Changing the Parameter Without Changing the Path Class
A path records a continuous journey through a space, with its parameter increasing from \(0\) to \(1\). A different parameter can make the journey run at a different speed, or pause at some points. If that change preserves the starting and finishing parameters, it should not change the path up to homotopy with endpoints fixed. The key is to interpolate the parameters before applying the path; this works even when the space itself has no linear structure.
Throughout, \(I=[0,1]\) has its usual subspace topology from \(\mathbb{R}\), and \(X\) is a topological space. A path in \(X\) is a continuous map \(\gamma:I\to X\). Its initial and terminal points are \(\gamma(0)\) and \(\gamma(1)\).
Thus the homotopy changes the path continuously while keeping its two endpoint values unchanged throughout. We will construct such a homotopy by changing only the input to \(\gamma\).
The Parameter Interpolation
Proof. The functions \((s,t)\mapsto 1-s\), \((s,t)\mapsto\phi(t)\), \((s,t)\mapsto s\), and \((s,t)\mapsto t\) are continuous on \(I\times I\). Products and sums of real-valued continuous functions are continuous, so \(\theta\) is continuous. Since \(s\) and \(1-s\) are nonnegative and sum to \(1\), \(\theta(s,t)\) is a weighted average of \(\phi(t)\in I\) and \(t\in I\); hence \(0\leq\theta(s,t)\leq1\). Substitution gives \(\theta(0,t)=\phi(t)\), \(\theta(1,t)=t\), \(\theta(s,0)=(1-s)\phi(0)=0\), and \(\theta(s,1)=(1-s)\phi(1)+s=1\). \(\square\)
Proof. Define \(H:I\times I\to X\) by \(H(s,t)=\gamma(\theta(s,t))\), where \(\theta(s,t)=(1-s)\phi(t)+st\). The lemma shows that \(\theta\) is a continuous map into \(I\). Since \(\gamma\) is continuous, the composition \(H=\gamma\circ\theta\) is continuous. At \(s=0\), \(H(0,t)=\gamma(\phi(t))\); at \(s=1\), \(H(1,t)=\gamma(t)\). At the endpoints of each intermediate path, \(H(s,0)=\gamma(0)\) and \(H(s,1)=\gamma(1)\), by the endpoint identities for \(\theta\). These are exactly the conditions for a homotopy with endpoints fixed. \(\square\)
The nondecreasing condition means that the parameter does not move backward: if \(t_1\leq t_2\), then \(\phi(t_1)\leq\phi(t_2)\). It allows pauses, where \(\phi\) is constant on an interval, and stretches, where it increases more quickly. The displayed homotopy proof in fact uses only continuity, the range \(\phi(I)\subseteq I\), and the endpoint conditions; monotonicity is not needed for that conclusion. It is included because it expresses the usual meaning of changing speed without reversing the order of traversal.
Worked Reparametrizations
Worked Example: A Slower Parameter on the Unit Interval
Take \(X=I\) with its usual topology, \(\gamma(t)=t^2\), and \(\phi(t)=t^2\). The map \(\gamma\) is a path because it is continuous and takes values in \(I\). The map \(\phi\) is continuous, nondecreasing on \(I\), and has \(\phi(0)=0^2=0\) and \(\phi(1)=1^2=1\). The reparametrized path is \((\gamma\circ\phi)(t)=\gamma(t^2)=(t^2)^2=t^4\).
The theorem's homotopy is \(H(s,t)=\gamma((1-s)t^2+st)=((1-s)t^2+st)^2\). At \(t=\tfrac12\), the reparametrized path has value \((\tfrac12)^4=\tfrac1{16}\), while the original path has value \((\tfrac12)^2=\tfrac14\). At the intermediate homotopy time \(s=\tfrac12\), the interpolated parameter is \((1-\tfrac12)(\tfrac12)^2+\tfrac12(\tfrac12)=\tfrac18+\tfrac14=\tfrac38\); therefore \(H(\tfrac12,\tfrac12)=(\tfrac38)^2=\tfrac9{64}\).
Check: For every \(s\in I\), \(H(s,0)=((1-s)0+s0)^2=0=\gamma(0)\) and \(H(s,1)=((1-s)1+s1)^2=1=\gamma(1)\). Also \(H(0,t)=(t^2)^2=t^4\) and \(H(1,t)=t^2\), so the two ends of the homotopy are the required paths.
Worked Example: Reaching the Endpoint Early
Let \(X=I\), \(\gamma(t)=t\), and set \(\phi(t)=\min(2t,1)\). Explicitly, \(\phi(t)=2t\) for \(0\leq t\leq\tfrac12\), and \(\phi(t)=1\) for \(\tfrac12\leq t\leq1\). The formulas agree at \(t=\tfrac12\), since \(2(\tfrac12)=1\), so \(\phi\) is continuous. It is nondecreasing, takes values in \(I\), and satisfies \(\phi(0)=0,\ \phi(1)=1\). The reparametrized path is \((\gamma\circ\phi)(t)=\min(2t,1)\); it reaches \(1\) at \(t=\tfrac12\) and then stays there.
Here the homotopy is \(H(s,t)=(1-s)\min(2t,1)+st\). For \(t=\tfrac14\), \(\phi(\tfrac14)=\min(\tfrac12,1)=\tfrac12\), so \(H(\tfrac12,\tfrac14)=\tfrac12\cdot\tfrac12+\tfrac12\cdot\tfrac14=\tfrac38\). For \(t=\tfrac34\), \(\phi(\tfrac34)=\min(\tfrac32,1)=1\), so \(H(\tfrac12,\tfrac34)=\tfrac12\cdot1+\tfrac12\cdot\tfrac34=\tfrac78\).
Check: At \(s=0\) the formula gives \(\min(2t,1)\), and at \(s=1\) it gives \(t\). At \(t=0\) it gives \(0\) for every \(s\); at \(t=1\) it gives \((1-s)1+s1=1\). Thus the endpoint values stay fixed even while the early arrival is gradually removed.
Worked Example: A Pause on a Circle
Give \(S^1=\{(x,y)\in\mathbb{R}^2:x^2+y^2=1\}\) its subspace topology from \(\mathbb{R}^2\). Define \(\gamma(t)=(\cos(2\pi t),\sin(2\pi t))\), a path in \(S^1\), and let \(\phi(t)=0\) for \(0\leq t\leq\tfrac13\), while \(\phi(t)=\tfrac{3t-1}{2}\) for \(\tfrac13\leq t\leq1\). At \(t=\tfrac13\), the second expression is \((1-1)/2=0\), so the pieces agree. The second expression increases from \(0\) to \((3-1)/2=1\), so \(\phi\) is continuous, nondecreasing, and fixes \(0\) and \(1\).
At \(t=\tfrac13\), the reparametrized path is \(\gamma(0)=(1,0)\): the path has paused at its initial point. At the same path parameter and homotopy time \(s=\tfrac12\), the interpolated parameter is \(\theta(\tfrac12,\tfrac13)=(1-\tfrac12)0+\tfrac12\cdot\tfrac13=\tfrac16\). Thus \(H(\tfrac12,\tfrac13)=\gamma(\tfrac16)=(\cos(\pi/3),\sin(\pi/3))=(\tfrac12,\tfrac{\sqrt3}{2})\).
For another check, at \(t=\tfrac23\) we have \(\phi(\tfrac23)=(2-1)/2=\tfrac12\). Hence \(\theta(\tfrac12,\tfrac23)=\tfrac12\cdot\tfrac12+\tfrac12\cdot\tfrac23=\tfrac14+\tfrac13=\tfrac7{12}\), and \(H(\tfrac12,\tfrac23)=\gamma(\tfrac7{12})=(\cos(7\pi/6),\sin(7\pi/6))=(-\tfrac{\sqrt3}{2},-\tfrac12)\). The endpoints are \(\gamma(0)=(1,0)\) and \(\gamma(1)=(1,0)\), and the theorem's formula fixes both throughout the homotopy.
Why the Formula Uses Parameters, Not Path Values
A tempting alternative is to interpolate directly between the two path values: \((1-s)\gamma(\phi(t))+s\gamma(t)\). That expression is not guaranteed to lie in \(X\). The correct construction interpolates the numbers \(\phi(t)\) and \(t\), which both lie in \(I\), and then applies \(\gamma\). Its values therefore remain on the path, in any topological space \(X\).
Common pitfall. This issue is visible even for a simple path in \(S^1\). Use the circle path \(\gamma(t)=(\cos(\pi t),\sin(\pi t))\), the continuous nondecreasing map \(\phi(t)=t^2\), and the direct value-interpolation formula at \(s=\tfrac12,\ t=\tfrac12\). The two path values are \(\gamma(\phi(\tfrac12))=\gamma(\tfrac14)=(\tfrac{\sqrt2}{2},\tfrac{\sqrt2}{2})\) and \(\gamma(\tfrac12)=(0,1)\). Their average is \((\tfrac{\sqrt2}{4},\tfrac{2+\sqrt2}{4})\). Its squared distance from the origin is \((\tfrac{\sqrt2}{4})^2+(\tfrac{2+\sqrt2}{4})^2 =\tfrac18+\tfrac{6+4\sqrt2}{16} =\tfrac{2+\sqrt2}{4}\ne1\). So this average is not in \(S^1\), and direct value interpolation does not define a homotopy into the circle. By contrast, parameter interpolation gives \(\theta(\tfrac12,\tfrac12)=\tfrac12\cdot\tfrac14+\tfrac12\cdot\tfrac12=\tfrac38\), and \(\gamma(\tfrac38)\) lies in \(S^1\) by its definition.
The endpoint assumptions also matter. For example, if \(\gamma(t)=t\) in \(I\) and \(\phi(t)=t/2\), then \(\phi(1)=1/2\), so \((\gamma\circ\phi)(1)=1/2\ne1=\gamma(1)\). No homotopy with endpoints fixed can join these two paths: the required terminal value would have to equal both \(1/2\) and \(1\). Continuity matters as well. The map equal to \(0\) for \(t<1/2\) and \(1\) for \(t\geq1/2\) fixes the endpoints and is nondecreasing, but is discontinuous; composing it with \(\gamma(t)=t\) does not produce a path.
The construction gives a useful principle: when a path is traversed in the same order with a continuous change of parameter fixing the endpoints, the specific speed profile is not an obstruction to endpoint-preserving homotopy. The proof does not require coordinates on \(X\), a metric on \(X\), or an inverse for \(\phi\); it uses only the interval parameters and continuity of the path.
Check Your Understanding
For each question, keep separate the map on the parameter interval and the path into the space.
- For \(\gamma(t)=t^3\) in \(I\) and \(\phi(t)=t^2\), write the reparametrized path and the endpoint-preserving homotopy given by the theorem.
- Verify directly that \(\phi(t)=\min(3t,1)\) is continuous, nondecreasing, and fixes the endpoints of \(I\).
- In the parameter-interpolation formula, why does \(\theta(s,t)\) remain in \(I\) for every \((s,t)\in I\times I\)?
- Can the homotopy theorem's conclusion hold for a continuous \(\phi:I\to I\) that is not nondecreasing but still fixes both endpoints? Explain using the formula.
- Give an example of two paths with different terminal points and explain why they cannot be homotopic with endpoints fixed.