A Checklist for Explicit Constructions
Homotopy equivalence, contraction, and deformation retraction are closely related, but their definitions ask for different data. In a worked problem, a formula alone is not enough: its domain and codomain must be right, its endpoint maps must be the required maps, and a deformation retraction must fix the target subspace at every intermediate time.
Throughout, \(I=[0,1]\) has its standard topology. A homotopy from \(f:X\to Y\) to \(g:X\to Y\) is a continuous map \(H:X\times I\to Y\) such that \(H(x,0)=f(x)\) and \(H(x,1)=g(x)\) for every \(x\in X\).
When writing these formulas, keep track of two different kinds of endpoint condition. A homotopy of maps must have the specified maps at \(t=0\) and \(t=1\). A strong deformation retraction must, in addition, leave every point of \(A\) fixed for all \(t\). A homotopy between paths with endpoints fixed has another condition: the path endpoints stay fixed as the homotopy parameter varies. That requirement is not part of the definition of an arbitrary homotopy of maps.
Homotopy Equivalence from a Strong Deformation Retraction
Proof. Since \(r(a)=a\) for every \(a\in A\), the composite \(r\circ i:A\to A\) is \(\operatorname{id}_A\). The homotopy in the definition of strong deformation retraction runs from \(\operatorname{id}_X\) to \(i\circ r\), so \(i\circ r\simeq\operatorname{id}_X\). These are exactly the two conditions for \(i\) and \(r\) to be homotopy inverses. \(\square\)
Worked Example: A Strip Retracts onto Its Bottom Edge
Give \(X=\mathbb{R}\times I\) and \(A=\mathbb{R}\times\{0\}\) their subspace topologies from the standard Euclidean plane. Define \(r:X\to A\) by \(r(x,y)=(x,0)\). This map is continuous, takes values in \(A\), and satisfies \(r(x,0)=(x,0)\), so it is a retraction. Define
This formula is continuous. At \(t=0\), \(H((x,y),0)=(x,y)\). At \(t=1\), \(H((x,y),1)=(x,0)=r(x,y)\). For \(0\leq t\leq1\) and \(0\leq y\leq1\), \(0\leq(1-t)y\leq1\), so the formula stays in \(X\). If \((x,y)\in A\), then \(y=0\) and \(H((x,0),t)=(x,0)\) for every \(t\). Thus this is a strong deformation retraction.
To check the resulting homotopy equivalence directly, identify \(A\) with \(\mathbb{R}\) and let \(i:\mathbb{R}\to X\) be \(i(x)=(x,0)\). Then \(r\circ i(x)=x\). The other composite is \(i\circ r(x,y)=(x,0)\), and the displayed \(H\) is a homotopy from \(\operatorname{id}_X\) to \(i\circ r\). For instance, at \(((3,1),1/2)\) the formula gives \(H((3,1),1/2)=(3,1/2)\), which remains in the strip; at \(t=1\) it gives \((3,0)\), as required.
Contractions and Homotopy Equivalence to a Point
The definition of contractibility is exactly the data needed for a homotopy equivalence with a one-point space. This observation is useful in both directions: an explicit contraction gives the equivalence, and an equivalence with a point supplies a contraction.
Proof. Suppose first that \(X\) is contractible, with a homotopy \(H:X\times I\to X\) from \(\operatorname{id}_X\) to \(c_{x_0}\). Define \(f:X\to P\) to be the constant map and \(g:P\to X\) by \(g(p)=x_0\). Both maps are continuous. The composite \(f\circ g\) is \(\operatorname{id}_P\), since \(P\) has only one point. The composite \(g\circ f\) is \(c_{x_0}\), homotopic to \(\operatorname{id}_X\) by \(H\); hence \(f\) and \(g\) are homotopy inverses.
Conversely, suppose \(f:X\to P\) and \(g:P\to X\) are homotopy inverses. Since \(P\) has one point, \(g\circ f\) is a constant map, with value \(x_0=g(p)\). By the homotopy-inverse condition, \(g\circ f\simeq\operatorname{id}_X\). Reversing the parameter of that homotopy, by replacing \(t\) with \(1-t\), gives a homotopy from \(\operatorname{id}_X\) to \(c_{x_0}\). Thus \(X\) is contractible. \(\square\)
Worked Example: Contracting a Three-Armed Subspace
Give the subspace \(T\) of the standard Euclidean plane the topology induced from \(\mathbb{R}^2\), where
Define \(H:T\times I\to T\) by \(H((x,y),t)=((1-t)x,(1-t)y)\). For a point on the right arm, \(0\leq x\leq2\), so \(0\leq(1-t)x\leq2\); its image remains on that arm. For a point on the left arm, \(-1\leq x\leq0\), so \(-1\leq(1-t)x\leq0\); its image remains on the left arm. For a point on the vertical arm, \(0\leq y\leq1\), so \(0\leq(1-t)y\leq1\); its image remains on that arm. Hence the formula is a well-defined map into \(T\), and it is continuous as the restriction of a continuous formula on the plane times \(I\).
At \(t=0\), \(H((x,y),0)=(x,y)\); at \(t=1\), \(H((x,y),1)=(0,0)\). For the point \((2,0)\), the intermediate value at \(t=1/4\) is \((3/2,0)\), which lies on the right arm. For \((0,1)\), it is \((0,3/4)\), which lies on the vertical arm. These checks illustrate the general inequalities above. Thus \(H\) is a contraction, and the proposition shows \(T\) is homotopy equivalent to a point.
Collapsing an Attached Segment
A retraction can collapse part of a space while leaving the rest fixed. For an explicit formula, it is important that the definitions on the pieces agree where those pieces meet; otherwise the proposed map or homotopy may fail to be continuous.
Worked Example: A Circle with an Attached Segment
Let \(C=\{(x,y)\in\mathbb{R}^2:x^2+y^2=1\}\) and \(L=[1,2]\times\{0\}\), with their subspace topologies from the standard Euclidean plane. Set \(X=C\cup L\) with its subspace topology, and take \(A=C\). The pieces meet only at \(a=(1,0)\). Define \(r:X\to C\) to be the identity on \(C\) and to send every point of \(L\) to \(a\).
This map is well-defined at the intersection: both rules send \(a\) to \(a\). It is continuous by the finite closed pasting argument: \(C\) and \(L\) are closed in \(X\), the two restrictions are continuous, and they agree on the intersection. Indeed, for any closed \(F\subseteq C\), the preimage \(r^{-1}(F)\) is the union of the closed preimage in \(C\) and the closed preimage in \(L\); agreement on \(C\cap L\) makes these restrictions a single well-defined map, and the finite union is closed in \(X\). Also \(r(c)=c\) for every \(c\in C\), so \(r\) is a retraction.
For \(s\in[1,2]\), define \(H\) on the two pieces by
At the intersection \(s=1\), the second formula gives \((1,0)\), matching the first, so the definition is unambiguous. Each restriction is continuous and the pieces \(C\times I\) and \(L\times I\) are closed in \(X\times I\); the same finite closed pasting argument proves continuity of \(H\). At \(t=0\), the segment formula gives \((1+s-1,0)=(s,0)\), so \(H(\cdot,0)=\operatorname{id}_X\). At \(t=1\), it gives \((1,0)=r(s,0)\), while on \(C\) the map stays the identity, so \(H(\cdot,1)\) is \(r\), viewed as a map into \(X\).
For an intermediate check, take \(s=3/2\) and \(t=1/2\). The formula gives \((1+(1/2)(1/2),0)=(5/4,0)\in L\). Every point of \(C\) stays fixed at all times. Therefore \(A=C\) is a strong deformation retract of \(X\). The inclusion \(i:C\to X\) and \(r:X\to C\) are homotopy inverses: \(r\circ i=\operatorname{id}_C\), and the displayed \(H\) joins \(\operatorname{id}_X\) to \(i\circ r\).
A Linear Retraction onto the Diagonal
Worked Example: Retracting a Square onto Its Diagonal
Give \(X=I\times I\) and \(A=\{(u,u):u\in I\}\) their subspace topologies from \(\mathbb{R}^2\). Define \(r:X\to A\) by \(r(x,y)=((x+y)/2,(x+y)/2)\). Since \(0\leq x,y\leq1\), their average \(m=(x+y)/2\) satisfies \(0\leq m\leq1\), so \(r(x,y)=(m,m)\) belongs to \(A\). If \((x,y)=(u,u)\in A\), then \(m=(u+u)/2=u\), so \(r(u,u)=(u,u)\). This continuous map is a retraction.
Use the straight-line homotopy
Both coordinates are convex combinations of numbers in \([0,1]\), and therefore remain in \([0,1]\); thus \(H\) maps into \(X\). The formula is continuous. Substitution gives \(H((x,y),0)=(x,y)\) and \(H((x,y),1)=(m,m)=r(x,y)\). If \((x,y)=(u,u)\), then \(m=u\), and both coordinates of \(H((u,u),t)\) equal \((1-t)u+tu=u\) for every \(t\). Hence this is a strong deformation retraction.
For example, at \((x,y)=(0,1)\), \(m=(0+1)/2=1/2\). At \(t=1/2\), the coordinates are \((1/2)0+(1/2)(1/2)=1/4\) and \((1/2)1+(1/2)(1/2)=3/4\), giving \(H((0,1),1/2)=(1/4,3/4)\in X\). At \(t=1\), the result is \((1/2,1/2)\in A\), as required.
A Common Mistake: A Formula Can Leave the Space
Common pitfall. For the circle \(C\), the formula \(H(z,t)=(1-t)z\) is sometimes offered as a contraction to the origin. It has the desired initial value \(H(z,0)=z\), but it is not a map \(C\times I\to C\). For example, with \(z=(1,0)\) and \(t=1/2\), it gives \((1/2,0)\), whose squared distance from the origin is \(1/4\), not \(1\). At \(t=1\) it gives \((0,0)\notin C\). So it does not define a homotopy in the circle.
The correction is not to ignore the codomain condition: a homotopy on \(C\) must take values in \(C\) at every time. The same formula does define a contraction of the closed disk \(D^2=\{z\in\mathbb{R}^2:\|z\|\leq1\}\), because \(\|(1-t)z\|=(1-t)\|z\|\leq1\). But that is a contraction of a different space, not of the circle. In any construction, check the image condition alongside the endpoint equations and, for a strong deformation retraction, the fixed-point condition on the retract.
Check Your Understanding
For each construction, identify the domain, codomain, endpoint maps, and any additional fixed-point condition that must be checked.
- For the strip retraction, verify directly that \(r\circ i=\operatorname{id}_{\mathbb{R}}\) and write the other composite as a map \(X\to X\).
- In the three-armed space, what inequalities show that the contraction remains in each arm?
- Why must the two formulas for the circle with an attached segment agree at \((1,0)\)?
- For the square diagonal retraction, compute the image of \((1,0)\) under \(r\), and verify that it lies in \(A\).
- What extra condition distinguishes a strong deformation retraction from a homotopy from \(\operatorname{id}_X\) to a retraction?