Composing a Homotopy
A homotopy is a continuous family of maps, with a parameter running through the unit interval. Composing the maps in that family with a fixed continuous map gives another family. There are two different operations: a map into the original domain can be used to precompose every map in the family, and a map out of the original codomain can be used to postcompose every map. Keeping track of the domain and codomain at each stage prevents a common notational error.
Let \(I=[0,1]\) have its standard subspace topology from \(\mathbb{R}\). Products are given the product topology. For topological spaces \(X\) and \(Y\), a homotopy from \(f:X\to Y\) to \(g:X\to Y\) is a continuous map \(H:X\times I\to Y\) satisfying the endpoint conditions below. Write \(H_t(x)=H(x,t)\) for the map at parameter \(t\).
Thus each \(H_t\) has domain \(X\) and codomain \(Y\). The continuity required in this definition is continuity of the single map \(H\) on the product \(X\times I\), not merely continuity of each map \(H_t\) separately.
Proof. First consider precomposition. The product map \(a\times\operatorname{id}_I:W\times I\to X\times I\) sends \((w,t)\) to \((a(w),t)\). To verify its continuity, take a basis rectangle \(U\times V\) in \(X\times I\), where \(U\) is open in \(X\) and \(V\) is open in \(I\). Its preimage is $$ (a\times\operatorname{id}_I)^{-1}(U\times V)=a^{-1}(U)\times V. $$ Because \(a\) is continuous, \(a^{-1}(U)\) is open in \(W\), so this preimage is open in \(W\times I\). Every open set in the product is a union of basis rectangles, and preimages preserve unions; therefore \(a\times\operatorname{id}_I\) is continuous. It follows that \(H\circ(a\times\operatorname{id}_I)\) is continuous. At \(t=0\), its value at \(w\) is \(H(a(w),0)=f(a(w))\); at \(t=1\), its value is \(H(a(w),1)=g(a(w))\). This proves the precomposition assertion.
For postcomposition, \(b\circ H\) is continuous because \(b\) and \(H\) are continuous. Its values at the two endpoints are \(b(H(x,0))=b(f(x))\) and \(b(H(x,1))=b(g(x))\). Thus it is a homotopy from \(b\circ f\) to \(b\circ g\). Applying the two assertions in succession gives the simultaneous formula, with the same endpoint calculation. These arguments also apply when a domain is empty: the continuity and endpoint conditions then hold on the empty domain. \(\square\)
The distinction between the two operations is visible in their formulas. Precomposition changes the domain from \(X\) to \(W\) but leaves the codomain \(Y\) unchanged. Postcomposition leaves the domain \(X\) unchanged but changes the codomain from \(Y\) to \(Z\). In either case the parameter is still \(t\in I\), and the endpoints are obtained by composing the original endpoint maps in the same way.
Restricting a Homotopy Along an Inclusion
Worked Example: Restricting to the Circle
Give \(\mathbb{R}^2\) its standard Euclidean topology, and give \(S^1=\{(x,y)\in\mathbb{R}^2:x^2+y^2=1\}\) the subspace topology. Consider the map \(H:\mathbb{R}^2\times I\to\mathbb{R}^2\) defined by $$ H((x,y),t)=(x,(1-t)y). $$ Its coordinate formulas are continuous, so \(H\) is continuous. At \(t=0\), \(H_0(x,y)=(x,y)\); at \(t=1\), \(H_1(x,y)=(x,0)\). Thus \(H\) is a homotopy from the identity map of \(\mathbb{R}^2\) to the map \(g(x,y)=(x,0)\).
Let \(i:S^1\to\mathbb{R}^2\) be the inclusion \(i(x,y)=(x,y)\), which is continuous by the definition of the subspace topology. Precomposing \(H\) by \(i\) gives $$ H(i(x,y),t)=(x,(1-t)y),\qquad (x,y)\in S^1. $$ This is a homotopy from \(i\) to \(g\circ i\), where \((g\circ i)(x,y)=(x,0)\). The codomain remains \(\mathbb{R}^2\): the intermediate points need not lie on \(S^1\).
Check: At \((x,y)=(3/5,4/5)\), which lies on \(S^1\) because \((3/5)^2+(4/5)^2=9/25+16/25=1\), and at \(t=1/4\), the value is $$ H((3/5,4/5),1/4)=(3/5,(3/4)(4/5))=(3/5,3/5). $$ At \(t=0\) the formula returns \((3/5,4/5)\), and at \(t=1\) it returns \((3/5,0)\), agreeing with the two endpoint maps.
This construction is useful whenever a homotopy on a space is to be examined only on a subspace. The inclusion supplies the required continuous map into the original domain; the composition theorem then supplies the restricted family without requiring a new continuity proof for every parameter value.
Applying a Projection
Worked Example: Taking a Coordinate of a Homotopy
Give \(\mathbb{R}\) and \(\mathbb{R}^2\) their standard topologies. Define \(H:\mathbb{R}\times I\to\mathbb{R}^2\) by $$ H(x,t)=(x,(1-t)x). $$ This map is continuous because both coordinate functions are continuous. Its endpoint maps are \(f(x)=(x,x)\) and \(g(x)=(x,0)\). Let \(\pi_2:\mathbb{R}^2\to\mathbb{R}\) be the second-coordinate projection, \(\pi_2(u,v)=v\). It is continuous in the product topology.
Postcomposition gives the homotopy $$ (\pi_2\circ H)(x,t)=(1-t)x $$ from \((\pi_2\circ f)(x)=x\) to \((\pi_2\circ g)(x)=0\). For a direct numerical check, substitute \(x=3\) and \(t=1/3\): $$ (\pi_2\circ H)(3,1/3)=(1-1/3)3=(2/3)3=2. $$ At \(t=0\) the same formula gives \((1-0)3=3\), the identity map's value at \(3\); at \(t=1\) it gives \((1-1)3=0\), the constant map's value at \(3\). The calculation confirms both the intermediate value and the endpoints.
For comparison, postcomposing by the first projection \(\pi_1(u,v)=u\) gives \((\pi_1\circ H)(x,t)=x\) for every \(t\). This is a homotopy from the identity map of \(\mathbb{R}\) to itself. A projection can therefore preserve a visible change in one coordinate while discarding a change in another.
Precomposition and a Coordinate Map Together
Worked Example: Combining Both Operations
Give \(\mathbb{R}\) and \(\mathbb{R}^2\) their standard topologies. Define \(H:\mathbb{R}\times I\to\mathbb{R}^2\) by $$ H(x,t)=(x+t,2x-t). $$ This is continuous. Its endpoint maps are \(f(x)=(x,2x)\) and \(g(x)=(x+1,2x-1)\). Define \(a:\mathbb{R}\to\mathbb{R}\) by \(a(u)=2u\), and define the continuous coordinate map \(c:\mathbb{R}^2\to\mathbb{R}\) by \(c(r,s)=r-s\).
The theorem gives the combined homotopy $$ (c\circ H\circ(a\times\operatorname{id}_I))(u,t) =c(H(2u,t)) =(2u+t)-(4u-t) =-2u+2t. $$ At \(t=0\), this equals \(-2u\), which also follows directly from $$ (c\circ f\circ a)(u)=c(2u,4u)=2u-4u=-2u. $$ At \(t=1\), it equals \(-2u+2\), and the other endpoint calculation is $$ (c\circ g\circ a)(u)=c(2u+1,4u-1)=(2u+1)-(4u-1)=-2u+2. $$ Thus the formula has exactly the required endpoint maps.
Check: Set \(u=3/2\) and \(t=1/4\). Then \(a(u)=3\), and $$ H(3,1/4)=(3+1/4,6-1/4)=(13/4,23/4). $$ Applying \(c\) gives \(13/4-23/4=-10/4=-5/2\). The combined formula gives \(-2(3/2)+2(1/4)=-3+1/2=-5/2\), the same value. Its endpoint values at this \(u\) are \(-3\) and \(-1\), respectively, matching \((c\circ f\circ a)(3/2)\) and \((c\circ g\circ a)(3/2)\).
In this example, the same parameter \(t\) is used throughout: first \(a\) sends \(u\) to the original domain, then \(H\) applies the homotopy, and finally \(c\) extracts a real-valued coordinate combination. The order matters because the domains and codomains must match at each composition.
Why Continuity Is Essential
The composition theorem depends on continuity of the map being composed. For postcomposition, take \(X\) to be a one-point space, take \(Y=\mathbb{R}\), and let \(H(*,t)=t\). This is a homotopy from the constant value \(0\) to the constant value \(1\). Define \(b:\mathbb{R}\to\mathbb{R}\) by \(b(s)=0\) for \(s\leq 1/2\) and \(b(s)=1\) for \(s>1/2\). This map is not continuous at \(1/2\). The proposed composite has value \(0\) for \(t\leq 1/2\) and \(1\) for \(t>1/2\), so it is not continuous at \(t=1/2\). It is therefore not a homotopy. This example shows why continuity of the postcomposing map is needed for the composition argument; it does not assert that the two resulting endpoint maps could not have some other homotopy.
A separate pitfall is to write \(H\circ a\) when \(a:W\to X\). This expression omits the homotopy parameter and has mismatched inputs: \(H\) takes inputs in \(X\times I\), not \(X\). The correct map is \(H\circ(a\times\operatorname{id}_I)\), which sends \((w,t)\) to \(H(a(w),t)\). Checking the domain of each map is a reliable way to catch this error.
The practical value of the composition theorem is that a homotopy can be transported to a new domain or codomain while its endpoint maps are tracked exactly. In applications, inclusions let us restrict attention to a subspace, and projections or coordinate maps let us retain selected information. The theorem gives continuity and endpoint control in one argument, rather than requiring a fresh construction from the start.
Check Your Understanding
For each question, identify the domains and codomains before computing endpoint maps.
- If \(H:X\times I\to Y\) is a homotopy from \(f\) to \(g\), and \(a:W\to X\) is continuous, write the formula for the resulting homotopy and state both endpoints.
- In Worked Example 1, what is the value of the restricted homotopy at \((0,1)\in S^1\) and \(t=1/2\)? Does it have to lie in \(S^1\)?
- For the homotopy in Worked Example 2, compute \((\pi_2\circ H)(-2,1/4)\) and compare it with its endpoint values at \(x=-2\).
- In Worked Example 3, compute the combined homotopy at \(u=0\) and \(t=3/4\), and verify it using the intermediate point \(H(a(u),t)\).
- Which continuity assumption ensures that \(a\times\operatorname{id}_I\) is continuous, and which ensures that \(b\circ H\) is continuous?