Associativity of Loop Concatenation

Associativity Is a Statement About Classes

Concatenating two paths requires assigning each path a portion of the parameter interval. With the usual convention, the first path uses the first half and the second path uses the second half. For three paths, the two possible parenthesizations assign different portions of the interval to the same paths. The resulting paths need not be equal as functions. Associativity of the fundamental-group operation is instead the claim that the two paths are endpoint-preservingly homotopic, so they represent the same based-loop class.

Let $X$ be a topological space and fix a point $x_0\in X$. A based loop at $x_0$ is a continuous map $\alpha:I\to X$, where $I=[0,1]$ has its usual topology, satisfying $\alpha(0)=\alpha(1)=x_0$. Two based loops $\alpha,\beta$ are endpoint-preservingly homotopic if there is a continuous map $H:I\times I\to X$, with the product topology on $I\times I$, such that $H(0,t)=\alpha(t)$, $H(1,t)=\beta(t)$, and $H(s,0)=H(s,1)=x_0$ for all $s,t\in I$. Their equivalence class is written $[\alpha]$.

Definition: For paths $\alpha,\beta:I\to X$ with $\alpha(1)=\beta(0)$, their concatenation is the path $$ (\alpha*\beta)(t)= \begin{cases} \alpha(2t),&0\le t\le \frac12,\\ \beta(2t-1),&\frac12\le t\le1. \end{cases} $$ The two formulas agree at $t=\frac12$, since both give $\alpha(1)=\beta(0)$. If $\alpha,\beta$ are based loops at $x_0$, then $\alpha*\beta$ is also a based loop at $x_0$.

The well-definedness of concatenation on based-loop classes means that replacing either loop by an endpoint-preservingly homotopic loop does not change the class of the concatenation. Thus the operation on classes is $[\alpha]\cdot[\beta]=[\alpha*\beta]$. The result proved here concerns associativity of this operation.

The Two Parenthesizations

For three based loops $\alpha,\beta,\gamma$, the path $(\alpha*\beta)*\gamma$ uses $\alpha,\beta,\gamma$ on intervals of lengths $\frac14,\frac14,\frac12$, respectively. The path $\alpha*(\beta*\gamma)$ uses them on intervals of lengths $\frac12,\frac14,\frac14$. Expanding the definitions gives the following formulas.

((\alpha*\beta)*\gamma)(t)= \begin{cases} \alpha(4t),&0\le t\le \frac14,\\ \beta(4t-1),&\frac14\le t\le \frac12,\\ \gamma(2t-1),&\frac12\le t\le1, \end{cases} \qquad (\alpha*(\beta*\gamma))(t)= \begin{cases} \alpha(2t),&0\le t\le \frac12,\\ \beta(4t-2),&\frac12\le t\le \frac34,\\ \gamma(4t-3),&\frac34\le t\le1. \end{cases}

For instance, on the first interval of $(\alpha*\beta)*\gamma$, the outer concatenation feeds $2t$ into $\alpha*\beta$; when $0\le t\le\frac14$, this lies in the first half of that inner concatenation, giving $\alpha(2(2t))=\alpha(4t)$. The other pieces follow by the same substitutions. The differing breakpoints suggest the homotopy: move them continuously from $(\frac14,\frac12)$ to $(\frac12,\frac34)$.

Theorem (Associativity of loop concatenation): If $\alpha,\beta,\gamma$ are based loops at $x_0$, then $(\alpha*\beta)*\gamma$ and $\alpha*(\beta*\gamma)$ are endpoint-preservingly homotopic. Consequently, $$ ([\alpha]\cdot[\beta])\cdot[\gamma]=[\alpha]\cdot([\beta]\cdot[\gamma]). $$

Proof. For $s\in I$, set $a(s)=\frac{1+s}{4}$ and $b(s)=\frac{2+s}{4}$. At $s=0$, these are $\frac14,\frac12$; at $s=1$, they are $\frac12,\frac34$. For every $s\in I$, $a(s)\ge\frac14$, $b(s)-a(s)=\frac14$, and $1-b(s)\ge\frac14$. In particular, all three intervals have positive length. Define $$ H(s,t)= \begin{cases} \alpha\!\left(\dfrac{t}{a(s)}\right),&0\le t\le a(s),\\[6pt] \beta\!\left(\dfrac{t-a(s)}{b(s)-a(s)}\right),&a(s)\le t\le b(s),\\[6pt] \gamma\!\left(\dfrac{t-b(s)}{1-b(s)}\right),&b(s)\le t\le1. \end{cases} $$

Each formula uses a parameter in $I$. On the first region, $0\le t/a(s)\le1$; on the second, $0\le (t-a(s))/(b(s)-a(s))\le1$; and on the third, $0\le(t-b(s))/(1-b(s))\le1$. At $t=a(s)$, the first two formulas give $\alpha(1)=x_0$ and $\beta(0)=x_0$. At $t=b(s)$, the second and third give $\beta(1)=x_0$ and $\gamma(0)=x_0$. Thus the pieces agree wherever their regions meet.

For continuity, the three regions in $I\times I$ defined by $t\le a(s)$, $a(s)\le t\le b(s)$, and $b(s)\le t$ are closed and cover $I\times I$. Each formula is continuous on its region: the denominators are positive, the displayed parameter expressions are continuous, and the paths are continuous. The finite closed-set pasting lemma says that continuous functions on a finite closed cover define a continuous function on the whole space if they agree on overlaps. This lemma follows by taking any closed set $F\subseteq X$: its preimage is the finite union of the preimages in the closed pieces, each closed in its piece and therefore closed in the whole domain; a finite union of closed sets is closed. Hence $H$ is continuous.

At $s=0$, the three intervals have endpoints $0,\frac14,\frac12,1$, so the formula for $H(0,t)$ is exactly the displayed formula for $(\alpha*\beta)*\gamma$. At $s=1$, their endpoints are $0,\frac12,\frac34,1$, giving $H(1,t)=\alpha*(\beta*\gamma)$. For every $s$, $H(s,0)=\alpha(0)=x_0$ and $H(s,1)=\gamma(1)=x_0$. Thus $H$ is an endpoint-preserving homotopy. The two loops have the same class, which proves the asserted equality of classes. $\square$

Worked Examples

Worked Example: The Paths Can Differ as Functions

Take $X=\mathbb{R}$ with its standard topology, $x_0=0$, and let all three loops be $\alpha(t)=\beta(t)=\gamma(t)=4t(1-t)$. The endpoint check is $\alpha(0)=0$ and $\alpha(1)=4(1)(0)=0$, and the same holds for $\beta,\gamma$.

At $t=\frac18$, the first parenthesization is still on its $\alpha$-piece, so $((\alpha*\beta)*\gamma)(\frac18)=\alpha(4\cdot\frac18)=\alpha(\frac12)=4\cdot\frac12\cdot\frac12=1$. The second parenthesization is also on its $\alpha$-piece, but $(\alpha*(\beta*\gamma))(\frac18)=\alpha(2\cdot\frac18)=\alpha(\frac14)=4\cdot\frac14\cdot\frac34=\frac34$. Since $1\ne\frac34$, these are not equal as paths. The theorem says they nevertheless represent the same based-loop class.

Check: at $t=0$, both paths have value $\alpha(0)=0$, and at $t=1$, both have value $\gamma(1)=0$. The mismatch at the interior point does not violate either path’s based-loop endpoint conditions.

Worked Example: Evaluating the Moving-Breakpoint Homotopy

Keep $X=\mathbb{R}$, and now choose $\alpha(t)=4t(1-t)$, $\beta(t)=8t(1-t)$, and $\gamma(t)=12t(1-t)$. Each starts and ends at $0$, since substituting $t=0$ or $t=1$ into each formula gives $0$. At $s=\frac12$, the breakpoints are $a(\frac12)=\frac38$ and $b(\frac12)=\frac58$.

At $t=\frac{3}{16}$, the first piece applies. Its path parameter is $(\frac{3}{16})/(\frac38)=\frac12$, so $H(\frac12,\frac{3}{16})=\alpha(\frac12)=4\cdot\frac12\cdot\frac12=1$.

At $t=\frac12$, the middle piece applies. Its parameter is $(\frac12-\frac38)/(\frac58-\frac38)=(\frac18)/(\frac14)=\frac12$. Therefore $H(\frac12,\frac12)=\beta(\frac12)=8\cdot\frac12\cdot\frac12=2$.

At $t=\frac{13}{16}$, the last piece applies. Its parameter is $(\frac{13}{16}-\frac58)/(1-\frac58)=(\frac{3}{16})/(\frac38)=\frac12$. Thus $H(\frac12,\frac{13}{16})=\gamma(\frac12)=12\cdot\frac12\cdot\frac12=3$.

Check: at $t=0$ the first piece gives $\alpha(0)=0$, and at $t=1$ the last gives $\gamma(1)=12\cdot1\cdot0=0$. The computed middle values come from the three loops at the same internal parameter $\frac12$, as expected from the chosen sample points.

Worked Example: Three Loops on the Circle

Give $S^1=\{z\in\mathbb{C}:|z|=1\}$ its subspace topology from $\mathbb{C}$, and use base point $1$. Consider $\alpha(t)=e^{2\pi i t}$, $\beta(t)=e^{4\pi i t}$, and $\gamma(t)=e^{6\pi i t}$. These are loops because at $t=0$ and $t=1$ their values are $1$, $e^{2\pi i}=1$, $e^{4\pi i}=1$, and $e^{6\pi i}=1$, respectively.

The two concatenations are again not necessarily the same path. At $t=\frac18$, the first gives $\alpha(\frac12)=e^{\pi i}=-1$, whereas the second gives $\alpha(\frac14)=e^{\pi i/2}=i$. At $s=\frac12$, the moving-breakpoint homotopy has $a=\frac38$, $b=\frac58$. At $t=\frac{3}{16}$ its value is $\alpha(\frac12)=e^{\pi i}=-1$; at $t=\frac12$ its value is $\beta(\frac12)=e^{2\pi i}=1$; and at $t=\frac{13}{16}$ its value is $\gamma(\frac12)=e^{3\pi i}=-1$.

Check: on the initial boundary $t=0$, the homotopy has value $\alpha(0)=1$; on the terminal boundary $t=1$, it has value $\gamma(1)=e^{6\pi i}=1$. Thus these computations are consistent with an endpoint-preserving homotopy in $S^1$, even though the two paths differ at $t=\frac18$.

Common Pitfall and Why the Homotopy Matters

Common pitfall. One might claim that concatenation is associative because the symbolic expression contains the same three loops in the same order. That confuses the order of traversal with the parameterization. The calculation in the first example is a concrete correction: at $t=\frac18$, the two paths have values $1$ and $\frac34$. They are unequal as functions. What the proof establishes is equality of their homotopy classes, using a homotopy that keeps both endpoints fixed throughout.

This distinction is essential when defining an algebraic operation from paths. An operation on classes must respect the equivalence relation, and its algebraic laws must hold for those classes. Here, the explicit homotopy supplies the required associativity law without asserting a false pointwise equality. The proof also shows why the shared base point is important at the moving boundaries: it is precisely $\alpha(1)=\beta(0)=x_0$ and $\beta(1)=\gamma(0)=x_0$ that make the pieces fit continuously while the breakpoints move.

Check Your Understanding

Use the piecewise formulas and the moving-breakpoint construction to check each claim.

For $(\alpha*\beta)*\gamma$, what are the three parameter intervals, and what formula is used on each?
Verify that $a(s)=\frac{1+s}{4}$ and $b(s)=\frac{2+s}{4}$ satisfy $0<a(s)<b(s)<1$ for every $s\in I$.
At $t=a(s)$, evaluate the two adjacent formulas for $H(s,t)$. Why do their values agree?
Explain why a homotopy between the two parenthesizations proves equality of classes but does not imply equality of paths.
For the circle loops in the third example, compute the value of the first parenthesization at $t=\frac18$ and compare it with the second.

Associativity of Loop Concatenation

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