Null-Homotopy and the Identity Class
A loop can be complicated as a parametrized path and still represent the identity element of a fundamental group. The relevant question is whether it can be continuously deformed to the constant loop without moving its basepoint. This requirement is important: a deformation that moves the endpoints is not an endpoint-preserving homotopy of based loops.
Let \((X,\mathcal{T})\) be a topological space, let \(x_0\in X\), and let \(I=[0,1]\) have its usual topology. A based loop at \(x_0\) is a continuous map \(\gamma:I\to X\) with \(\gamma(0)=\gamma(1)=x_0\). The constant loop \(c_{x_0}\) is defined by \(c_{x_0}(t)=x_0\) for every \(t\in I\). Two based loops are endpoint-preservingly homotopic if a continuous map \(H:I\times I\to X\) deforms one to the other while keeping both endpoints at \(x_0\) for every value of the homotopy parameter.
The fundamental group \(\pi_1(X,x_0)\) consists of endpoint-preserving homotopy classes of based loops, with multiplication induced by concatenation. Write \([\gamma]\) for the class of \(\gamma\). The identity element is \([c_{x_0}]\), since the constant loop represents the identity class.
Proof. By definition, \([\gamma]\) is the identity element exactly when \([\gamma]=[c_{x_0}]\). Equality of these equivalence classes means that \(\gamma\) and \(c_{x_0}\) are endpoint-preservingly homotopic. In the stated direction, that homotopy is precisely a continuous \(H:I\times I\to X\) satisfying \(H(0,t)=\gamma(t)\), \(H(1,t)=c_{x_0}(t)\), and \(H(s,0)=H(s,1)=x_0\). These conditions say exactly that \(\gamma\) is based null-homotopic. Conversely, any based null-homotopy satisfies those same conditions, so it is an endpoint-preserving homotopy from \(\gamma\) to \(c_{x_0}\). Therefore \([\gamma]=[c_{x_0}]\), the identity element. \(\square\)
The criterion is an equivalence, not a shortcut that removes the need to check a proposed deformation. To establish that a loop represents the identity, give a homotopy whose values stay in \(X\), whose formulas fit together continuously, and whose two endpoint tracks remain at \(x_0\).
Contractions in Convex Subsets of Euclidean Space
A useful source of based null-homotopies is a convex subset of \(\mathbb{R}^n\). Here \(\mathbb{R}^n\) has its Euclidean topology, and a subset \(C\subseteq\mathbb{R}^n\) has the subspace topology. Convexity means that whenever \(u,v\in C\) and \(0\leq\lambda\leq1\), the point \((1-\lambda)u+\lambda v\) belongs to \(C\).
Proof. Define \(H:I\times I\to\mathbb{R}^n\) by $$ H(s,t)=(1-s)\gamma(t)+s x_0. $$ For each \((s,t)\), the coefficients \(1-s\) and \(s\) are nonnegative and sum to \(1\); convexity therefore gives \(H(s,t)\in C\). The map \(H\) is continuous because \(\gamma\) is continuous and the displayed formula uses continuous scalar multiplication and addition in \(\mathbb{R}^n\). At \(s=0\), \(H(0,t)=\gamma(t)\). At \(s=1\), \(H(1,t)=x_0=c_{x_0}(t)\). For every \(s\), the loop endpoint conditions give $$ H(s,0)=(1-s)x_0+s x_0=x_0,\qquad H(s,1)=(1-s)x_0+s x_0=x_0. $$ Thus \(H\) is a based null-homotopy of \(\gamma\). \(\square\)
Worked Example: The Circle Loop Contracted in the Disk
Let \(D^2=\{(x,y)\in\mathbb{R}^2:x^2+y^2\leq1\}\) have the subspace topology from \(\mathbb{R}^2\). Consider the loop \(\gamma:I\to D^2\) given by \(\gamma(t)=(\cos(2\pi t),\sin(2\pi t))\). Its basepoint is \(x_0=(1,0)\), since \(\gamma(0)=\gamma(1)=(1,0)\). Define $$ H(s,t)=(1-s)(\cos(2\pi t),\sin(2\pi t))+s(1,0). $$
To check that this formula stays in \(D^2\), use the Euclidean norm and the triangle inequality: $$ \|H(s,t)\|\leq (1-s)\|(\cos(2\pi t),\sin(2\pi t))\|+s\|(1,0)\| =(1-s)\cdot1+s\cdot1=1. $$ Thus \(H(s,t)\in D^2\) for every \(s,t\). The starting loop is \(H(0,t)=\gamma(t)\), and the final loop is \(H(1,t)=(1,0)\). At either endpoint \(t=0\) or \(t=1\), the trigonometric vector is \((1,0)\), so \(H(s,0)=H(s,1)=(1,0)\) for every \(s\).
Check: at \(s=\tfrac12\) and \(t=\tfrac12\), the trigonometric vector is \((-1,0)\), and $$ H\left(\tfrac12,\tfrac12\right)=\tfrac12(-1,0)+\tfrac12(1,0)=(0,0). $$ Its norm is \(0\leq1\), as required. The theorem therefore gives a based null-homotopy in \(D^2\); this example makes no claim that the same formula takes values in the circle \(S^1\).
A Loop That Goes Out and Comes Back
Some null-homotopies come directly from the way a loop is built. If a path is followed by its reverse, the resulting loop can contract by shortening the portion of the path that it traverses.
Worked Example: A Path Followed by Its Reverse
Take \(X=\mathbb{R}^2\) with the Euclidean topology, \(x_0=(0,0)\), and the path \(\alpha:I\to X\) given by \(\alpha(u)=(u,0)\). The loop that follows \(\alpha\) and then reverses it is $$ \gamma(t)= \begin{cases} (2t,0),&0\leq t\leq\tfrac12,\\ (2-2t,0),&\tfrac12\leq t\leq1. \end{cases} $$ The formulas agree at \(t=\tfrac12\), where both give \((1,0)\); also \(\gamma(0)=\gamma(1)=(0,0)\). Define \(r(t)=2t\) for \(0\leq t\leq\tfrac12\) and \(r(t)=2-2t\) for \(\tfrac12\leq t\leq1\). Then \(r\) is continuous because its formulas agree at \(\tfrac12\), and \(\gamma(t)=(r(t),0)\). Set $$ H(s,t)=((1-s)r(t),0). $$
For every \(s,t\), this point lies in \(\mathbb{R}^2\). The formula is continuous on each of the two parameter regions, and the formulas agree at their shared edge \(t=\tfrac12\); hence \(H\) is continuous on \(I\times I\). At \(s=0\), \(H(0,t)=\gamma(t)\); at \(s=1\), \(H(1,t)=(0,0)\). Since \(r(0)=r(1)=0\), both endpoint tracks satisfy \(H(s,0)=H(s,1)=(0,0)\).
Check: at \(s=\tfrac12\), \(t=\tfrac14\), we have \(r(t)=2(\tfrac14)=\tfrac12\), so \(H(\tfrac12,\tfrac14)=(\tfrac14,0)\). At \(t=\tfrac34\), \(r(t)=2-2(\tfrac34)=\tfrac12\), and \(H(\tfrac12,\tfrac34)=(\tfrac14,0)\) as well. At \(s=1\), both values become \((0,0)\). This verifies the shortening in both halves of the loop and the constant final loop.
A Piecewise Linear Null-Homotopy
Worked Example: Contracting a Triangular Loop
Again let \(X=\mathbb{R}^2\) have its Euclidean topology and basepoint \(x_0=(0,0)\). Define a loop by traversing the triangle with vertices \((0,0)\), \((1,0)\), and \((0,1)\): $$ \gamma(t)= \begin{cases} (3t,0),&0\leq t\leq\tfrac13,\\ (2-3t,3t-1),&\tfrac13\leq t\leq\tfrac23,\\ (0,3-3t),&\tfrac23\leq t\leq1. \end{cases} $$ At \(t=\tfrac13\), the first two formulas both give \((1,0)\). At \(t=\tfrac23\), the last two formulas both give \((0,1)\). The endpoints are \((0,0)\), so \(\gamma\) is a loop based at \(x_0\). Define \(H(s,t)=(1-s)\gamma(t)\).
Each piece of \(\gamma\) is continuous and the pieces agree at their shared endpoints, so \(\gamma\), and therefore \(H\), is continuous. Every value of \(H\) lies in \(\mathbb{R}^2\). The initial loop is \(H(0,t)=\gamma(t)\); the final loop is \(H(1,t)=(0,0)\); and \(H(s,0)=H(s,1)=(0,0)\) because \(\gamma(0)=\gamma(1)=(0,0)\).
Check: at \(t=\tfrac12\), the middle formula gives \(\gamma(\tfrac12)=(2-\tfrac32,\tfrac32-1)=(\tfrac12,\tfrac12)\). Therefore \(H(\tfrac12,\tfrac12)=(\tfrac14,\tfrac14)\). At \(s=1\), the same input gives \((0,0)\), and at \(t=0\) the value is \((0,0)\) for every \(s\). These substitutions check the contraction and the fixed basepoint directly.
Common Pitfall: Leaving the Space or Moving the Basepoint
A formula that contracts a loop in one space need not define a homotopy in a smaller space. For example, the circle loop in the first worked example has values in \(S^1=\{(x,y)\in\mathbb{R}^2:x^2+y^2=1\}\). The attempted formula \(K(s,t)=(1-s)\gamma(t)\) has \(K(1,t)=(0,0)\), but \((0,0)\notin S^1\). It therefore is not a homotopy into \(S^1\). Moreover, at \(s=\tfrac12\) and \(t=0\), it gives \(K(\tfrac12,0)=(\tfrac12,0)\), not the basepoint \((1,0)\). So it fails both the requirement to stay in the circle and the requirement to fix the basepoint.
The correction is not to ignore either condition: choose a homotopy whose image lies in the space in question and whose endpoint tracks are constant. The first worked example succeeds in \(D^2\) because the disk contains the straight segments used by its formula, and the formula was arranged to keep the endpoints at \((1,0)\). A contraction in \(D^2\) is not automatically a contraction in its subspace \(S^1\).
Null-homotopy is useful because it translates a statement about an element of \(\pi_1(X,x_0)\) into a concrete task: construct a continuous map on the square and verify its boundary conditions. The identity-class criterion says exactly when that task succeeds. A proposed deformation proves the identity-class claim only after its continuity, image, initial and final loops, and fixed endpoints have all been checked.
Check Your Understanding
For each question, state the space and basepoint involved, and distinguish a based homotopy from a deformation that merely moves the loop.
- Write the four conditions that a based null-homotopy \(H:I\times I\to X\) must satisfy.
- Why does the formula \(H(s,t)=(1-s)\gamma(t)+s x_0\) stay in a convex subset of \(\mathbb{R}^n\)? Which two endpoint equalities must also be checked?
- For the path \(\alpha(u)=(u,0)\), compute the loop \(\alpha\) followed by its reverse on each half of \(I\), and verify that the two formulas agree at \(t=\tfrac12\).
- In the triangular-loop example, evaluate \(\gamma(\tfrac12)\) and \(H(\tfrac12,\tfrac12)\).
- Why is \(K(s,t)=(1-s)\gamma(t)\) not a based null-homotopy of the circle loop as a map into \(S^1\)?