Loops in an Interval
A loop may travel back and forth many times and still represent the same fundamental-group element as a constant loop. For an interval, the reason is especially direct: at each stage of a homotopy, move every point of the loop along the real line toward its base point. The interval property keeps the moving points inside the space. The endpoint check is essential: a homotopy of loops must keep the base point fixed, not merely end at a constant map.
Let \(J\subseteq\mathbb{R}\) be a nonempty interval, with the subspace topology inherited from the standard topology on \(\mathbb{R}\), and choose a base point \(x_0\in J\). Here, “interval” means that whenever \(u,v\in J\) and \(u\leq w\leq v\), then \(w\in J\). This includes open, closed, half-open, bounded, and unbounded intervals, as well as a one-point interval.
In a based homotopy, \(t\) marks a position along the loop, while \(s\) marks the stage of the deformation. Thus \(H(0,t)\) is the starting loop and \(H(1,t)\) is the ending loop. The two equations involving \(H(s,0)\) and \(H(s,1)\) require the endpoints to remain at \(x_0\) at every stage, including intermediate stages.
The Linear Contraction
For a based loop \(\gamma\), define the homotopy by taking a weighted average of the loop value and the base point:
At \(s=0\), the formula gives the original loop. At \(s=1\), it gives the constant loop at \(x_0\). For intermediate \(s\), it moves each value \(\gamma(t)\) toward \(x_0\) along the real line. Two facts make this formula work: an interval contains every point between any two of its points, and the loop has both endpoint values equal to \(x_0\).
Proof. We check the range, continuity, initial and final maps, and fixed endpoints. Since \(\gamma(t)\in J\) and \(x_0\in J\), the number \(H(s,t)\) is a convex combination of two points of \(J\): its coefficients \(1-s\) and \(s\) are nonnegative and sum to \(1\). Therefore \(H(s,t)\) lies between \(\gamma(t)\) and \(x_0\), so the interval property gives \(H(s,t)\in J\). The function \((s,t)\mapsto (1-s)\gamma(t)+s x_0\) is continuous because \(\gamma\), the coordinate functions, and addition and multiplication of real numbers are continuous. Substituting \(s=0\) gives \(H(0,t)=\gamma(t)\), and substituting \(s=1\) gives \(H(1,t)=x_0=c_{x_0}(t)\). Finally, for every \(s\in I\), the loop endpoint conditions give $$ H(s,0)=(1-s)x_0+s x_0=x_0 \quad\text{and}\quad H(s,1)=(1-s)x_0+s x_0=x_0. $$ Thus \(H\) is a continuous homotopy in \(J\) that fixes both endpoints throughout. \(\square\)
Computing the Fundamental Group
Proof. Let \(\gamma\) be any based loop at \(x_0\). By the proposition, \(\gamma\) is based-homotopic to the constant loop \(c_{x_0}\). Therefore every based loop represents the identity class \([c_{x_0}]\). Since every element of \(\pi_1(J,x_0)\) is represented by a based loop, this group has exactly one element. A group with one element is the trivial group. This reasoning also covers a one-point interval: its only possible loop is already constant. \(\square\)
The calculation does not depend on whether the interval includes its endpoints. The homotopy never needs a point beyond the interval: every value stays between a loop value and the chosen base point, both of which belong to \(J\). Nor does the argument require that the base point be an endpoint of the interval. It works for every \(x_0\in J\).
Worked Example: A Triangular Loop Based at 0
Take \(J=I=[0,1]\), \(x_0=0\), and define $$ \gamma(t)= \begin{cases} 2t,&0\leq t\leq \frac12,\\ 2-2t,&\frac12\leq t\leq1. \end{cases} $$ The two formulas agree at \(t=\frac12\), where both give \(1\); also \(\gamma(0)=0\) and \(\gamma(1)=0\). The loop rises to \(1\) and returns to \(0\).
The linear contraction is \(H(s,t)=(1-s)\gamma(t)\). Written on each part of the loop, it is $$ H(s,t)= \begin{cases} 2t(1-s),&0\leq t\leq \frac12,\\ (2-2t)(1-s),&\frac12\leq t\leq1. \end{cases} $$ At the joining point \(t=\frac12\), both expressions give \(1-s\). At \(s=0\), these pieces give the original formulas for \(\gamma\); at \(s=1\), both give \(0\). For any \(s,t\in I\), \(0\leq 1-s\leq1\) and \(0\leq\gamma(t)\leq1\), so \(0\leq H(s,t)\leq1\), as required for the range to lie in \(I\).
Endpoint check: \(H(s,0)=2(0)(1-s)=0\), and \(H(s,1)=(2-2)(1-s)=0\), for every \(s\in I\). For a sample intermediate value, \(H(\frac12,\frac14)=2(\frac14)(1-\frac12)=\frac14\), while \(H(\frac12,\frac12)=1(1-\frac12)=\frac12\). Thus at the halfway stage the loop has moved halfway toward the base point, without moving either endpoint.
Check: At \(s=0\), \(H(0,\frac14)=2(\frac14)(1)=\frac12=\gamma(\frac14)\). At \(s=1\), \(H(1,\frac14)=2(\frac14)(0)=0=c_0(\frac14)\). The formula has the asserted starting and ending loops.
Worked Example: A Loop Based at an Interior Point
Let \(J=[0,1]\), take \(x_0=\frac12\), and define a loop that goes from \(\frac12\) to \(1\), then to \(0\), and then back to \(\frac12\): $$ \gamma(t)= \begin{cases} \frac12+\frac32t,&0\leq t\leq\frac13,\\ 2-3t,&\frac13\leq t\leq\frac23,\\ \frac32t-1,&\frac23\leq t\leq1. \end{cases} $$ The successive values at the breakpoints are \(\gamma(0)=\frac12\), \(\gamma(\frac13)=1\), \(\gamma(\frac23)=0\), and \(\gamma(1)=\frac12\). The adjacent formulas agree at both breakpoints, and all values lie in \([0,1]\).
Its based contraction is $$ H(s,t)=(1-s)\gamma(t)+\frac{s}{2}. $$ For every \(t\), this is between \(\gamma(t)\in[0,1]\) and \(\frac12\), so it lies in \([0,1]\). At the start and end of the loop, $$ H(s,0)=(1-s)\frac12+\frac{s}{2}=\frac12, \qquad H(s,1)=(1-s)\frac12+\frac{s}{2}=\frac12. $$ At the halfway stage \(s=\frac12\), the contraction gives \(H(\frac12,t)=\frac12\gamma(t)+\frac14\). For \(t=0,\frac13,\frac23,1\), respectively, these values are \(\frac12,\frac34,\frac14,\frac12\). In particular, the loop has not yet become constant, but its endpoints remain fixed.
Check: At \(s=0\), \(H(0,t)=\gamma(t)\); at \(s=1\), \(H(1,t)=\frac12\) for every \(t\). The endpoint values at the halfway stage are both \(\frac12\), matching the base point.
Worked Example: A Smooth Oscillating Loop
Again take \(J=[0,1]\) and \(x_0=\frac12\), and set $$ \gamma(t)=\frac12+\frac14\sin(2\pi t). $$ Since \(-1\leq\sin(2\pi t)\leq1\), the loop values lie in \([\frac14,\frac34]\subseteq[0,1]\). Also \(\gamma(0)=\gamma(1)=\frac12\), so this is a based loop. Its contraction is $$ H(s,t)=(1-s)\left(\frac12+\frac14\sin(2\pi t)\right)+\frac{s}{2} =\frac12+\frac{1-s}{4}\sin(2\pi t). $$ The sine bound gives \(\frac12-\frac{1-s}{4}\leq H(s,t)\leq\frac12+\frac{1-s}{4}\). Because \(0\leq1-s\leq1\), this implies \(\frac14\leq H(s,t)\leq\frac34\), so the contraction stays inside \(J\).
At \(s=\frac12\), the formula is \(H(\frac12,t)=\frac12+\frac18\sin(2\pi t)\). Using the sine values at \(t=0,\frac14,\frac12,\frac34,1\), namely \(0,1,0,-1,0\), the corresponding contraction values are \(\frac12,\frac58,\frac12,\frac38,\frac12\). For every \(s\), the endpoint calculation is $$ H(s,0)=\frac12+\frac{1-s}{4}(0)=\frac12, \qquad H(s,1)=\frac12+\frac{1-s}{4}(0)=\frac12. $$ The oscillation shrinks to zero at \(s=1\), giving the constant loop.
A Common Mistake: Moving the Base Point
The endpoint check cannot be omitted. A formula that contracts values toward \(0\) may work for a loop based at \(0\), but it need not define a based homotopy for a loop with a different base point. For example, take the loop from the previous example, based at \(x_0=\frac12\), and try \(H_{\mathrm{wrong}}(s,t)=(1-s)\gamma(t)\). At the initial endpoint this gives $$ H_{\mathrm{wrong}}\left(\frac12,0\right) =\left(1-\frac12\right)\gamma(0) =\frac12\cdot\frac12 =\frac14, $$ which is not the required base point \(\frac12\). The same failure occurs at \(t=1\). This is a homotopy of maps into \([0,1]\), but it is not a based homotopy.
The correction is to move toward the actual base point, not automatically toward \(0\): $$ H(s,t)=(1-s)\gamma(t)+s\frac12. $$ Now \(H(s,0)=(1-s)\frac12+s\frac12=\frac12\), and \(H(s,1)=(1-s)\frac12+s\frac12=\frac12\) for every \(s\). In general, the safe formula is \((1-s)\gamma(t)+s x_0\), and its endpoint conditions should be checked explicitly.
The computation of \(\pi_1(J,x_0)\) rests on that distinction. A free deformation that ends at a constant map is not enough: the loop must be contracted while its base point remains fixed. For an interval, the linear formula meets both requirements, so every based loop belongs to the identity class.
Check Your Understanding
Use the linear contraction formula and check the endpoint conditions in each case.
- For a loop \(\gamma:I\to J\) based at \(x_0\), write out \(H(0,t)\), \(H(1,t)\), \(H(s,0)\), and \(H(s,1)\) for \(H(s,t)=(1-s)\gamma(t)+s x_0\).
- Explain why the interval property guarantees that \(H(s,t)\in J\), even when \(J\) is open or unbounded.
- For a loop based at \(x_0=2\), what does the incorrect formula \(H(s,t)=(1-s)\gamma(t)\) give at \(t=0\)? What formula corrects it?
- What is \(\pi_1(J,x_0)\) when \(J\) consists of a single point? Explain how this case fits the theorem.
- Why does showing that each loop is homotopic to some constant loop not suffice by itself unless that constant is the loop’s base point and the homotopy fixes the endpoints?