Based Loops and Endpoint-Preserving Homotopies

Loops with a Distinguished Point

A loop is a path that starts and ends at the same point. In the study of loops, that common endpoint can be kept fixed throughout a homotopy, or it can be allowed to move. These are different requirements: fixing the endpoint gives a based homotopy, while allowing it to move gives a free homotopy. The distinction is expressed by boundary conditions on a homotopy map.

Let $I=[0,1]$ have its usual subspace topology from $\mathbb{R}$. A path in a topological space $(X,\mathcal{T})$ is a continuous map $\gamma:I\to X$. Its initial point is $\gamma(0)$, and its final point is $\gamma(1)$.

Definition: Given $x_0\in X$, a loop based at $x_0$ is a continuous map $\gamma:I\to X$ such that $\gamma(0)=\gamma(1)=x_0$. A based homotopy from a loop $\gamma$ based at $x_0$ to a loop $\delta$ based at $x_0$ is a continuous map $H:I\times I\to X$, where $I\times I$ has the product topology, satisfying $$ H(s,0)=\gamma(s),\qquad H(s,1)=\delta(s),\qquad H(0,u)=H(1,u)=x_0 $$ for all $s,u\in I$.

The first coordinate $s$ runs along a loop; the second coordinate $u$ runs through the homotopy. At $u=0$, the loop is $\gamma$, and at $u=1$, it is $\delta$. The conditions on $s=0$ and $s=1$ require both endpoints to stay at $x_0$ for every intermediate value of $u$. In particular, at each fixed $u$, the map $s\mapsto H(s,u)$ is itself a loop based at $x_0$.

Definition: A free homotopy of loops from $\gamma$ to $\delta$ is a continuous map $H:I\times I\to X$ with $H(s,0)=\gamma(s)$, $H(s,1)=\delta(s)$, and $H(0,u)=H(1,u)$ for every $u\in I$. The common endpoint may depend on $u$; it is not required to remain at a particular point of $X$.

Thus a free homotopy still gives a loop at every intermediate stage: its two ends agree. What changes is the location of that common endpoint. A based homotopy is a special case of a free homotopy, because the fixed-endpoint conditions imply that the two ends agree.

What the Endpoint Conditions Force

Proposition: If $H$ is a based homotopy from $\gamma$ to $\delta$, then $\gamma$ and $\delta$ have the same initial point and the same final point. In particular, based homotopies can relate loops only when their basepoints agree. Every based homotopy is also a free homotopy.

Proof. For a based homotopy, the boundary conditions give $\gamma(0)=H(0,0)=x_0$ and $\delta(0)=H(0,1)=x_0$. They also give $\gamma(1)=H(1,0)=x_0$ and $\delta(1)=H(1,1)=x_0$. Hence both loops have the same initial and final point. For each $u$, the same conditions give $H(0,u)=x_0=H(1,u)$, so $H(0,u)=H(1,u)$. This is the endpoint condition for a free homotopy. $\square$

The proposition is a necessary-condition test: before trying to construct a based homotopy, check that the loops have the same basepoint. It does not say that any two loops with the same basepoint are based-homotopic. That depends on the space and on the loops.

Worked Example: A Based Homotopy in the Plane

Give a based homotopy between $\gamma(t)=(4t(1-t),0)$ and the constant loop $\delta(t)=(0,0)$ in $\mathbb{R}^2$ with its Euclidean topology.

At $t=0$, $4t(1-t)=4\cdot0\cdot1=0$, and at $t=1$, $4t(1-t)=4\cdot1\cdot0=0$. Both paths therefore start and end at $x_0=(0,0)$. Define $$ H(s,u)=\bigl((1-u)4s(1-s),0\bigr). $$

For $u=0$, $H(s,0)=(4s(1-s),0)=\gamma(s)$. For $u=1$, $H(s,1)=(0,0)=\delta(s)$. For every $u\in I$, $H(0,u)=((1-u)4\cdot0\cdot1,0)=(0,0)$ and $H(1,u)=((1-u)4\cdot1\cdot0,0)=(0,0)$. The formula is continuous because its coordinates are polynomials in $s$ and $u$, and hence it is a based homotopy.

Check: At $s=\tfrac12$ and $u=\tfrac12$, the first coordinate is $(1-\tfrac12)4(\tfrac12)(1-\tfrac12)=\tfrac12$. Thus this intermediate loop passes through $(\tfrac12,0)$; at $u=1$, it is the constant loop as required.

Convex Targets Allow Straight-Line Homotopies

The preceding example uses straight-line interpolation in the plane. This construction works whenever the target is convex. A subset $X\subseteq\mathbb{R}^n$ is convex if, for any $p,q\in X$ and any $\lambda\in I$, the point $(1-\lambda)p+\lambda q$ belongs to $X$. Give $X$ the subspace topology inherited from the usual Euclidean topology on $\mathbb{R}^n$.

Theorem: Let $X\subseteq\mathbb{R}^n$ be convex, and let $\gamma,\delta:I\to X$ be loops based at the same point $x_0$. Then they are based-homotopic by the formula $H(s,u)=(1-u)\gamma(s)+u\delta(s)$.

Proof. Fix $s,u\in I$. Since $\gamma(s)$ and $\delta(s)$ belong to $X$, and $X$ is convex, their convex combination $(1-u)\gamma(s)+u\delta(s)$ belongs to $X$. Thus the formula defines a map into $X$. It is continuous as a map into $\mathbb{R}^n$, because $\gamma$ and $\delta$ are continuous and addition and scalar multiplication in $\mathbb{R}^n$ are continuous. A continuous map into $\mathbb{R}^n$ whose values lie in $X$ is continuous as a map into the subspace $X$. At the homotopy ends, $H(s,0)=\gamma(s)$ and $H(s,1)=\delta(s)$. At the loop endpoints, $\gamma(0)=\delta(0)=x_0$ and $\gamma(1)=\delta(1)=x_0$, so $$ H(0,u)=(1-u)x_0+ux_0=x_0,\qquad H(1,u)=(1-u)x_0+ux_0=x_0. $$ All the required conditions hold, so $H$ is a based homotopy. $\square$

Worked Example: Two Based Loops in the Unit Interval

In $X=[0,1]$ with the subspace topology from $\mathbb{R}$, consider loops based at $0$: $$ \gamma(t)=4t(1-t),\qquad \delta(t)=2t(1-t). $$ Construct a based homotopy and check that all its values remain in $X$.

Set $H(s,u)=(1-u)\gamma(s)+u\delta(s)$. Substituting the two formulas gives $$ H(s,u)=(1-u)4s(1-s)+u\,2s(1-s)=(4-2u)s(1-s). $$

Since $0\leq u\leq1$, $2\leq4-2u\leq4$. Also $s(1-s)\geq0$, and $s(1-s)\leq\tfrac14$, because $(s-\tfrac12)^2\geq0$ implies $s-s^2\leq\tfrac14$. Consequently $0\leq H(s,u)\leq4\cdot\tfrac14=1$, so $H(s,u)\in[0,1]$ everywhere. At $u=0$ the formula is $4s(1-s)=\gamma(s)$, and at $u=1$ it is $2s(1-s)=\delta(s)$. At $s=0$ and $s=1$, the factor $s(1-s)$ is zero, so both endpoints stay at $0$.

Check: For $s=\tfrac12$ and $u=\tfrac12$, the value is $(4-2\cdot\tfrac12)\cdot\tfrac12\cdot\tfrac12=3\cdot\tfrac14=\tfrac34$, which lies in $[0,1]$. At $u=1$ and $s=\tfrac12$, it is $2\cdot\tfrac14=\tfrac12=\delta(\tfrac12)$.

Free Homotopy Can Move the Basepoint

Worked Example: A Free Homotopy on the Circle

Let $S^1=\{z\in\mathbb{C}:|z|=1\}$ have the subspace topology inherited from the usual topology on $\mathbb{C}$. Consider $$ \gamma(s)=e^{2\pi i s},\qquad \delta(s)=e^{2\pi i(s+1/2)}. $$ Both are loops: their values at $s=0$ and $s=1$ agree. Their basepoints differ, since $\gamma(0)=e^0=1$, while $\delta(0)=e^{\pi i}=\cos(\pi)+i\sin(\pi)=-1$.

Define $H(s,u)=e^{2\pi i(s+u/2)}$. This map is continuous and takes values in $S^1$, since the modulus of every complex exponential with purely imaginary exponent is $1$. At $u=0$, it is $\gamma(s)$; at $u=1$, it is $\delta(s)$. For every $u$, $$ H(0,u)=e^{\pi i u}=H(1,u), $$ because $H(1,u)=e^{2\pi i(1+u/2)}=e^{2\pi i}e^{\pi i u}=e^{\pi i u}$. Thus $H$ is a free homotopy. Its moving basepoint starts at $1$ and ends at $-1$.

These loops cannot be joined by a based homotopy: such a homotopy would require their initial points to agree, but $1\ne-1$. The example shows why the free-homotopy endpoint condition is equality of the two ends at each stage, not constancy of the common endpoint.

A Common Pitfall: Interpolation May Leave the Space

The straight-line formula in the convex-target theorem depends on convexity. It cannot be applied without checking that the intermediate values remain in the target. For a concrete warning, take the loop $\gamma(s)=e^{2\pi i s}$ in $S^1$ and the constant loop $\delta(s)=1$, also based at $1$. A tempting formula is $$ H(s,u)=(1-u)e^{2\pi i s}+u. $$

At $s=\tfrac12$, $e^{2\pi i s}=e^{\pi i}=-1$. At $u=\tfrac12$, the proposed value is $$ H\left(\tfrac12,\tfrac12\right)=\left(1-\tfrac12\right)(-1)+\tfrac12=\;0. $$ But $0\notin S^1$, since its modulus is $0$, not $1$. So this formula does not define a homotopy into $S^1$, even though it gives the intended starting and ending loops and fixes their endpoints. This calculation invalidates this particular construction; by itself, it does not decide whether some other based homotopy exists.

When checking a proposed homotopy, verify both kinds of requirements: the boundary conditions and membership of every intermediate value in the target space. The formulas may look appropriate on the boundary and still fail in the interior of $I\times I$.

Check Your Understanding

For each question, write down the endpoint conditions explicitly before deciding whether a proposed homotopy is based or free.

State the three boundary conditions that define a based homotopy from $\gamma$ to $\delta$ at $x_0$.
Can a based homotopy join loops whose basepoints are different? Explain using the endpoint conditions.
In the free homotopy on $S^1$ above, compute the common endpoint when $u=\tfrac12$.
Why does convexity ensure that the straight-line homotopy in the theorem takes its values in $X$?
For the proposed interpolation between $e^{2\pi i s}$ and the constant loop at $1$, compute the value at $(s,u)=(\tfrac12,\tfrac12)$ and explain why it causes a problem.