Reversing Paths and Concatenating Segments

Reversing and Joining Paths

A path records a continuous journey through a topological space: its parameter begins at 0 and ends at 1. Reversing a path changes the direction of travel, while concatenating paths joins journeys whose endpoints match. The formulas must account for the parameter interval as well as the points in the space. In particular, each part of a concatenated path must be rescaled to the portion of $[0,1]$ assigned to it.

Throughout, $I=[0,1]$ has its usual topology as a subspace of $\mathbb{R}$. A path in a topological space $(X,\mathcal{T})$ is a continuous map $\gamma:I\to X$. Its initial point is $\gamma(0)$, and its terminal point is $\gamma(1)$. The paths below need not lie in a metric space; $\mathbb{R}$ has its standard topology and $\mathbb{R}^2$ its Euclidean topology when those spaces are used.

Definition: For a path $\gamma:I\to X$, its reverse is the map $\gamma^{\mathrm{rev}}:I\to X$ defined by $\gamma^{\mathrm{rev}}(t)=\gamma(1-t)$. If $\gamma(1)=\eta(0)$ for paths $\gamma,\eta:I\to X$, their concatenation is the map $\gamma*\eta:I\to X$ defined by $$ (\gamma*\eta)(t)= \begin{cases} \gamma(2t),&0\le t\le \frac12,\\ \eta(2t-1),&\frac12\le t\le 1. \end{cases} $$

The two formulas for concatenation use different parameters in the original paths. As $t$ runs from 0 to $1/2$, $2t$ runs from 0 to 1, so the first path is completed. As $t$ runs from $1/2$ to 1, $2t-1$ runs from 0 to 1, so the second path is completed. At the joining parameter $t=1/2$, the first formula gives $\gamma(1)$, and the second gives $\eta(0)$. The matching-endpoint condition makes these values equal.

Reversal: Continuity and Endpoints

Theorem: If $\gamma:I\to X$ is a path, then $\gamma^{\mathrm{rev}}$ is a path, with $\gamma^{\mathrm{rev}}(0)=\gamma(1)$ and $\gamma^{\mathrm{rev}}(1)=\gamma(0)$.

Proof. The map $r:I\to I$ defined by $r(t)=1-t$ is continuous, since it is the restriction of a continuous affine map on $\mathbb{R}$. The reverse is the composition $\gamma\circ r$, so it is continuous. Substitution at the endpoints gives $\gamma^{\mathrm{rev}}(0)=\gamma(1-0)=\gamma(1)$ and $\gamma^{\mathrm{rev}}(1)=\gamma(1-1)=\gamma(0)$. Thus it is a path with the stated endpoints. $\square$

Worked Example: Reversing a Path in the Plane

In $\mathbb{R}^2$ with the Euclidean topology, let $\gamma(t)=(2t,t^2)$ for $0\le t\le1$. Its initial point is $\gamma(0)=(0,0)$, and its terminal point is $\gamma(1)=(2,1)$. Find and check its reverse.

Substitute $1-t$ for the parameter in both coordinates: $$ \gamma^{\mathrm{rev}}(t)=\gamma(1-t)=(2(1-t),(1-t)^2)=(2-2t,1-2t+t^2). $$ At $t=0$, this is $(2-0,1-0+0)=(2,1)=\gamma(1)$. At $t=1$, it is $(2-2,1-2+1)=(0,0)=\gamma(0)$. Each coordinate is a polynomial in $t$, hence continuous, so this formula is a continuous path.

Check by reversing again: $(\gamma^{\mathrm{rev}})^{\mathrm{rev}}(t)=\gamma^{\mathrm{rev}}(1-t)=\gamma(1-(1-t))=\gamma(t)$. Thus reversing twice recovers the original path at every parameter.

Concatenation: Continuity and Endpoints

The two pieces in a concatenation are defined on the closed subintervals $[0,1/2]$ and $[1/2,1]$. We use the following pasting fact, and include its proof to account for continuity at their shared endpoint.

Lemma (Closed-set pasting): Suppose a topological space $Z$ is the union of two closed subsets $A$ and $B$. If $f_A:A\to Y$ and $f_B:B\to Y$ are continuous and agree on $A\cap B$, then the map $f:Z\to Y$ defined by the two maps on their respective subsets is continuous.

Proof. Agreement on $A\cap B$ makes $f$ well-defined. Let $F$ be any closed subset of $Y$. By the continuity of the two maps, $f_A^{-1}(F)$ is closed in $A$ and $f_B^{-1}(F)$ is closed in $B$. Since $A$ and $B$ are closed in $Z$, a relatively closed subset of either is closed in $Z$. Also, $$ f^{-1}(F)=f_A^{-1}(F)\cup f_B^{-1}(F), $$ which is a finite union of closed subsets of $Z$, hence closed. Therefore preimages of closed sets under $f$ are closed. Taking complements shows that preimages of open sets are open, so $f$ is continuous. $\square$

Theorem: If $\gamma,\eta:I\to X$ are paths with $\gamma(1)=\eta(0)$, then $\gamma*\eta$ is a path, and $$ (\gamma*\eta)(0)=\gamma(0),\qquad (\gamma*\eta)(1)=\eta(1). $$

Proof. The intervals $A=[0,1/2]$ and $B=[1/2,1]$ are closed in $I$ and cover $I$. On $A$, the formula is the continuous composition $t\mapsto\gamma(2t)$; on $B$, it is the continuous composition $t\mapsto\eta(2t-1)$. At the only point in $A\cap B$, the two values are $\gamma(2\cdot\frac12)=\gamma(1)$ and $\eta(2\cdot\frac12-1)=\eta(0)$, which agree by hypothesis. The closed-set pasting lemma therefore gives continuity on all of $I$. At the endpoints, the applicable formulas give $(\gamma*\eta)(0)=\gamma(0)$ and $(\gamma*\eta)(1)=\eta(1)$. Hence the concatenation is a path with the stated endpoints. $\square$

Worked Example: Joining Two Line Segments

In $\mathbb{R}^2$, let $\gamma(t)=(2t,0)$ and $\eta(t)=(2,3t)$, for $t\in I$. The first path goes from $(0,0)$ to $(2,0)$; the second goes from $(2,0)$ to $(2,3)$. Their endpoints match, so their concatenation is defined.

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

Check at the join: the first branch at $t=1/2$ is $(4\cdot\frac12,0)=(2,0)$; the second is $(2,6\cdot\frac12-3)=(2,0)$. At $t=0$ the value is $(0,0)$, and at $t=1$ it is $(2,6-3)=(2,3)$. These are exactly the initial point of $\gamma$ and terminal point of $\eta$. The theorem also guarantees continuity at the join.

Joining Three Paths

For three paths $\gamma_1,\gamma_2,\gamma_3:I\to X$ with $\gamma_1(1)=\gamma_2(0)$ and $\gamma_2(1)=\gamma_3(0)$, one may assign one third of the parameter interval to each path. On the $j$-th subinterval $[(j-1)/3,j/3]$, the rescaled parameter is $3t-(j-1)$, which runs from 0 to 1. This gives the following piecewise formula; the matching endpoints ensure agreement at the joins.

Worked Example: Concatenating Three Segments

In $\mathbb{R}^2$, define $\gamma_1(s)=(s,0)$, $\gamma_2(s)=(1,s)$, and $\gamma_3(s)=(1-s,1-s)$ for $s\in I$. Their successive endpoints are $(0,0)$ to $(1,0)$, then $(1,0)$ to $(1,1)$, then $(1,1)$ to $(0,0)$. Thus the required endpoints match.

On the first third, use $s=3t$, giving $\gamma_1(3t)=(3t,0)$. On the second third, use $s=3t-1$, giving $\gamma_2(3t-1)=(1,3t-1)$. On the final third, use $s=3t-2$, giving $\gamma_3(3t-2)=(1-(3t-2),1-(3t-2))=(3-3t,3-3t)$. Therefore the concatenated path is $$ \Gamma(t)= \begin{cases} (3t,0),&0\le t\le\frac13,\\ (1,3t-1),&\frac13\le t\le\frac23,\\ (3-3t,3-3t),&\frac23\le t\le1. \end{cases} $$

Check the first join: at $t=1/3$, the first branch gives $(3\cdot\frac13,0)=(1,0)$, and the second gives $(1,3\cdot\frac13-1)=(1,0)$. Check the second join: at $t=2/3$, the second branch gives $(1,3\cdot\frac23-1)=(1,1)$, and the third gives $(3-3\cdot\frac23,3-3\cdot\frac23)=(1,1)$. The endpoint values are $\Gamma(0)=(0,0)$ and $\Gamma(1)=(3-3,3-3)=(0,0)$, as required. Each branch is continuous, and the formulas agree at both joins, so applying the closed-set pasting lemma to the three closed subintervals shows that $\Gamma$ is continuous.

A Common Rescaling Error

When using a path on a subinterval, the rescaled parameter must start at 0 at the left endpoint and finish at 1 at the right endpoint. For the second half of a two-path concatenation, that parameter is $2t-1$, not $2t$.

Common pitfall. In the two-segment example, using $\eta(2t)$ on $1/2\le t\le1$ would give $\eta(2\cdot\frac12)=\eta(1)=(2,3)$ at the join. The first segment ends there at $(2,0)$, so this incorrect formula does not define a continuous joined path. The correction is to use $\eta(2t-1)$: at $t=1/2$ its parameter is $2\cdot\frac12-1=0$, and its value is $\eta(0)=(2,0)$, matching the first segment. At $t=1$, its parameter is $2-1=1$, so it completes the second path at $(2,3)$.

Reversal and concatenation are useful for constructing paths with prescribed directions and successive stages. The endpoint calculations are part of the construction, not a formality: they show both that the pieces join and that the resulting path has the intended initial and terminal points.

Check Your Understanding

For each question, use the definitions and check the endpoint or joining values by substitution.

If $\gamma(t)=(t,2t)$ in $\mathbb{R}^2$, write its reverse and calculate both endpoint values.
Suppose $\gamma(1)\ne\eta(0)$. Which condition in the definition or continuity proof of $\gamma*\eta$ fails?
For paths $\gamma,\eta$ with $\gamma(1)=\eta(0)$, evaluate the two branch formulas for $\gamma*\eta$ at $t=1/2$.
For three matching paths assigned successive thirds of $I$, what rescaled parameter is used on the third subinterval $[2/3,1]$? Check its values at both ends.
Explain why continuity of each formula on its own subinterval does not, by itself, establish continuity of the piecewise-defined map at a join.