Concatenation and Reversal of Loops

Paths, Loops, and Their Endpoints

Concatenation joins one path to another when the first path ends where the second begins. Reversal changes the direction in which a path is traversed. Both operations change the parametrization, so their formulas make the roles of the endpoints explicit.

Throughout, $I=[0,1]$ has its usual topology as a subspace of $\mathbb{R}$, and $(X,\mathcal{T})$ is a topological space. A path in $X$ is a continuous map $\gamma:I\to X$. Its starting point is $\gamma(0)$, and its ending point is $\gamma(1)$. A loop based at $x_0$ is a path $\gamma$ such that $\gamma(0)=\gamma(1)=x_0$.

These are definitions for actual parametrized paths. When we write down an operation, the resulting map must still be continuous to be a path. We will verify continuity, not just calculate the endpoints.

Concatenation

Definition: Let $\gamma,\delta:I\to X$ be paths with $\gamma(1)=\delta(0)$. Their concatenation $\gamma*\delta:I\to X$ is defined by $$ (\gamma*\delta)(t)= \begin{cases} \gamma(2t), & 0\le t\le \frac12,\\ \delta(2t-1), & \frac12\le t\le 1. \end{cases} $$ The first path is traversed on the first half of $I$, and the second path on the second half. The two formulas agree at $t=\frac12$, since both give $\gamma(1)=\delta(0)$.

Theorem: If $\gamma$ and $\delta$ are paths in $X$ and $\gamma(1)=\delta(0)$, then $\gamma*\delta$ is a path from $\gamma(0)$ to $\delta(1)$.

Proof. The two formulas defining $\gamma*\delta$ are continuous on their respective closed intervals $[0,\frac12]$ and $[\frac12,1]$: each is a composition of a given continuous path with a continuous affine map. They agree at the shared endpoint because $\gamma(2\cdot\frac12)=\gamma(1)=\delta(0)=\delta(2\cdot\frac12-1)$. To verify continuity of the combined map, let $F$ be closed in $X$. On each of the two closed subintervals, the preimage of $F$ is closed relative to that subinterval, hence closed in $I$. The preimage $(\gamma*\delta)^{-1}(F)$ is the union of those two closed sets, so is closed in $I$. The closed-set test for continuity now shows that $\gamma*\delta$ is continuous. Finally, $(\gamma*\delta)(0)=\gamma(0)$ and $(\gamma*\delta)(1)=\delta(1)$, proving the endpoint claim. $\square$

Worked Example: Joining Two Paths in the Real Line

Use the standard topology on $\mathbb{R}$. Let $\gamma(t)=2-3t$ and $\delta(t)=-1+4t$ for $t\in I$. The endpoints are $\gamma(0)=2$, $\gamma(1)=2-3=-1$, $\delta(0)=-1+0=-1$, and $\delta(1)=-1+4=3$. Since $\gamma(1)=\delta(0)$, concatenation is defined.

For $0\le t\le\frac12$, substitute $2t$ into $\gamma$: $(\gamma*\delta)(t)=2-3(2t)=2-6t$. For $\frac12\le t\le1$, substitute $2t-1$ into $\delta$: $(\gamma*\delta)(t)=-1+4(2t-1)=-1+8t-4=8t-5$. Thus $$ (\gamma*\delta)(t)= \begin{cases} 2-6t, & 0\le t\le \frac12,\\ 8t-5, & \frac12\le t\le 1. \end{cases} $$

Check: At the join, the first expression gives $2-6(\frac12)=2-3=-1$, and the second gives $8(\frac12)-5=4-5=-1$. The endpoints are $(\gamma*\delta)(0)=2-0=2$ and $(\gamma*\delta)(1)=8-5=3$, as required. This concatenated path starts where $\gamma$ starts and ends where $\delta$ ends.

If $\gamma$ and $\delta$ are both loops based at the same point $x_0$, then $\gamma(1)=x_0=\delta(0)$, so their concatenation is defined. The endpoint calculation gives $(\gamma*\delta)(0)=x_0$ and $(\gamma*\delta)(1)=x_0$, so it is again a loop based at $x_0$. The condition that the base points match matters: loops based at different points cannot be concatenated in this order unless the ending point of the first is the starting point of the second.

Reversal

Definition: For a path $\gamma:I\to X$, its reversal $\overline{\gamma}:I\to X$ is $$ \overline{\gamma}(t)=\gamma(1-t). $$ It traverses the same parametrized path in the opposite direction.

Theorem: If $\gamma$ is a path from $x_0$ to $x_1$, then $\overline{\gamma}$ is a path from $x_1$ to $x_0$. In particular, reversal takes a loop based at $x_0$ to a loop based at $x_0$.

Proof. The map $r:I\to I$ given by $r(t)=1-t$ is continuous, so $\overline{\gamma}=\gamma\circ r$ is continuous as a composition of continuous maps. Its starting point is $\overline{\gamma}(0)=\gamma(1-0)=x_1$, and its ending point is $\overline{\gamma}(1)=\gamma(1-1)=x_0$. If $x_1=x_0$, these endpoints are both $x_0$, as claimed. $\square$

Worked Example: Reversing and Concatenating Circle Loops

Give $S^1=\{(x,y)\in\mathbb{R}^2:x^2+y^2=1\}$ the subspace topology from $\mathbb{R}^2$. Consider the loop $\alpha(t)=(\cos(2\pi t),\sin(2\pi t))$, based at $(1,0)$, and the loop $\beta(t)=(\cos(4\pi t),\sin(4\pi t))$, also based at $(1,0)$. Their endpoints match, so $\alpha*\beta$ is defined.

The reversal of $\alpha$ is $\overline{\alpha}(t)=(\cos(2\pi(1-t)),\sin(2\pi(1-t)))$. Using $\cos(2\pi-2\pi t)=\cos(2\pi t)$ and $\sin(2\pi-2\pi t)=-\sin(2\pi t)$, this becomes $\overline{\alpha}(t)=(\cos(2\pi t),-\sin(2\pi t))$. For instance, $\overline{\alpha}(0)=(\cos0,-\sin0)=(1,0)$, and $\overline{\alpha}(\frac14)=(\cos\frac{\pi}{2},-\sin\frac{\pi}{2})=(0,-1)$. The reversed loop still begins and ends at $(1,0)$.

For the concatenation, the formula on the first half is $\alpha(2t)=(\cos(4\pi t),\sin(4\pi t))$. On the second half it is $\beta(2t-1)=(\cos(8\pi t-4\pi),\sin(8\pi t-4\pi))$. Therefore $$ (\alpha*\beta)(t)= \begin{cases} (\cos(4\pi t),\sin(4\pi t)), & 0\le t\le\frac12,\\ (\cos(8\pi t-4\pi),\sin(8\pi t-4\pi)), & \frac12\le t\le1. \end{cases} $$

Check: At $t=\frac12$, the first expression is $(\cos2\pi,\sin2\pi)=(1,0)$; the second is $(\cos0,\sin0)=(1,0)$. At $t=0$ the value is $(\cos0,\sin0)=(1,0)$, and at $t=1$ it is $(\cos4\pi,\sin4\pi)=(1,0)$. At $t=\frac14$, the value is $(\cos\pi,\sin\pi)=(-1,0)$; at $t=\frac58$, it is $(\cos\pi,\sin\pi)=(-1,0)$. These values show the first loop is completed on the first half of the parameter interval and the second loop is traversed on the second half.

Tracking a Concatenation Through Its Parameter

Because each component receives a rescaled part of the interval, a parameter value in the concatenation does not usually mean the same thing as that value in either original path. For $t\le\frac12$, the first path is evaluated at $2t$; for $t\ge\frac12$, the second is evaluated at $2t-1$. The join is at $t=\frac12$, regardless of the shapes or speeds of the original paths.

Worked Example: Concatenating Two Loops on the Real Line

Give $\mathbb{R}$ its standard topology. Define two loops based at $0$ by $$ \gamma(t)= \begin{cases} 2t, & 0\le t\le\frac12,\\ 2-2t, & \frac12\le t\le1, \end{cases} \qquad \delta(t)= \begin{cases} -2t, & 0\le t\le\frac12,\\ -2+2t, & \frac12\le t\le1. \end{cases} $$ For $\gamma$, the two values at $t=\frac12$ are $2(\frac12)=1$ and $2-2(\frac12)=1$; its endpoints are $0$ and $2-2=0$. For $\delta$, the two values at $t=\frac12$ are $-2(\frac12)=-1$ and $-2+2(\frac12)=-1$; its endpoints are $0$ and $-2+2=0$. Thus both are continuous loops based at $0$, and $\gamma*\delta$ is defined.

Substitution into the concatenation formula gives $$ (\gamma*\delta)(t)= \begin{cases} 4t, & 0\le t\le\frac14,\\ 2-4t, & \frac14\le t\le\frac12,\\ 2-4t, & \frac12\le t\le\frac34,\\ 4t-4, & \frac34\le t\le1. \end{cases} $$ For example, on the third interval the argument $2t-1$ lies between $0$ and $\frac12$, so the first formula for $\delta$ gives $\delta(2t-1)=-2(2t-1)=2-4t$. On the fourth interval the argument lies between $\frac12$ and $1$, so the second formula gives $\delta(2t-1)=-2+2(2t-1)=4t-4$.

Check: The values at $t=\frac14,\frac12,\frac34$ from the adjacent formulas are, respectively, $4(\frac14)=1$ and $2-4(\frac14)=1$; $2-4(\frac12)=0$ and $2-4(\frac12)=0$; and $2-4(\frac34)=-1$ and $4(\frac34)-4=-1$. The full path has values $0$ at $t=0$ and $t=1$, so its endpoints are the original base point. At the sample parameters $t=\frac14,\frac12,\frac34$, its values are $1,0,-1$, respectively.

A Common Pitfall: Matching the Join

The piecewise formula alone does not guarantee a path. Its two pieces must agree at the join. For a concrete failure in $\mathbb{R}$, let $\gamma(t)=t$ and $\delta(t)=2+t$. The first path ends at $\gamma(1)=1$, while the second begins at $\delta(0)=2$. If one nevertheless applies the two formulas, the value approaching the join from the first half is $\gamma(1)=1$, but the value from the second formula at the join is $\delta(0)=2$. The resulting map is discontinuous at $t=\frac12$, so it is not a path.

The correction is to check the matching condition before defining concatenation. For example, replacing $\delta$ by $\delta(t)=1+t$ gives $\delta(0)=1=\gamma(1)$. The formulas then agree at the join: the first gives $1$, and the second gives $1$. The concatenation is a path from $0$ to $2$. Reversal has its own endpoint check: it swaps the start and end, so reversing a path is not generally a loop unless its original endpoints agree.

Concatenation and reversal are operations on parametrized paths, with endpoint conditions that can be checked directly. The calculations above establish that they produce continuous paths and determine their endpoints. Further algebraic claims about the corresponding equivalence classes require additional arguments; endpoint calculations alone do not establish those claims.

Check Your Understanding

For each question, write down the endpoint condition before applying a formula.

Let $\gamma(t)=3-2t$ and $\delta(t)=1+5t$ in $\mathbb{R}$. Is $\gamma*\delta$ defined? If so, calculate its piecewise formula and its endpoints.
If $\gamma$ is a path from $a$ to $b$, what are the starting and ending points of $\overline{\gamma}$?
Let $\alpha(t)=(\cos(\pi t),\sin(\pi t))$ in $S^1$. Find the endpoints of $\alpha$ and determine whether $\alpha*\alpha$ is defined.
For two loops based at different points, what condition would be needed for their concatenation in a specified order to be defined?
Explain why agreeing values at the joining parameter are necessary for the piecewise concatenation formula to define a continuous path.