Constant Loops as the Identity

Why the Constant Loop Acts as an Identity

Concatenating two based loops means traversing the first loop and then the second, with each assigned half of the parameter interval. If one of the loops is constant, this definition appears to insert an extra pause at the base point. The pause changes the parametrized path, but it does not change its based homotopy class. We will show this by explicitly deforming the concatenation into the original loop while keeping both endpoints fixed throughout.

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 $\alpha:I\to X$ satisfying $\alpha(0)=\alpha(1)=x_0$. The constant loop at $x_0$ is $c_{x_0}:I\to X$, defined by $c_{x_0}(t)=x_0$ for every $t\in I$.

Definition: For based loops $\alpha,\beta$ at $x_0$, their concatenation $\alpha*\beta$ is the loop that follows $\alpha$ on the first half of $I$ and $\beta$ on the second half: $$ (\alpha*\beta)(t)= \begin{cases} \alpha(2t),&0\le t\le \frac12,\\ \beta(2t-1),&\frac12\le t\le 1. \end{cases} $$ A based homotopy from $\gamma_0$ to $\gamma_1$ is a continuous map $H:I\times I\to X$ such that $H(0,t)=\gamma_0(t)$, $H(1,t)=\gamma_1(t)$, and $H(s,0)=H(s,1)=x_0$ for every $s,t\in I$. The based homotopy class of $\alpha$ is written $[\alpha]$.

The two concatenations to consider have different parameter formulas. In $c_{x_0}*\alpha$, the path stays at $x_0$ for the first half of the interval and then traverses $\alpha$. In $\alpha*c_{x_0}$, it traverses $\alpha$ during the first half and then stays at $x_0$. The proof uses a parameter function that gradually changes each of these schedules into the ordinary schedule $t\mapsto t$.

A Parameter-Change Construction

Lemma: Define $$ \phi_L(t)=\max\{0,2t-1\}, \qquad \phi_R(t)=\min\{2t,1\} $$ for $t\in I$. For either choice $\phi=\phi_L$ or $\phi=\phi_R$, the map $$ \phi_s(t)=(1-s)\phi(t)+st $$ is continuous as a function of $(s,t)\in I\times I$, takes values in $I$, and satisfies $\phi_s(0)=0$ and $\phi_s(1)=1$ for all $s\in I$.

Proof. Each of $\phi_L$ and $\phi_R$ is piecewise linear: its formulas on either side of $t=\frac12$ agree at $t=\frac12$, so each is continuous. Both take values in $[0,1]$, with value $0$ at $t=0$ and value $1$ at $t=1$. For $s,t\in I$, $\phi_s(t)$ is a weighted average of $\phi(t)\in[0,1]$ and $t\in[0,1]$, with nonnegative weights $1-s$ and $s$ whose sum is $1$. Thus $\phi_s(t)\in[0,1]$. The formula for $\phi_s(t)$ is continuous in $(s,t)$, because $\phi$, the coordinate functions, and sums and products of continuous real-valued functions are continuous. Finally, $\phi_s(0)=(1-s)0+s0=0$ and $\phi_s(1)=(1-s)1+s1=1$. $\square$

Theorem (Constant loop identity): For every based loop $\alpha$ at $x_0$, the loops $c_{x_0}*\alpha$ and $\alpha*c_{x_0}$ are based homotopic to $\alpha$. Consequently, $$ [c_{x_0}]\cdot[\alpha]=[\alpha] \qquad\text{and}\qquad [\alpha]\cdot[c_{x_0}]=[\alpha], $$ where the product of loop classes is defined by concatenation.

Proof. For the left concatenation define $H_L:I\times I\to X$ by $H_L(s,t)=\alpha((1-s)\phi_L(t)+st)$. The function inside $\alpha$ is continuous by the lemma, so $H_L$ is continuous as a composition of continuous maps. At $s=0$, if $0\le t\le\frac12$, then $\phi_L(t)=0$; if $\frac12\le t\le1$, then $\phi_L(t)=2t-1$. Therefore $H_L(0,t)=x_0$ on the first half and $H_L(0,t)=\alpha(2t-1)$ on the second half, exactly $c_{x_0}*\alpha$. At $s=1$, $H_L(1,t)=\alpha(t)$. For every $s$, the lemma gives $H_L(s,0)=\alpha(0)=x_0$ and $H_L(s,1)=\alpha(1)=x_0$. Thus this is a based homotopy from $c_{x_0}*\alpha$ to $\alpha$.

For the right concatenation define $H_R(s,t)=\alpha((1-s)\phi_R(t)+st)$. This map is continuous for the same reason. At $s=0$, $\phi_R(t)=2t$ when $0\le t\le\frac12$, and $\phi_R(t)=1$ when $\frac12\le t\le1$. Hence $H_R(0,t)=\alpha(2t)$ on the first half and $H_R(0,t)=x_0$ on the second half, exactly $\alpha*c_{x_0}$. At $s=1$, $H_R(1,t)=\alpha(t)$. At every stage the endpoints are $H_R(s,0)=\alpha(0)=x_0$ and $H_R(s,1)=\alpha(1)=x_0$. This is a based homotopy from $\alpha*c_{x_0}$ to $\alpha$. The two homotopies give the asserted equalities of classes by the definition of based homotopy class and the concatenation product. $\square$

The important feature is that the homotopy changes only the speed at which the loop is traversed. It never changes the loop’s image by applying an operation to points of $X$; this is why the construction works for an arbitrary topological space and an arbitrary based loop in it.

Worked Homotopies in Specific Spaces

Worked Example: A Loop in the Real Line

Give explicit left and right identity homotopies for $\alpha(t)=t(1-t)$ in $\mathbb{R}$ with its standard topology. Its values lie in $[0,\frac14]$, and $\alpha(0)=0=\alpha(1)$, so it is a based loop at $0$. Substituting into the left concatenation gives $$ (c_0*\alpha)(t)= \begin{cases} 0,&0\le t\le\frac12,\\ (2t-1)(1-(2t-1)),&\frac12\le t\le1. \end{cases} $$ For the right concatenation, the result is $2t(1-2t)$ on $0\le t\le\frac12$ and $0$ on $\frac12\le t\le1$.

The homotopies are $H_L(s,t)=\alpha((1-s)\phi_L(t)+st)$ and $H_R(s,t)=\alpha((1-s)\phi_R(t)+st)$. For a numerical check of the left homotopy, take $s=\frac12$ and $t=\frac14$. Then $\phi_L(\frac14)=\max\{0,\frac12-1\}=0$, so the new parameter is $(1-\frac12)0+\frac12\cdot\frac14=\frac18$. The path value is $\alpha(\frac18)=\frac18(1-\frac18)=\frac{7}{64}$. For the right homotopy at $s=\frac12,t=\frac34$, $\phi_R(\frac34)=\min\{\frac32,1\}=1$, so the new parameter is $\frac12\cdot1+\frac12\cdot\frac34=\frac78$, and the path value is $\alpha(\frac78)=\frac78(1-\frac78)=\frac{7}{64}$.

Check: At $t=0$, both homotopy parameters are $0$, giving $\alpha(0)=0$. At $t=1$, both parameters are $1$, giving $\alpha(1)=0$. At $s=1$, each parameter is $t$, so both homotopies end at $\alpha(t)$, as required.

Worked Example: A Once-Traversed Circle

Take $X=S^1=\{z\in\mathbb{C}:|z|=1\}$ with the subspace topology from $\mathbb{C}$, base point $1$, and $\alpha(t)=e^{2\pi i t}$. This is a based loop since $\alpha(0)=e^0=1$ and $\alpha(1)=e^{2\pi i}=1$. The left homotopy is $$ H_L(s,t)=e^{2\pi i((1-s)\max\{0,2t-1\}+st)}. $$ At $s=0$, it is $1$ for $0\le t\le\frac12$, then $e^{2\pi i(2t-1)}$ for $\frac12\le t\le1$; this is $c_1*\alpha$. At $s=1$, it is $e^{2\pi it}=\alpha(t)$.

For a direct intermediate-value check, set $s=\frac12,t=\frac14$. The inner parameter is $(1-\frac12)\cdot0+\frac12\cdot\frac14=\frac18$, so $H_L(\frac12,\frac14)=e^{2\pi i/8}=e^{\pi i/4}=(\frac{\sqrt2}{2},\frac{\sqrt2}{2})$, using the identification of $\mathbb{C}$ with $\mathbb{R}^2$. For the right homotopy, use $\phi_R$ instead. At $s=0$ this gives $\alpha*c_1$, while at $s=1$ it again gives $\alpha$. At $s=\frac12,t=\frac34$, its inner parameter is $(1-\frac12)\cdot1+\frac12\cdot\frac34=\frac78$, so its value is $e^{2\pi i(7/8)}=e^{7\pi i/4}=(\frac{\sqrt2}{2},-\frac{\sqrt2}{2})$.

Check: For both homotopies, the inner parameter equals $0$ at $t=0$ and $1$ at $t=1$. The corresponding circle values are $e^0=1$ and $e^{2\pi i}=1$, so the base point remains fixed.

Worked Example: A Triangular Loop in the Plane

Use $\mathbb{R}^2$ with its Euclidean topology and base point $(0,0)$. Define the loop $$ \alpha(t)= \begin{cases} (3t,0),&0\le t\le\frac13,\\ (2-3t,3t-1),&\frac13\le t\le\frac23,\\ (0,3-3t),&\frac23\le t\le1. \end{cases} $$ The formulas agree at the joins: at $t=\frac13$, the first gives $(1,0)$ and the second gives $(2-1,1-1)=(1,0)$; at $t=\frac23$, the second gives $(2-2,2-1)=(0,1)$ and the third gives $(0,3-2)=(0,1)$. The endpoint values are $(0,0)$, so this is a continuous based loop.

For the left identity homotopy, $H_L(s,t)=\alpha((1-s)\phi_L(t)+st)$. At $s=0$, its parameter is $0$ for $t\le\frac12$ and $2t-1$ for $t\ge\frac12$, giving exactly $c_{(0,0)}*\alpha$; at $s=1$, its parameter is $t$, giving $\alpha$. At the intermediate point $s=\frac12,t=\frac12$, the parameter is $(1-\frac12)\cdot0+\frac12\cdot\frac12=\frac14$. Since $\frac14\le\frac13$, $\alpha(\frac14)=(3\cdot\frac14,0)=(\frac34,0)$.

For the right identity homotopy, use $H_R(s,t)=\alpha((1-s)\phi_R(t)+st)$. At $s=0$, its parameter is $2t$ for $t\le\frac12$, and $1$ for $t\ge\frac12$, so it gives $\alpha*c_{(0,0)}$; at $s=1$, it gives $\alpha$. At $s=\frac12,t=\frac12$, the parameter is $(1-\frac12)\cdot1+\frac12\cdot\frac12=\frac34$. Since $\frac23\le\frac34\le1$, $\alpha(\frac34)=(0,3-3\cdot\frac34)=(0,\frac34)$.

Check: In each homotopy, the parameters at the two endpoints are $0$ and $1$, independent of $s$. The triangular loop takes the value $(0,0)$ at both, so each stage fixes the base point.

A Common Pitfall: Moving an Endpoint

A formula can look like a plausible shortening homotopy and still fail to be based. For example, one might try to deform the left concatenation using $\widetilde H(s,t)=\alpha((1-s)\phi_L(t))$. At $s=0$ this does start at $c_{x_0}*\alpha$, but at $t=1$ it gives $\widetilde H(s,1)=\alpha(1-s)$, which is not generally $x_0$. Thus the endpoint can move during the proposed deformation.

For the circle loop in the preceding example, at $s=\frac12,t=1$ this faulty formula gives $\alpha(\frac12)=e^{\pi i}=-1$, not the base point $1$. The correction is to interpolate between two parameter functions that both send the endpoints $0,1$ to themselves: $\phi_L(t)$ and $t$. Their weighted average $(1-s)\phi_L(t)+st$ preserves both endpoints at every stage. The same endpoint check is essential whenever constructing a based homotopy.

The result identifies the class of the constant loop as a two-sided identity for concatenation. It does not say that the constant loop is the only loop in that class, or that every loop is homotopic to a constant loop. It says precisely that inserting a constant segment at either end does not alter a loop’s based homotopy class.

Check Your Understanding

Use the endpoint-preserving parameter construction to answer the following.

Write $\phi_L$ and $\phi_R$ as piecewise linear formulas on the two halves of $I$.
For a based loop $\alpha$, verify directly that $H_L(0,t)=(c_{x_0}*\alpha)(t)$ on each half of $I$.
Explain why the equality $\phi_s(1)=1$ is necessary for a based homotopy, and check it from the interpolation formula.
For $\alpha(t)=e^{2\pi it}$, calculate $H_R(\frac12,\frac14)$.
Identify the endpoint failure in the proposed formula $\alpha((1-s)\phi_R(t))$ for a right identity homotopy.

What You'll Learn