Why Reversing a Loop Cancels It
A loop may travel through a space in a complicated way, but if it immediately retraces that route, the resulting loop can be contracted to a point. The important condition here is that the contraction keeps the base point fixed throughout. We will construct such a contraction explicitly, both when the loop comes first and when its reversal comes first.
Let \(X\) be a topological space, let \(x_0\in X\), and let \(\alpha:I\to X\) be a loop based at \(x_0\), meaning that \(\alpha\) is continuous and \(\alpha(0)=\alpha(1)=x_0\). The constant loop at \(x_0\) is \(c_{x_0}(t)=x_0\) for every \(t\in I\). The reversal of \(\alpha\) is the path \(\bar{\alpha}(t)=\alpha(1-t)\). For paths with matching endpoints, use the concatenation convention \(\alpha*\beta\) traverses \(\alpha\) on the first half of \(I\) and \(\beta\) on the second half.
The Two Contractions
We need a parameter that starts at \(0\), rises to \(1\), and returns to \(0\) as \(t\) runs from \(0\) to \(1\). Set \(m(t)=1-|2t-1|\). This function is continuous, takes values in \([0,1]\), and satisfies \(m(0)=m(1)=0\), \(m(\frac12)=1\). More explicitly, \(m(t)=2t\) when \(0\leq t\leq\frac12\), and \(m(t)=2(1-t)\) when \(\frac12\leq t\leq1\).
Proof. First consider \(\alpha*\bar{\alpha}\). Define \(H_+(s,t)=\alpha((1-s)m(t))\). The map \((s,t)\mapsto(1-s)m(t)\) is continuous and takes values in \(I\), so \(H_+\) is continuous as a composition of continuous maps. At \(s=0\), for \(0\leq t\leq\frac12\), we have \(H_+(0,t)=\alpha(2t)\); for \(\frac12\leq t\leq1\), we have \(H_+(0,t)=\alpha(2-2t)=\bar{\alpha}(2t-1)\). Thus \(H_+(0,t)=(\alpha*\bar{\alpha})(t)\). At \(s=1\), \(H_+(1,t)=\alpha(0)=x_0\), so this is the constant loop. At either endpoint \(t=0\) or \(t=1\), \(m(t)=0\), giving \(H_+(s,0)=H_+(s,1)=\alpha(0)=x_0\) for every \(s\).
For the other order define \(H_-(s,t)=\alpha(1-(1-s)m(t))\). Its parameter also lies in \(I\): since \(0\leq(1-s)m(t)\leq1\), we have \(0\leq1-(1-s)m(t)\leq1\). It is a continuous function of \((s,t)\), so \(H_-\) is continuous. At \(s=0\), on the first half of \(I\) it is \(\alpha(1-2t)=\bar{\alpha}(2t)\); on the second half it is \(\alpha(2t-1)\). Hence \(H_-(0,t)=(\bar{\alpha}*\alpha)(t)\). At \(s=1\), \(H_-(1,t)=\alpha(1)=x_0\). When \(t=0\) or \(t=1\), \(m(t)=0\), so \(H_-(s,0)=H_-(s,1)=\alpha(1)=x_0\). Both homotopies therefore have the required fixed endpoints and final constant loop. \(\square\)
The formulas express the same geometric operation in two parameterizations. For \(\alpha*\bar{\alpha}\), the point reached at the midpoint is \(\alpha(1-s)\); as \(s\) increases, the farthest parameter reached along \(\alpha\) moves back toward \(0\). For \(\bar{\alpha}*\alpha\), the midpoint is \(\alpha(s)\); the excursion moves forward from \(\alpha(0)\) toward \(\alpha(1)\), which is the base point. The formulas work for any space \(X\); no geometric shape of the path or additional property of the space is required.
Worked Example: A Loop Around the Unit Circle
Give the first contraction for the loop \(\alpha:I\to S^1\subseteq\mathbb{C}\), where \(S^1\) has the subspace topology from \(\mathbb{C}\) and \(\alpha(t)=(\cos(2\pi t),\sin(2\pi t))\). The base point is \(x_0=(1,0)\). The contraction is \(H_+(s,t)=\alpha((1-s)m(t))\).
At \(s=0\), the loop is \(\alpha*\bar{\alpha}\). For example, at \(t=\frac14\), \(m(t)=\frac12\), so \(H_+(0,\frac14)=\alpha(\frac12)=(\cos\pi,\sin\pi)=(-1,0)\). At \(t=\frac34\), \(m(t)=\frac12\) again, so the value is also \((-1,0)\). At the midpoint \(t=\frac12\), \(m(t)=1\), giving \(H_+(0,\frac12)=\alpha(1)=(1,0)\), the point where the concatenated loop completes its first traversal and begins the reversal.
Now take \(s=\frac12\). At \(t=\frac14\), the parameter of \(\alpha\) is \((1-\frac12)\frac12=\frac14\), so \(H_+(\frac12,\frac14)=\alpha(\frac14)=(\cos\frac{\pi}{2},\sin\frac{\pi}{2})=(0,1)\). At \(t=\frac12\), the parameter is \(\frac12\), and the value is \(\alpha(\frac12)=(-1,0)\). At \(t=\frac34\), the parameter is again \(\frac14\), and the value is \((0,1)\). The excursion has become smaller, while both endpoints remain \((1,0)\).
Check: At \(s=1\), the parameter is \(0\) for every \(t\), so \(H_+(1,t)=\alpha(0)=(1,0)\). At \(t=0\) and \(t=1\), \(m(t)=0\) for every \(s\), so these endpoints also remain \((1,0)\).
Worked Example: Reversal Followed by the Original Loop
Let \(X=\mathbb{R}^2\) with its Euclidean topology. Define a loop based at \((0,0)\) by \(\alpha(t)=(3t,0)\) for \(0\leq t\leq\frac13\), \(\alpha(t)=(1,3t-1)\) for \(\frac13\leq t\leq\frac23\), and \(\alpha(t)=(3-3t,3-3t)\) for \(\frac23\leq t\leq1\). At the joining parameters, the formulas agree: at \(t=\frac13\) they give \((1,0)\), and at \(t=\frac23\) they give \((1,1)\). At \(t=0\) and \(t=1\) they give \((0,0)\), so this is a continuous loop.
Use the second contraction \(H_-(s,t)=\alpha(1-(1-s)m(t))\). At \(s=0\), the parameters of \(\alpha\) at \(t=\frac14,\frac12,\frac34\) are, respectively, \(1-\frac12=\frac12\), \(1-1=0\), and \(1-\frac12=\frac12\). The corresponding values are \(\alpha(\frac12)=(1,\frac12)\), \(\alpha(0)=(0,0)\), and \(\alpha(\frac12)=(1,\frac12)\), as required for \(\bar{\alpha}*\alpha\).
At \(s=\frac12\), those three parameters are \(1-\frac12\cdot\frac12=\frac34\), \(1-\frac12\cdot1=\frac12\), and \(1-\frac12\cdot\frac12=\frac34\). The definition of \(\alpha\) gives \(\alpha(\frac34)=(3-\frac94,3-\frac94)=(\frac34,\frac34)\), and \(\alpha(\frac12)=(1,\frac12)\). Thus the three values of the homotopy are \((\frac34,\frac34)\), \((1,\frac12)\), and \((\frac34,\frac34)\).
Check: At \(s=1\), the parameter is \(1\) for every \(t\), so \(H_-(1,t)=\alpha(1)=(0,0)\). At \(t=0\) and \(t=1\), it is also \(1\), confirming that the homotopy fixes both endpoints.
Worked Example: A Loop in the Real Line
Give the first contraction for \(X=\mathbb{R}\) with its standard topology and \(\alpha(t)=4t(1-t)\). This is a loop based at \(0\), since \(\alpha(0)=4\cdot0\cdot1=0\) and \(\alpha(1)=4\cdot1\cdot0=0\). Its contraction is \(H_+(s,t)=\alpha((1-s)m(t))\).
At \(s=0\) and \(t=\frac14\), the parameter is \(m(\frac14)=\frac12\), and the value is \(\alpha(\frac12)=4\cdot\frac12\cdot\frac12=1\). At \(t=\frac12\), the parameter is \(1\), so the value is \(\alpha(1)=0\); at \(t=\frac34\), the parameter is again \(\frac12\), giving \(1\). These values display the loop followed by its reversal.
At \(s=\frac14\), the parameter at \(t=\frac12\) is \((1-\frac14)\cdot1=\frac34\). Therefore \(H_+(\frac14,\frac12)=\alpha(\frac34)=4\cdot\frac34\cdot(1-\frac34)=4\cdot\frac34\cdot\frac14=\frac34\). At \(t=\frac14\), the parameter is \(\frac34\cdot\frac12=\frac38\), and \(\alpha(\frac38)=4\cdot\frac38\cdot\frac58=\frac{15}{16}\). At \(t=\frac34\), the parameter is again \(\frac38\), so the value is again \(\frac{15}{16}\).
Check: The endpoint values are \(H_+(\frac14,0)=H_+(\frac14,1)=\alpha(0)=0\). When \(s=1\), \(H_+(1,t)=\alpha(0)=0\) for every \(t\). Thus the computed intermediate values belong to a homotopy that keeps its endpoints fixed and ends at the constant loop.
A Common Parameterization Pitfall
A tempting but incorrect contraction is \(K(s,t)=\alpha((1-s)t)\). It shrinks the parameter on a single pass through \(\alpha\), not on the out-and-back path that defines \(\alpha*\bar{\alpha}\). Worse, it need not keep the endpoint fixed: \(K(s,1)=\alpha(1-s)\), which can differ from \(x_0\) when \(0<s<1\).
For a concrete check, take the unit-circle loop from the first example, \(s=0\), and \(t=\frac34\). The actual concatenation has value \((\alpha*\bar{\alpha})(\frac34)=\bar{\alpha}(\frac12)=\alpha(\frac12)=(-1,0)\). The proposed formula gives \(K(0,\frac34)=\alpha(\frac34)=(0,-1)\), which is different. The correction is to use a parameter that rises and falls, namely \(m(t)\), so the loop retraces its route and its endpoints remain at the base point during the contraction.
These homotopies give the explicit cancellation needed when a loop is paired with its reversal. Their key feature is not merely that the final loop is constant: the base point stays fixed at both ends for every intermediate loop. That is why the constructions apply to based loops and establish the reversal as an inverse in the fundamental group.
Check Your Understanding
For each question, check both the loop parameter and the endpoint conditions.
- For \(H_+(s,t)=\alpha((1-s)m(t))\), compute the argument of \(\alpha\) when \(s=\frac13\) and \(t=\frac12\).
- Write the parameter of \(\alpha\) in \(H_-(s,t)\) on each half of \(I\) when \(s=0\).
- Why does \(m(0)=m(1)=0\) ensure that \(H_+\) fixes the base point at both endpoints?
- For \(\alpha(t)=4t(1-t)\), compute \(H_+(\frac12,\frac12)\).
- Which required condition can fail for \(K(s,t)=\alpha((1-s)t)\), even if its final loop is constant?