Homotopies of Concatenated Paths

Concatenating Paths and Homotopies

A path made from two successive pieces must spend one part of its parameter interval traversing the first piece and the remaining part traversing the second. The change of parameter has to preserve the direction of travel and match the pieces at their common endpoint. The same issue appears when the paths vary through homotopies: at every stage, the two moving pieces must still meet. We will give the parameter formulas and verify continuity, including at the point where the pieces join.

Let $I=[0,1]$. A path in a topological space $X$ is a continuous map $\alpha:I\to X$. Suppose $\alpha(1)=\beta(0)$, so the endpoint of $\alpha$ is the starting point of $\beta$.

Definition: The concatenation of $\alpha$ and $\beta$, denoted $\alpha*\beta:I\to X$, is $$ (\alpha*\beta)(t)= \begin{cases} \alpha(2t),&0\leq t\leq \frac12,\\ \beta(2t-1),&\frac12\leq t\leq 1. \end{cases} $$ The first path is traversed on the first half of $I$, and the second path on the second half.

At $t=\frac12$, the first formula gives $\alpha(1)$, while the second gives $\beta(0)$. The assumption that these are equal is precisely what makes the two formulas agree at the join. The linear changes $2t$ and $2t-1$ each run through the full interval $I$ on their respective halves.

We use the following elementary continuity fact. If a space $Z$ is the union of finitely many closed subsets $F_1,\ldots,F_n$, and a function $f:Z\to X$ is continuous on each $F_j$ with its subspace topology, then $f$ is continuous on $Z$, provided the restrictions agree where the subsets overlap. Indeed, for any closed set $C\subseteq X$, each $f^{-1}(C)\cap F_j$ is closed in $F_j$, hence closed in $Z$ because $F_j$ is closed; their finite union is $f^{-1}(C)$, which is closed. This proves continuity by the closed-set definition.

Apply this fact to $Z=I$, $F_1=[0,\frac12]$, and $F_2=[\frac12,1]$. Each piece of the concatenation is continuous as a composition of continuous maps, and the two values at $t=\frac12$ agree. Thus $\alpha*\beta$ is a path. Its initial point is $\alpha(0)$, and its terminal point is $\beta(1)$.

The Homotopy Formula

A homotopy between paths $\alpha_0$ and $\alpha_1$ is a continuous map $H:I\times I\to X$ such that $H(0,t)=\alpha_0(t)$ and $H(1,t)=\alpha_1(t)$. It is endpoint-preserving when there are fixed points $a,b\in X$ with $H(s,0)=a$ and $H(s,1)=b$ for every $s\in I$. The first variable $s$ records the stage of the homotopy; the second variable records position along each path.

Theorem: Suppose $H:I\times I\to X$ is an endpoint-preserving homotopy from $\alpha_0$ to $\alpha_1$, with fixed endpoints $a,b$. Suppose $K:I\times I\to X$ is an endpoint-preserving homotopy from $\beta_0$ to $\beta_1$, with fixed endpoints $b,c$. Then $\alpha_0*\beta_0$ and $\alpha_1*\beta_1$ are homotopic by the continuous map $L:I\times I\to X$ given by $$ L(s,t)= \begin{cases} H(s,2t),&0\leq t\leq \frac12,\\ K(s,2t-1),&\frac12\leq t\leq 1. \end{cases} $$ Moreover, $L(s,0)=a$ and $L(s,1)=c$ for every $s\in I$.

Proof. The hypotheses give $\alpha_i(1)=b=\beta_i(0)$ for $i=0,1$, so both concatenations in the statement are defined. On the closed rectangle $I\times[0,\frac12]$, the first formula for $L$ is the composition of $H$ with the continuous map $(s,t)\mapsto(s,2t)$, so it is continuous there. On $I\times[\frac12,1]$, the second formula is the composition of $K$ with the continuous map $(s,t)\mapsto(s,2t-1)$, so it is continuous there. On the overlap $I\times\{\frac12\}$, the first formula is $H(s,1)=b$, and the second is $K(s,0)=b$. The formulas agree on the overlap. The closed-set pasting argument above therefore proves that $L$ is continuous on $I\times I$.

At $s=0$, the first half of $L(0,t)$ is $\alpha_0(2t)$ and the second half is $\beta_0(2t-1)$, so $L(0,t)=(\alpha_0*\beta_0)(t)$. At $s=1$, the same calculation gives $L(1,t)=(\alpha_1*\beta_1)(t)$. Finally, $L(s,0)=H(s,0)=a$ and $L(s,1)=K(s,1)=c$. Thus $L$ is the claimed homotopy and fixes the concatenation’s endpoints throughout. $\square$

The matching endpoint $b$ is essential twice: it makes each concatenation a path, and it makes the two formulas for $L$ agree along the entire seam, for every $s$. The theorem does not require the paths to lie in a metric space or to have any special form. It uses continuity of the given homotopies and the fixed-endpoint conditions.

Worked Example: Two Paths Deformed in the Real Line

Give explicit paths from $0$ to $1$ and from $1$ to $2$, and deform both while fixing their endpoints. Let $X=\mathbb{R}$ with its standard topology and define, for $s,u,v\in I$, $$ H(s,u)=u+s\,u(1-u),\qquad K(s,v)=1+v-s\,v(1-v). $$

At $u=0$, $H(s,0)=0+s\cdot0\cdot1=0$; at $u=1$, $H(s,1)=1+s\cdot1\cdot0=1$. At $v=0$, $K(s,0)=1+0-s\cdot0\cdot1=1$; at $v=1$, $K(s,1)=1+1-s\cdot1\cdot0=2$. Both maps are polynomials in their variables, hence continuous, and their endpoints match at $1$.

The induced homotopy is $$ L(s,t)= \begin{cases} 2t+2st(1-2t),&0\leq t\leq \frac12,\\ 2t-2s(2t-1)(1-t),&\frac12\leq t\leq 1. \end{cases} $$ For the first piece, substitute $u=2t$ into $H$: $H(s,2t)=2t+s(2t)(1-2t)$. For the second, substitute $v=2t-1$ into $K$: $K(s,2t-1)=1+(2t-1)-s(2t-1)(1-(2t-1))$, which simplifies to the displayed expression.

Check at an interior point of the first piece: take $s=1$, $t=\frac14$. Then $2t=\frac12$, and $L(1,\frac14)=2(\frac14)+2(1)(\frac14)(1-2(\frac14))=\frac12+\frac12(\frac12)=\frac34$. This agrees with $H(1,\frac12)=\frac12+1\cdot\frac12(1-\frac12)=\frac34$.

Check at an interior point of the second piece: take $s=1$, $t=\frac34$. Then $2t-1=\frac12$, and $L(1,\frac34)=\frac32-2(1)(\frac12)(\frac14)=\frac32-\frac14=\frac54$. Direct substitution gives $K(1,\frac12)=1+\frac12-\frac12(1-\frac12)=\frac32-\frac14=\frac54$.

Check the seam and endpoints: at $t=\frac12$, the first formula gives $1+2s(\frac12)(0)=1$ and the second gives $1-2s(0)(\frac12)=1$. At $t=0$, the first formula gives $0$; at $t=1$, the second gives $2-2s(1)(0)=2$. At $s=0$, both pieces reduce to $2t$, the straight path from $0$ to $2$. These substitutions verify the formula at the join and at both endpoints.

Worked Example: Deforming Only the Second Piece

Let $X=\mathbb{R}^2$ with its Euclidean topology. Keep the first path fixed as $\alpha(u)=(u,0)$, from $(0,0)$ to $(1,0)$, and let the second path vary as $\beta_s(v)=(1+v,s\,v(1-v))$, from $(1,0)$ to $(2,0)$. The homotopy of the first path is $H(s,u)=\alpha(u)$, and the homotopy of the second is $K(s,v)=\beta_s(v)$. Each is continuous, and the displayed endpoint values show that the endpoints remain fixed.

The concatenated homotopy is $$ L(s,t)= \begin{cases} (2t,0),&0\leq t\leq\frac12,\\ (2t,\ s(2t-1)(2-2t)),&\frac12\leq t\leq1. \end{cases} $$ For the first piece, $\alpha(2t)=(2t,0)$. For the second, $v=2t-1$, so $1+v=2t$ and $v(1-v)=(2t-1)(2-2t)$.

For $s=1$ and $t=\frac34$, the second piece has $2t=\frac32$, $2t-1=\frac12$, and $2-2t=\frac12$. Thus $L(1,\frac34)=(\frac32,\frac14)$, which equals $\beta_1(\frac12)=(1+\frac12,\frac12(1-\frac12))=(\frac32,\frac14)$. At the seam, the first piece gives $(1,0)$; the second gives $(1,s\cdot0\cdot1)=(1,0)$. At $t=0$ the value is $(0,0)$, and at $t=1$ it is $(2,s\cdot1\cdot0)=(2,0)$. The checks confirm that the first piece stays fixed while the second changes.

Finitely Many Pieces

The same construction works when several paths are homotoped at once. To make the parameter choice explicit, divide $I$ into equal subintervals, one for each path. Each subinterval is rescaled linearly to $I$. The fixed endpoints of consecutive homotopies ensure agreement where adjacent pieces meet.

Corollary: Let $n\in\mathbb{N}$, and suppose $H_j:I\times I\to X$, for $1\leq j\leq n$, is an endpoint-preserving homotopy from $\alpha_{j,0}$ to $\alpha_{j,1}$, with fixed endpoints $x_{j-1},x_j$. Then the concatenation of $\alpha_{1,0},\ldots,\alpha_{n,0}$ is homotopic to the concatenation of $\alpha_{1,1},\ldots,\alpha_{n,1}$, through the formula $$ L(s,t)=H_j(s,nt-(j-1)) \quad\text{when}\quad \frac{j-1}{n}\leq t\leq\frac{j}{n}. $$ The homotopy fixes $x_0$ and $x_n$.

Proof. On each closed strip $I\times[\frac{j-1}{n},\frac{j}{n}]$, the map $(s,t)\mapsto(s,nt-(j-1))$ is continuous and its second coordinate runs from $0$ to $1$. Thus the formula is continuous on that strip. At the common boundary $t=\frac{j}{n}$ of strips $j$ and $j+1$, the formula from strip $j$ is $H_j(s,1)=x_j$, while the formula from strip $j+1$ is $H_{j+1}(s,0)=x_j$. They agree for every $s$. The finitely many strips are closed and cover $I\times I$, so the closed-set pasting argument proves continuity of $L$. At $s=0$ and $s=1$, its pieces are exactly the corresponding concatenated paths, by the equal-subinterval rescaling. At $t=0$, the value is $H_1(s,0)=x_0$; at $t=1$, it is $H_n(s,1)=x_n$. $\square$

Worked Example: Two Based Loops on the Circle

Let $S^1=\{z\in\mathbb{C}:|z|=1\}$ with the subspace topology inherited from $\mathbb{C}$. Define two homotopies of loops based at $1$ by $$ H(s,u)=e^{2\pi i s u(1-u)},\qquad K(s,v)=e^{-2\pi i s v(1-v)}. $$ The exponential values have modulus $1$, so both maps take values in $S^1$. At $u=0,1$, the exponent in $H$ is $0$; at $v=0,1$, the exponent in $K$ is $0$. Hence every path in both homotopies begins and ends at $e^0=1$. The maps are continuous as compositions of continuous real-valued functions with the complex exponential.

The homotopy of concatenations uses $H(s,2t)$ on the first half and $K(s,2t-1)$ on the second. At $s=1,t=\frac14$, the first path parameter is $2t=\frac12$. Its exponent is $2\pi i\cdot\frac12(1-\frac12)=\frac{\pi i}{2}$, so the value is $e^{\pi i/2}=i$. At $s=1,t=\frac34$, the second path parameter is $2t-1=\frac12$. Its exponent is $-2\pi i\cdot\frac12(1-\frac12)=-\frac{\pi i}{2}$, so the value is $e^{-\pi i/2}=-i$.

At the seam, both pieces give $1$, because $H(s,1)=1=K(s,0)$. At $s=0$, each exponential is $e^0=1$ at every parameter, so the concatenation is the constant loop at $1$. The example illustrates that the two halves can move in different directions while the joining point and the based-loop endpoints stay fixed.

A Common Parameter Mistake

A frequent error is to use $\beta(2t)$ on the second half instead of $\beta(2t-1)$. The second-half parameter must start at $0$ when $t=\frac12$ and finish at $1$ when $t=1$. The incorrect choice starts at $1$, so it does not traverse the second path in the intended direction from its initial point.

For a concrete failure, let $\alpha(u)=u$ and $\beta(v)=1+v$ in $\mathbb{R}$. They meet at $\alpha(1)=1=\beta(0)$. The correct concatenation has value $1$ at $t=\frac12$ from both pieces. If the second piece is incorrectly written as $\beta(2t)$, then at $t=\frac12$ it gives $\beta(1)=2$, while the first piece gives $\alpha(1)=1$. The proposed function is discontinuous at the seam, since its two formulas disagree there. The correction is to substitute $2t-1$: at $t=\frac12$, this is $0$, and at $t=1$, it is $1$.

A related error is to assume that matching endpoints only at the start and finish of the homotopy suffices. For the pasted map to be continuous, the two moving paths must match at every homotopy stage $s$. The fixed-endpoint hypotheses guarantee exactly this: $H(s,1)=b=K(s,0)$ for every $s\in I$.

Check Your Understanding

Use the parameter formulas and endpoint conditions to justify each answer.

If $\alpha(1)=\beta(0)$, what values do the two pieces of $\alpha*\beta$ give at $t=\frac12$?
In the homotopy formula, why must $H(s,1)=K(s,0)$ hold for every $s$, rather than just for $s=0$ and $s=1$?
For $H(s,u)=u+s\,u(1-u)$, compute $H(1,\frac14)$ and verify the result by substituting directly into the formula.
Write the concatenation formula for three paths on three equal subintervals of $I$.
Using $\alpha(u)=u$ and $\beta(v)=1+v$, compute the two proposed seam values if the incorrect second-piece parameter $2t$ is used.

Homotopies of Concatenated Paths

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