Changing the Timing of a Path
A path records both where it goes and when it gets there. The parameter \(t\in[0,1]\) supplies the timing: at parameter \(t\), the path is at a particular point of its space. A new parameter map can make the path move at a different pace, or pause for an interval, without changing the ordered journey. The key conditions are that the new parameter map is continuous, nondecreasing, and covers the whole unit interval.
Throughout, \(I=[0,1]\) has the subspace topology inherited from the standard topology on \(\mathbb{R}\), and \(X\) is a topological space with topology \(\mathcal{T}\). A path in \(X\) is a continuous map \(\gamma:I\to X\). Its initial point is \(\gamma(0)\), its terminal point is \(\gamma(1)\), and its trace is the subset \(\gamma(I)=\{\gamma(t):t\in I\}\) of \(X\).
Continuity ensures that the new map is still a path. The nondecreasing condition prevents the parameter from moving backward along the original path, while surjectivity ensures that every original parameter value is still reached. These conditions play separate roles; the examples below show what can go wrong when one is omitted.
The Endpoint and Trace Theorem
Proof. First, \(\gamma\circ\phi\) is continuous. If \(U\) is open in \(X\), then \(\gamma^{-1}(U)\) is open in \(I\) because \(\gamma\) is continuous, and \(\phi^{-1}(\gamma^{-1}(U))\) is open in \(I\) because \(\phi\) is continuous. This preimage is exactly \((\gamma\circ\phi)^{-1}(U)\).
Since \(\phi\) is nondecreasing and surjective onto \(I\), its least value is \(0\) and its greatest value is \(1\). In detail, surjectivity gives some \(t_0,t_1\in I\) with \(\phi(t_0)=0\) and \(\phi(t_1)=1\). Nondecreasingness gives \(\phi(0)\le\phi(t_0)=0\) and \(\phi(1)\ge\phi(t_1)=1\). Since \(\phi(0),\phi(1)\in I\), these inequalities imply \(\phi(0)=0\) and \(\phi(1)=1\). Therefore \((\gamma\circ\phi)(0)=\gamma(0)\) and \((\gamma\circ\phi)(1)=\gamma(1)\).
For the trace, every value of \(\gamma\circ\phi\) is a value of \(\gamma\), because \(\phi(t)\in I\) for all \(t\in I\). Thus \((\gamma\circ\phi)(I)\subseteq\gamma(I)\). Conversely, for any \(s\in I\), surjectivity gives \(t\in I\) with \(\phi(t)=s\), so \(\gamma(s)=(\gamma\circ\phi)(t)\). Hence \(\gamma(I)\subseteq(\gamma\circ\phi)(I)\). The two traces are equal. \(\square\)
The theorem does not say that the two paths have the same value at each parameter. It says that they start and finish at the same points and visit the same set of points. Their timing can differ, and a nondecreasing map can hold the parameter still over an interval.
Worked Example: A Path That Pauses
Let \(X=\mathbb{R}\) with its standard topology and take \(\gamma(t)=t\) on \(I\). Define \[ \phi(t)= \begin{cases} \frac{3}{2}t,&0\le t\le\frac{1}{3},\\ \frac{1}{2},&\frac{1}{3}\le t\le\frac{2}{3},\\ \frac{3}{2}t-\frac{1}{2},&\frac{2}{3}\le t\le1. \end{cases} \]
At \(t=\frac{1}{3}\), the first formula gives \(\frac{3}{2}\cdot\frac{1}{3}=\frac{1}{2}\), agreeing with the middle formula. At \(t=\frac{2}{3}\), the last formula gives \(\frac{3}{2}\cdot\frac{2}{3}-\frac{1}{2}=1-\frac{1}{2}=\frac{1}{2}\), again agreeing with the middle formula. Thus \(\phi\) is continuous at the joins; it is continuous on each of the three intervals as well. Its slopes on those intervals are \(\frac{3}{2}\), \(0\), and \(\frac{3}{2}\), so it is nondecreasing. The first interval gives all values from \(0\) to \(\frac{1}{2}\), and the last gives all values from \(\frac{1}{2}\) to \(1\); hence \(\phi\) is surjective onto \(I\).
The reparametrized path is \((\gamma\circ\phi)(t)=\phi(t)\). It remains at \(\frac{1}{2}\) for every \(t\in[\frac{1}{3},\frac{2}{3}]\), so it pauses there. Its endpoints are \(\phi(0)=0\) and \(\phi(1)=\frac{3}{2}-\frac{1}{2}=1\). Its trace is \([0,\frac{1}{2}]\cup\{\frac{1}{2}\}\cup[\frac{1}{2},1]=[0,1]\), the trace of \(\gamma\).
Check: at \(t=\frac{1}{6}\), \(\phi(t)=\frac{3}{2}\cdot\frac{1}{6}=\frac{1}{4}\), so the new path is at the point \(\frac{1}{4}\). At \(t=\frac{5}{6}\), \(\phi(t)=\frac{3}{2}\cdot\frac{5}{6}-\frac{1}{2}=\frac{5}{4}-\frac{1}{2}=\frac{3}{4}\). These values lie in the original trace, and the two pieces of the formula cover the whole interval of original parameter values.
Nonlinear Timing
A reparametrization need not contain a pause. For example, \(t\mapsto t^2\) moves slowly near \(0\) and more quickly later, while still reaching every parameter in \(I\). The next example checks the trace directly by matching each new parameter to an original one.
Worked Example: A Nonlinear Reparametrization of a Parabola
Use \(X=\mathbb{R}^2\) with its standard topology and let \(\gamma(t)=(t,t^2)\). Set \(\phi(t)=t^2\). The map \(\phi\) is continuous. If \(s\le t\) in \(I\), then \(t^2-s^2=(t-s)(t+s)\ge0\), so it is nondecreasing. For each \(u\in I\), \(t=\sqrt{u}\) belongs to \(I\) and satisfies \(\phi(t)=u\), so \(\phi\) is surjective.
Substitution gives \[ (\gamma\circ\phi)(t)=\gamma(t^2)=(t^2,(t^2)^2)=(t^2,t^4). \] At \(t=0\), this is \((0,0)=\gamma(0)\); at \(t=1\), it is \((1,1)=\gamma(1)\). For any \(u\in I\), choose \(t=\sqrt{u}\). Then \[ (\gamma\circ\phi)(t)=((\sqrt{u})^2,(\sqrt{u})^4)=(u,u^2)=\gamma(u). \] Thus every point of the original trace is attained by the new path; the reverse inclusion also follows directly because \(t^2\in I\) for every \(t\in I\).
Check: at \(t=\frac{1}{2}\), the new path has value \((\frac{1}{4},\frac{1}{16})\). The original path at parameter \(\frac{1}{4}\) has value \((\frac{1}{4},(\frac{1}{4})^2)=(\frac{1}{4},\frac{1}{16})\), as required. The timing changed, but the point lies on the same parabola segment.
Composing Reparametrizations
Proof. The composition \(\psi\circ\phi\) is continuous because both maps are continuous. If \(s\le t\), then \(\phi(s)\le\phi(t)\), and the nondecreasing property of \(\psi\) gives \(\psi(\phi(s))\le\psi(\phi(t))\). Thus the composition is nondecreasing. To prove surjectivity, take any \(u\in I\). Since \(\psi\) is surjective, there is \(v\in I\) with \(\psi(v)=u\); since \(\phi\) is surjective, there is \(t\in I\) with \(\phi(t)=v\). Consequently \((\psi\circ\phi)(t)=u\). Finally, associativity of function composition gives \((\gamma\circ\psi)\circ\phi=\gamma\circ(\psi\circ\phi)\), and the endpoint and trace conclusions follow from the theorem. \(\square\)
This result lets us combine timing changes without checking the resulting path from scratch: the combined parameter map still satisfies all three defining conditions. A related local consequence explains the word “pause.” If \(\phi\) is constant on an interval \([a,b]\subseteq I\), say \(\phi(t)=c\) there, then for every \(t\in[a,b]\), \((\gamma\circ\phi)(t)=\gamma(c)\). The reparametrized path is therefore constant throughout that interval.
Why Surjectivity Matters
Worked Example: A Continuous Map That Misses Part of the Trace
Keep \(X=\mathbb{R}\) with its standard topology and \(\gamma(t)=t\), but use \(\psi(t)=\frac{t}{2}\). This map is continuous and nondecreasing, and its values range from \(0\) to \(\frac{1}{2}\). In particular, it is not surjective onto \(I\), since it never takes the value \(1\).
The composite is \((\gamma\circ\psi)(t)=\frac{t}{2}\), so its trace is \([0,\frac{1}{2}]\), not \([0,1]\). Its initial point is \(0\), but its terminal point is \(\frac{1}{2}\), not \(\gamma(1)=1\). For instance, the original path reaches \(\frac{3}{4}\) at parameter \(\frac{3}{4}\), while the new path would need \(t=\frac{3}{2}\) to reach \(\frac{3}{4}\), and \(\frac{3}{2}\notin I\).
Common pitfall. A tempting but incorrect calculation is: “\(\psi\) is continuous and nondecreasing, so \((\gamma\circ\psi)(I)=\gamma(I)=[0,1]\).” The computation above shows the error: the composite’s values are \(\{t/2:t\in[0,1]\}=[0,\frac{1}{2}]\). Continuity and nondecreasingness alone do not guarantee that every original parameter value is reached. Surjectivity is essential to the trace conclusion, and in this example its failure also prevents preservation of the terminal point.
When checking a proposed reparametrization, verify each condition separately: continuity, nondecreasingness, and surjectivity onto the entire interval. Once all three hold, the theorem guarantees that the path remains continuous, has the same endpoints, and has exactly the same trace. A flat portion is allowed; skipping part of the parameter interval is not.
Check Your Understanding
For each question, identify which condition on the parameter map is being used.
- Let \(\phi(t)=3t-2t^2\) on \(I\). Is it continuous and nondecreasing on \(I\), and does it map onto \(I\)? Explain how you determine each property.
- For \(\gamma(t)=t\) in \(\mathbb{R}\), find the endpoints and trace of \(\gamma\circ\phi\) when \(\phi(t)=t^3\).
- Suppose \(\phi\) is a reparametrization map and \(\phi(t)=\frac{2}{5}\) for all \(t\in[\frac{1}{4},\frac{3}{4}]\). What does \(\gamma\circ\phi\) do on that interval?
- Give a continuous nondecreasing map \(I\to I\) that is not surjective, and use \(\gamma(t)=t\) to show that the trace can change.
- If \(\phi\) and \(\psi\) are reparametrization maps, state the combined parameter map for applying \(\psi\) first and then \(\phi\), and explain why its image is all of \(I\).