Path Homotopy With Endpoints Fixed

Paths and Fixed Endpoints

A path records a continuous journey through a space, with a specified starting point and finishing point. When comparing two paths, it matters whether those endpoints are allowed to move during the comparison. A path homotopy with endpoints fixed requires every intermediate path to start and finish at the same points as the original paths.

Let $I=[0,1]$ have its usual topology, and give $I\times I$ the product topology. A path in a topological space $(X,\mathcal{T})$ is a continuous map $\alpha:I\to X$. Its initial point is $\alpha(0)$, and its terminal point is $\alpha(1)$. Two paths can be joined by a homotopy with endpoints fixed only if they have the same initial point and the same terminal point.

Definition: Let $\alpha,\beta:I\to X$ be paths with $\alpha(0)=\beta(0)=x_0$ and $\alpha(1)=\beta(1)=x_1$. A path homotopy with endpoints fixed from $\alpha$ to $\beta$ is a continuous map $H:I\times I\to X$ such that, for every $s,t\in I$, $$ H(s,0)=\alpha(s),\qquad H(s,1)=\beta(s),\qquad H(0,t)=x_0,\qquad H(1,t)=x_1. $$ If such a map exists, the paths are path-homotopic with endpoints fixed.

The first variable $s$ runs along a path; the second variable $t$ moves from the first path to the second. The two equations involving $s=0$ and $s=1$ are essential: they require the endpoints to remain fixed throughout the homotopy. Continuity alone, or agreement of endpoints only at $t=0$ and $t=1$, is not enough.

Basic Properties of the Relation

Theorem: For fixed $x_0,x_1\in X$, path homotopy with endpoints fixed is an equivalence relation on the paths $\alpha:I\to X$ satisfying $\alpha(0)=x_0$ and $\alpha(1)=x_1$. If there are no such paths, the relation on that empty collection is still an equivalence relation.

Proof. We check reflexivity, symmetry, and transitivity. For reflexivity, given a path $\alpha$, set $H(s,t)=\alpha(s)$. This map is continuous because it is the composite of the projection $(s,t)\mapsto s$ and $\alpha$. It has $H(s,0)=H(s,1)=\alpha(s)$, $H(0,t)=x_0$, and $H(1,t)=x_1$.

For symmetry, suppose $H$ is a path homotopy from $\alpha$ to $\beta$. Define $K(s,t)=H(s,1-t)$. The map $(s,t)\mapsto(s,1-t)$ is continuous, so $K$ is continuous. Its values at $t=0$ and $t=1$ are $\beta(s)$ and $\alpha(s)$, respectively, and $K(0,t)=x_0$, $K(1,t)=x_1$. Thus $K$ is a path homotopy from $\beta$ to $\alpha$.

For transitivity, suppose $H$ joins $\alpha$ to $\beta$, and $G$ joins $\beta$ to $\gamma$, with all three paths having endpoints $x_0,x_1$. Define $$ K(s,t)= \begin{cases} H(s,2t),&0\le t\le \tfrac12,\\ G(s,2t-1),&\tfrac12\le t\le1. \end{cases} $$ At $t=\tfrac12$, the first formula gives $H(s,1)=\beta(s)$, and the second gives $G(s,0)=\beta(s)$, so the formulas agree. Each formula is continuous on its closed half of the square. The finite pasting lemma says that continuous maps on finitely many closed subsets that cover a space combine to a continuous map when they agree on overlaps. This lemma follows from the closed-set test for continuity: the preimage of a closed set is a finite union of sets closed in those closed subsets, hence is closed in the whole space. Therefore $K$ is continuous. It starts at $\alpha$, ends at $\gamma$, and has $K(0,t)=x_0$, $K(1,t)=x_1$ on both halves. This proves transitivity. If the collection of paths is empty, each equivalence-relation condition is vacuous. $\square$

This theorem concerns paths with a prescribed endpoint pair. It does not say that arbitrary paths with different endpoints can be compared while fixing both endpoints. In fact, if a map $H$ satisfies the definition, then evaluating its boundary equations at $t=0$ and $t=1$ forces $\alpha(0)=\beta(0)$ and $\alpha(1)=\beta(1)$.

Straight-Line Homotopies in Convex Sets

Theorem: Let $C$ be a convex subset of $\mathbb{R}^n$, with the subspace topology from the Euclidean topology. If paths $\alpha,\beta:I\to C$ have the same endpoints, then they are path-homotopic with endpoints fixed by $$ H(s,t)=(1-t)\alpha(s)+t\beta(s). $$

Proof. The maps $\alpha$ and $\beta$, regarded as maps into $\mathbb{R}^n$, are continuous because $C$ has the subspace topology. Addition and scalar multiplication in $\mathbb{R}^n$ are continuous, so the displayed formula defines a continuous map $I\times I\to\mathbb{R}^n$. For each $(s,t)$, the coefficients $1-t$ and $t$ are nonnegative and sum to $1$. Convexity of $C$ therefore gives $H(s,t)\in C$. At $t=0$, $H(s,0)=\alpha(s)$, and at $t=1$, $H(s,1)=\beta(s)$. Since the paths share endpoints, $$ H(0,t)=(1-t)x_0+tx_0=x_0,\qquad H(1,t)=(1-t)x_1+tx_1=x_1. $$ Thus $H$ satisfies every condition in the definition. $\square$

Worked Example: Two Paths in the Unit Interval

Give $I=[0,1]$ its usual subspace topology from $\mathbb{R}$. Consider $\alpha(s)=s$ and $\beta(s)=s^2$. Both are continuous paths in $I$: for $s\in[0,1]$, both $s$ and $s^2$ lie in $[0,1]$. Their endpoints are $\alpha(0)=\beta(0)=0$ and $\alpha(1)=\beta(1)=1$.

Use the straight-line formula $$ H(s,t)=(1-t)s+ts^2. $$ It is continuous as a polynomial expression in $s,t$. Since $s,s^2\in[0,1]$ and $0\le t\le1$, $H(s,t)$ is a weighted average of two numbers in $[0,1]$, so it also lies in $[0,1]$. At the boundaries, $H(s,0)=s=\alpha(s)$, $H(s,1)=s^2=\beta(s)$, $H(0,t)=0$, and $H(1,t)=1$.

Check: at $s=\tfrac12$ and $t=\tfrac25$, $$ H\left(\tfrac12,\tfrac25\right) =\left(1-\tfrac25\right)\tfrac12+\tfrac25\left(\tfrac12\right)^2 =\tfrac35\cdot\tfrac12+\tfrac25\cdot\tfrac14 =\tfrac3{10}+\tfrac1{10} =\tfrac25. $$ This value is in $I$, as required. The boundary checks also recover the original paths: $H(s,0)=s$ and $H(s,1)=s^2$. Thus the formula gives a path homotopy with endpoints fixed.

Worked Example: Paths in a Convex Triangle

Let $C=\{(x,y)\in\mathbb{R}^2:x\ge0,\ y\ge0,\ x+y\le1\}$, with the subspace topology. This set is convex: if two points satisfy the three displayed linear inequalities, every weighted average of them with weights $1-t,t$, $t\in[0,1]$, satisfies those inequalities as well.

Define $\alpha(s)=(s,0)$ and $\beta(s)=(s,s(1-s))$. For $\alpha$, the coordinates are nonnegative and their sum is $s\le1$. For $\beta$, both coordinates are nonnegative, and $$ s+s(1-s)=2s-s^2\le1 $$ because $1-(2s-s^2)=(1-s)^2\ge0$. Both paths start at $(0,0)$ and end at $(1,0)$.

The straight-line homotopy is $$ H(s,t)=\bigl(s,\;t\,s(1-s)\bigr). $$ Indeed, it equals $(1-t)\alpha(s)+t\beta(s)$. Convexity guarantees that its values remain in $C$, and continuity follows from the polynomial coordinate functions. Directly, $H(s,0)=(s,0)=\alpha(s)$, $H(s,1)=(s,s(1-s))=\beta(s)$, $H(0,t)=(0,0)$, and $H(1,t)=(1,0)$.

Check: at $s=\tfrac12$, $t=\tfrac25$, the second coordinate is $$ \tfrac25\cdot\tfrac12\left(1-\tfrac12\right) =\tfrac25\cdot\tfrac14=\tfrac1{10}. $$ Thus $H(\tfrac12,\tfrac25)=(\tfrac12,\tfrac1{10})$, whose coordinates are nonnegative and sum to $\tfrac35\le1$. This confirms the target condition at the chosen parameter values; the general target condition follows from convexity.

An Explicit Homotopy in the Circle

The circle $S^1=\{z\in\mathbb{C}:|z|=1\}$ has the subspace topology from $\mathbb{C}$. A linear interpolation of complex numbers does not automatically stay on $S^1$, so a homotopy formula must be checked against its target space, as well as against the endpoint conditions.

Worked Example: Two Loops in the Circle

Define paths in $S^1$ by $$ \alpha(s)=e^{2\pi i s},\qquad \beta(s)=e^{2\pi i\left(s+\frac16\sin(\pi s)\right)}. $$ Both maps are continuous, and their values have modulus $1$, so they take values in $S^1$. At $s=0$, both equal $1$. At $s=1$, $\alpha(1)=e^{2\pi i}=1$, and $\beta(1)=e^{2\pi i}=1$, since $\sin(\pi)=0$.

Consider $$ H(s,t)=e^{2\pi i\left(s+\frac{t}{6}\sin(\pi s)\right)}. $$ The exponential and the expression in its exponent are continuous, and $|H(s,t)|=1$, so $H$ is a continuous map into $S^1$. Substituting $t=0$ gives $H(s,0)=\alpha(s)$; substituting $t=1$ gives $H(s,1)=\beta(s)$. At $s=0$, the sine term is $0$, so $H(0,t)=1$. At $s=1$, it is again $0$, so $H(1,t)=e^{2\pi i}=1$. Hence this is a path homotopy with endpoints fixed.

Check: at $s=\tfrac12$, $\alpha(\tfrac12)=e^{\pi i}=-1$. For $t=1$, the exponent angle for $\beta(\tfrac12)$ is $$ 2\pi\left(\tfrac12+\tfrac16\sin\left(\tfrac\pi2\right)\right) =2\pi\left(\tfrac12+\tfrac16\right)=\tfrac{4\pi}{3}. $$ Thus $\beta(\tfrac12)=e^{4\pi i/3}=-\tfrac12-\tfrac{\sqrt3}{2}i$, which has modulus $1$. At $t=0$, the formula gives $-1$, as it must. The endpoints remain $1$ for every $t$, not merely at the start and end of the homotopy.

Common Pitfall: Checking Only the Two Boundary Paths

A proposed formula must meet three separate requirements: it must be continuous, its values must stay in $X$, and it must keep both endpoints fixed for every value of the homotopy parameter. Checking only $H(s,0)$ and $H(s,1)$ misses the last two requirements.

For a concrete target-space error, consider the two paths in $S^1$ given by $\alpha(s)=e^{2\pi i s}$ and $\beta(s)=e^{4\pi i s}$. Both begin and end at $1$. Their pointwise linear interpolation in $\mathbb{C}$ is $$ L(s,t)=(1-t)e^{2\pi i s}+te^{4\pi i s}. $$ At $s=\tfrac12$, the two path values are $-1$ and $1$. Consequently, $$ L\left(\tfrac12,\tfrac12\right) =\tfrac12(-1)+\tfrac12(1)=0. $$ But $0\notin S^1$, since $|0|=0\ne1$. So this formula does not define a map into $S^1$, even though it joins the two paths as maps into $\mathbb{C}$.

There is a separate endpoint issue with the angle interpolation $$ J(s,t)=e^{2\pi i((1-t)s+2ts)}. $$ It has the correct paths at $t=0$ and $t=1$. But at the terminal endpoint $s=1$, it gives $J(1,t)=e^{2\pi i(1+t)}$, which is not always $1$. For example, $J(1,\tfrac12)=e^{3\pi i}=-1$. Thus this formula does not keep the terminal endpoint fixed. These failures show why each condition in the definition must be checked independently.

For paths in a convex subset of Euclidean space, the straight-line formula supplies a homotopy because convexity keeps every interpolated value in the space. For paths in a space such as the circle, one must verify the target condition directly; a formula that works in the surrounding vector space may leave the space being studied.

Check Your Understanding

For each question, check the path endpoints and every condition in the definition.

State the four equations that a path homotopy with endpoints fixed must satisfy.
For paths $\alpha,\beta:I\to C$ with the same endpoints, explain why $(1-t)\alpha(s)+t\beta(s)$ remains in $C$ when $C$ is convex.
For the interval paths $\alpha(s)=s$ and $\beta(s)=s^2$, compute $H(\tfrac14,\tfrac12)$ using the straight-line homotopy.
Why does the formula $L$ in the circle example fail to be a path homotopy into $S^1$? Give the parameter values where its value is $0$.
In the transitivity proof, why must the two piecewise formulas agree at $t=\tfrac12$?