Why Convexity Contracts Loops
A loop in a convex subset of Euclidean space can be contracted by moving each of its points along the straight segment to the base point. Convexity ensures that these segments stay in the space. Choosing the base point as the destination is essential: the homotopy must keep the two endpoints of the loop fixed at every stage.
Let \(C\subseteq\mathbb{R}^n\) have the subspace topology inherited from the standard topology on \(\mathbb{R}^n\). A subset \(C\) is convex if, whenever \(u,v\in C\) and \(0\leq\lambda\leq1\), the point \((1-\lambda)u+\lambda v\) also belongs to \(C\). The set may be open, closed, or neither; the argument below uses convexity, not closedness or boundedness.
The Straight-Line Contraction
Proof. Let \(\gamma:I\to C\) be any continuous loop based at \(x_0\). Define \[ H(s,t)=(1-s)\gamma(t)+s x_0,\qquad s,t\in I. \] For each \(s,t\), the coefficients \(1-s\) and \(s\) are nonnegative and sum to \(1\). Since \(\gamma(t)\in C\), \(x_0\in C\), and \(C\) is convex, \(H(s,t)\in C\). The formula is continuous as a map into \(\mathbb{R}^n\), because \(\gamma\) and the coordinate operations are continuous. Since its image lies in \(C\), it is also continuous as a map into \(C\) with the subspace topology.
At \(s=0\), \(H(0,t)=\gamma(t)\); at \(s=1\), \(H(1,t)=x_0\). For every \(s\in I\), \[ H(s,0)=(1-s)x_0+s x_0=x_0,\qquad H(s,1)=(1-s)x_0+s x_0=x_0. \] Thus \(H\) is a based homotopy to the constant loop. The loop \(\gamma\) was arbitrary, so every element of \(\pi_1(C,x_0)\) is the identity class. Hence the group has exactly one element. \(\square\)
The proof gives more than a conclusion about a group: it gives a specific homotopy for each loop. The parameter \(s\) measures the progress of the contraction, while \(t\) runs around the loop. At each fixed \(t\), the point travels along the segment from \(\gamma(t)\) to \(x_0\).
Proof. Take any \(u,v\in C\). Define \(\alpha:I\to C\) by \(\alpha(t)=(1-t)u+tv\). Convexity implies \(\alpha(t)\in C\) for every \(t\in I\), and the formula is continuous. Also \(\alpha(0)=u\) and \(\alpha(1)=v\). Thus any two points of \(C\) can be joined by a path in \(C\), which is the definition of path-connectedness. \(\square\)
Worked Contractions
Worked Example: A Loop in an Interval
Let \(C=[-2,3]\subseteq\mathbb{R}\), with its subspace topology, and let \(x_0=1\). Consider \[ \gamma(t)=1+\sin(2\pi t),\qquad t\in I. \] Because \(-1\leq\sin(2\pi t)\leq1\), we have \(0\leq\gamma(t)\leq2\), so the loop lies in \(C\). Its endpoint values are \(\gamma(0)=1+\sin(0)=1\) and \(\gamma(1)=1+\sin(2\pi)=1\).
The straight-line contraction is \[ H(s,t)=(1-s)\gamma(t)+s =(1-s)\bigl(1+\sin(2\pi t)\bigr)+s =1+(1-s)\sin(2\pi t). \] Since \(0\leq1-s\leq1\), the sine term lies between \(-1\) and \(1\), so \(0\leq H(s,t)\leq2\). Thus every value remains in \(C\). At \(s=0\), this is \(\gamma(t)\); at \(s=1\), it is \(1\). At both \(t=0\) and \(t=1\), the sine is zero, so \(H(s,0)=H(s,1)=1\) for every \(s\).
Check: At \(s=\tfrac12\) and \(t=\tfrac14\), \(\sin(2\pi t)=\sin(\pi/2)=1\), so \(H(\tfrac12,\tfrac14)=1+(1-\tfrac12)1=\tfrac32\), which lies in \([-2,3]\). At \(s=1\), the formula gives \(H(1,t)=1\) for every \(t\), as required.
Worked Example: A Loop in the Closed Unit Disk
Let \(C=D^2=\{(x,y)\in\mathbb{R}^2:x^2+y^2\leq1\}\), with the subspace topology, and take \(x_0=(\tfrac12,0)\). Consider the loop \[ \gamma(t)=\left(\tfrac12\cos(2\pi t),\tfrac12\sin(2\pi t)\right). \] At \(t=0\) and \(t=1\), the cosine is \(1\) and the sine is \(0\), so both endpoints equal \(x_0\). For all \(t\), \(\|\gamma(t)\|=\tfrac12\), so the loop lies in \(D^2\).
The contraction is \[ H(s,t)=(1-s)\gamma(t)+s\left(\tfrac12,0\right) =\left(\tfrac{1-s}{2}\cos(2\pi t)+\tfrac{s}{2}, \frac{1-s}{2}\sin(2\pi t)\right). \] The Euclidean norm satisfies the triangle inequality. Since \(\|\gamma(t)\|=\tfrac12\) and \(\|x_0\|=\tfrac12\), \[ \|H(s,t)\|\leq(1-s)\|\gamma(t)\|+s\|x_0\| =(1-s)\tfrac12+s\tfrac12=\tfrac12\leq1. \] Therefore \(H(s,t)\in D^2\). The formula has \(H(0,t)=\gamma(t)\), \(H(1,t)=x_0\), and \(H(s,0)=H(s,1)=x_0\).
Check: At \(t=\tfrac14\), the trigonometric values are \(\cos(\pi/2)=0\) and \(\sin(\pi/2)=1\), so \(\gamma(\tfrac14)=(0,\tfrac12)\). At \(s=\tfrac12\), this gives \(H(\tfrac12,\tfrac14)=(\tfrac14,\tfrac14)\), whose squared norm is \((\tfrac14)^2+(\tfrac14)^2=\tfrac18\leq1\). At \(s=1\), both coordinates in the formula reduce to \(\tfrac12\) and \(0\), respectively.
Worked Example: A Polygonal Loop in a Triangle
Let \[ C=\{(x,y)\in\mathbb{R}^2:x\geq0,\ y\geq0,\ x+y\leq1\}, \] with its subspace topology, and choose the base point \(x_0=(0,0)\). This set is convex: a convex combination of two points with nonnegative coordinates still has nonnegative coordinates, and its coordinate sum is a convex combination of sums at most \(1\).
Define a loop by \[ \gamma(t)= \begin{cases} (3t,0),&0\leq t\leq\tfrac13,\\ (2-3t,3t-1),&\tfrac13\leq t\leq\tfrac23,\\ (0,3-3t),&\tfrac23\leq t\leq1. \end{cases} \] On the first interval, \(0\leq3t\leq1\); on the second, both coordinates are nonnegative and their sum is \((2-3t)+(3t-1)=1\); on the third, \(0\leq3-3t\leq1\). Thus every segment lies in \(C\). The formulas agree at the joins: at \(t=\tfrac13\) they give \((1,0)\), and at \(t=\tfrac23\) they give \((0,1)\). The endpoints at \(t=0,1\) are both \((0,0)\), so \(\gamma\) is a continuous based loop.
Since the base point is the origin, the contraction is \(H(s,t)=(1-s)\gamma(t)\), where the scalar multiplies both coordinates. If \(\gamma(t)=(x,y)\in C\), then the coordinates of \(H(s,t)\) are nonnegative and their sum is \((1-s)(x+y)\leq1\). Hence \(H(s,t)\in C\). At \(s=0\) it is \(\gamma\), at \(s=1\) it is the constant loop at the origin, and at \(t=0,1\) it remains the origin.
Check: At \(t=\tfrac12\), the middle formula gives \(\gamma(\tfrac12)=(2-\tfrac32,\tfrac32-1)=(\tfrac12,\tfrac12)\). At \(s=\tfrac12\), \(H(\tfrac12,\tfrac12)=(\tfrac14,\tfrac14)\); its coordinates are nonnegative and sum to \(\tfrac12\leq1\). This verifies membership in the triangle at an interior point of the contraction.
What the Convexity Hypothesis Does—and Does Not—Say
The theorem applies to every nonempty convex subset of Euclidean space, whether bounded or unbounded and whether open or closed. For example, it applies to an open ball, a closed half-space, a line segment, and a single point, each with its Euclidean subspace topology. In the single-point case there is only the constant loop, and the fundamental group is the one-element group. The nonempty condition matters because a fundamental group requires a chosen base point; the empty set has none.
Convexity supplies the particular straight segments used in the proof. Without convexity, the same formula need not define a homotopy in the space. This observation is about the method; it does not assert that every nonconvex space has a nontrivial fundamental group.
Common pitfall. Contracting to a point other than the base point can fail to be a based homotopy. For a concrete calculation, take \(C=[0,1]\) with its usual topology, base point \(x_0=0\), and the constant loop \(\gamma(t)=0\). If one tries to contract toward \(p=1\), the formula becomes \[ H(s,t)=(1-s)\gamma(t)+s p=s. \] For example, \(H(\tfrac12,0)=\tfrac12\ne0=x_0\). The loop endpoint has moved, so this formula does not satisfy the definition of based homotopy. Contracting instead to \(x_0=0\) gives \(H(s,t)=(1-s)0+s0=0\), which keeps both endpoints fixed.
A nonconvex set also shows why the image condition must be checked rather than assumed. For the unit circle \(S^1\subseteq\mathbb{R}^2\), the loop \(\gamma(t)=(\cos(2\pi t),\sin(2\pi t))\) is based at \((1,0)\). At \(t=\tfrac12\), its value is \((-1,0)\). The straight segment from that point to the base point passes through \((0,0)\), which is not in \(S^1\). Thus the direct straight-line formula leaves the space. The convex-set theorem makes no claim about this nonconvex example.
For a convex set, however, no separate calculation of loop classes is needed: the displayed homotopy works for every continuous based loop, regardless of its shape or parametrization. Once a base point \(x_0\in C\) is selected, the computation is always \(\pi_1(C,x_0)=\{1\}\).
Check Your Understanding
Use the straight-line formula and the based-homotopy conditions to answer the following.
- For a nonempty convex \(C\subseteq\mathbb{R}^n\), write the contraction of a loop \(\gamma\) based at \(x_0\), and verify the two endpoint conditions.
- Why does the formula \(H(s,t)=(1-s)\gamma(t)+s p\) fail to be based when \(p\ne x_0\)?
- Give a reason, using the definition of convexity, that an open Euclidean ball is covered by the theorem.
- For the triangle in the worked example, check directly that scaling any \((x,y)\in C\) by a number in \([0,1]\) keeps it in \(C\).
- Does the straight-line contraction argument apply to every nonconvex subset of Euclidean space? Explain what must be checked about the homotopy's image.