpha])\Phi_\gamma([\beta])]$ is represented by $\bar{\gamma}*\alpha*\gamma*\bar{\gamma}*\beta*\gamma$. The middle pair $\gamma*\bar{\gamma}$ is a loop at $x_0$ that travels along $\gamma$ and immediately retraces it. It is based homotopic to the constant loop at $x_0$, as shown below. Replacing that pair by the constant loop leaves the class $[\bar{\gamma}*\alpha*\beta*\gamma]$. This is $\Phi_\gamma([\alpha][\beta])$, so $\Phi_\gamma$ preserves multiplication. $\square$

Retracing a Path Contracts to a Constant Loop

Lemma: For any path $\delta$ from $a$ to $b$, the loop $\delta*\bar{\delta}$ is based homotopic at $a$ to the constant loop at $a$. Also, $\bar{\delta}*\delta$ is based homotopic at $b$ to the constant loop at $b$.

Proof. Write $c=\delta*\bar{\delta}$, with the stated half-interval parametrization. For $0\leq u,t\leq1$, define $$ H(u,t)= \begin{cases} \delta\bigl(2(1-u)t\bigr),&0\leq t\leq \tfrac12,\\ \delta\bigl(2(1-u)(1-t)\bigr),&\tfrac12\leq t\leq1. \end{cases} $$ At $t=\tfrac12$, both formulas give $\delta(1-u)$, so the pieces agree and $H$ is continuous. At $t=0$ and $t=1$, the value is $\delta(0)=a$ for every $u$, so the homotopy fixes the base point. When $u=0$, the formula is exactly $\delta*\bar{\delta}$; when $u=1$, both pieces are constantly $\delta(0)$. This proves the first assertion. Apply the same construction to the path $\bar{\delta}$, which starts at $b$, to prove the second. $\square$

The Reversed Path Gives an Inverse

Since $\bar{\gamma}$ is a path from $x_1$ to $x_0$, it defines a map in the opposite direction. For a loop $\beta$ at $x_1$, transport back by first travelling from $x_0$ to $x_1$, then traversing $\beta$, and then returning to $x_0$.

Theorem: The map $\Phi_\gamma:\pi_1(X,x_0)\to\pi_1(X,x_1)$ is a group isomorphism. Its inverse is $$ \Phi_{\bar{\gamma}}:\pi_1(X,x_1)\longrightarrow\pi_1(X,x_0), \qquad \Phi_{\bar{\gamma}}([\beta])=[\gamma*\beta*\bar{\gamma}]. $$

Proof. The same well-definedness and homomorphism arguments apply to $\Phi_{\bar{\gamma}}$. For a class $[\alpha]\in\pi_1(X,x_0)$, associativity gives \[ \Phi_{\bar{\gamma}}(\Phi_\gamma([\alpha])) =[\gamma*\bar{\gamma}*\alpha*\gamma*\bar{\gamma}]. \] The two outer retraced-path pairs $\gamma*\bar{\gamma}$ are loops based at $x_0$, and each is based homotopic to the constant loop at $x_0$ by the lemma. Thus this class equals $[\alpha]$.

For a class $[\beta]\in\pi_1(X,x_1)$, associativity gives \[ \Phi_\gamma(\Phi_{\bar{\gamma}}([\beta])) =[\bar{\gamma}*\gamma*\beta*\bar{\gamma}*\gamma]. \] Here each pair $\bar{\gamma}*\gamma$ is a loop based at $x_1$ and contracts to the constant loop at $x_1$. Hence this class equals $[\beta]$. The two maps are inverses, so $\Phi_\gamma$ is an isomorphism. $\square$

Worked Example: A Square Loop in the Plane

Give $\mathbb{R}^2$ its Euclidean topology. Let $x_0=(0,0)$, $x_1=(1,1)$, and $\gamma(t)=(t,t)$. Let $\alpha$ be the square loop at $x_0$ given by $$ \alpha(t)= \begin{cases} (0,4t),&0\leq t\leq\tfrac14,\\ (4t-1,1),&\tfrac14\leq t\leq\tfrac12,\\ (1,3-4t),&\tfrac12\leq t\leq\tfrac34,\\ (4-4t,0),&\tfrac34\leq t\leq1. \end{cases} $$

At $t=\tfrac14$, the first two pieces both give $(0,1)$. At $t=\tfrac12$, the second and third both give $(1,1)$. At $t=\tfrac34$, the third and fourth both give $(1,0)$. Finally, $\alpha(0)=(0,0)=\alpha(1)$, so these pieces form a continuous loop based at $x_0$. The reversal is $\bar{\gamma}(t)=(1-t,1-t)$, which starts at $(1,1)$ and ends at $(0,0)$.

The transported loop $\bar{\gamma}*\alpha*\gamma$ starts at $x_1$, reaches $x_0$, traverses the square, and returns to $x_1$. Each joining point matches: $\bar{\gamma}(1)=(0,0)=\alpha(0)$, and $\alpha(1)=(0,0)=\gamma(0)$. Its initial and final points are $\bar{\gamma}(0)=(1,1)=\gamma(1)$. These endpoint checks verify directly that the transport formula gives a loop at the new base point.

Check: the inverse transport begins with $\gamma$, which runs from $(0,0)$ to $(1,1)$, then traverses a loop based at $(1,1)$, and finishes with $\bar{\gamma}$, which returns to $(0,0)$. Applying it after the forward transport produces the retraced pieces $\gamma*\bar{\gamma}$ on either side of $\alpha$; the lemma contracts each while fixing $(0,0)$.

Worked Example: Transporting a Loop on the Circle

Give $S^1=\{z\in\mathbb{C}:|z|=1\}$ the subspace topology inherited from $\mathbb{C}$. Let $x_0=1$, $x_1=i$, and $\gamma(t)=e^{i\pi t/2}$. Then $\gamma(0)=1$, $\gamma(1)=e^{i\pi/2}=i$, and $\bar{\gamma}(t)=e^{i\pi(1-t)/2}$. Take the loop $\alpha(t)=e^{2\pi i t}$ at $1$. Its midpoint is $\alpha(1/2)=e^{i\pi}=-1$, and its endpoint is $\alpha(1)=e^{2\pi i}=1$.

With the three pieces each assigned one third of the parameter interval, the transported loop is $$ \eta(t)= \begin{cases} \bar{\gamma}(3t),&0\leq t\leq\tfrac13,\\ \alpha(3t-1),&\tfrac13\leq t\leq\tfrac23,\\ \gamma(3t-2),&\tfrac23\leq t\leq1. \end{cases} $$ At $t=\tfrac13$, the first piece ends at $\bar{\gamma}(1)=1$, and the second starts at $\alpha(0)=1$. At $t=\tfrac23$, the second ends at $\alpha(1)=1$, and the third starts at $\gamma(0)=1$. Also, $\eta(0)=\bar{\gamma}(0)=i$ and $\eta(1)=\gamma(1)=i$. Thus $\eta$ is a loop based at $i$, and the inverse transport uses $\gamma*\eta*\bar{\gamma}$ to return to a loop based at $1$.

A Common Composition Error

Common pitfall. With the left-to-right convention, the order of the paths cannot be rearranged. In the circle example, $\gamma$ runs from $1$ to $i$, while $\alpha$ starts and ends at $1$. Therefore $\gamma*\alpha$ is not defined, because $\gamma(1)=i\ne1=\alpha(0)$. Also, $\alpha*\bar{\gamma}$ is not defined, because $\alpha(1)=1\ne i=\bar{\gamma}(0)$. In particular, $\gamma*\alpha*\bar{\gamma}$ is not a loop assembled from these paths. The valid expression is $\bar{\gamma}*\alpha*\gamma$: each endpoint equals the next initial point, and the resulting loop is based at $i$.

Worked Example: Checking the Formula on the Real Line

Give $\mathbb{R}$ its standard topology. Set $x_0=0$, $x_1=2$, and $\gamma(t)=2t$. Let $$ \alpha(t)= \begin{cases} 2t,&0\leq t\leq\tfrac12,\\ 2-2t,&\tfrac12\leq t\leq1. \end{cases} $$ The pieces agree at $t=\tfrac12$, where both equal $1$, and $\alpha(0)=\alpha(1)=0$. Thus $\alpha$ is a loop at $0$.

Write $\bar{\gamma}(t)=2-2t$. The transported loop, using three equal parameter intervals, is $$ \eta(t)= \begin{cases} 2-6t,&0\leq t\leq\tfrac13,\\ 6t-2,&\tfrac13\leq t\leq\tfrac12,\\ 4-6t,&\tfrac12\leq t\leq\tfrac23,\\ 6t-4,&\tfrac23\leq t\leq1. \end{cases} $$ At $t=\tfrac13$, the first value is $2-6(\tfrac13)=0$, and the second value is $6(\tfrac13)-2=0$. At $t=\tfrac12$, the adjacent values are $6(\tfrac12)-2=1$ and $4-6(\tfrac12)=1$. At $t=\tfrac23$, the adjacent values are $4-6(\tfrac23)=0$ and $6(\tfrac23)-4=0$. The endpoints are $\eta(0)=2$ and $\eta(1)=6-4=2$, as required for a loop at $x_1$.

For a direct inverse check, let $\beta$ be the loop at $2$ that rises to $3$ and returns: $\beta(t)=2+2t$ for $0\leq t\leq\tfrac12$, and $\beta(t)=4-2t$ for $\tfrac12\leq t\leq1$. The inverse formula $\gamma*\beta*\bar{\gamma}$ starts at $0$, ends at $0$, and has successive joining values $2$, $3$, and $2$: the first leg ends at $\gamma(1)=2=\beta(0)$, the middle loop ends at $\beta(1)=2=\bar{\gamma}(0)$, and the final leg ends at $\bar{\gamma}(1)=0$. These computations confirm the direction and endpoint requirements for both transport formulas.

The isomorphism result uses only the path from one base point to the other and its reversal. Its inverse proof depends on the specific cancellation of a path followed by its reverse; it does not require the two base points to coincide or the path to be a loop. The essential checks are that every concatenation is composable, both resulting maps land in the stated fundamental groups, and the two retraced-path pairs contract while their base points remain fixed.

Check Your Understanding

For each question, use the left-to-right concatenation convention and state the base point of every loop you form.

If $\gamma$ runs from $x_0$ to $x_1$, write the formula for transporting a loop at $x_0$ to a loop at $x_1$, and check the endpoints of its three pieces.
Write an explicit based homotopy contracting $\delta*\bar{\delta}$ while keeping its base point fixed.
For $\gamma:x_0\to x_1$, write the inverse base-point-change formula and identify the base point of its output.
Explain why $\gamma*\alpha*\bar{\gamma}$ is generally not composable when $\alpha$ is a loop at $x_0$.
In the circle example, evaluate $\gamma(0)$, $\gamma(1)$, $\alpha(0)$, and $\alpha(1)$, and use them to verify the joining points in $\bar{\gamma}*\alpha*\gamma$.