What Must a Homotopy Do?
A retraction is a continuous map from a space onto a subspace that leaves every point of that subspace unchanged. But being a retraction is an endpoint condition on a map; it does not by itself provide a continuous deformation from the identity map to that retraction. To test a particular retraction, keep track of three requirements: where the homotopy starts, where it ends, and what it fixes throughout.
Let \((X,\mathcal{T})\) be a topological space and \(A\subseteq X\) with the subspace topology. Write \(i:A\to X\) for the inclusion map.
The endpoint condition is specific to the retraction \(r\) under consideration. It is not enough to find a homotopy ending at some retraction onto \(A\). The fixed-subspace condition is also imposed at every intermediate time, not only at \(t=0\) and \(t=1\).
Necessary Checks and a Useful Criterion
Proof. Fix \(x\in X\). The endpoint equations in the definition give \(H(x,0)=x\) and \(H(x,1)=i(r(x))\), so \(t\mapsto H(x,t)\) has the stated endpoints. If \(a\in A\), the fixed-subspace equation gives \(H(a,t)=a\) for all \(t\in I\). Finally, continuity of the family as \((x,t)\) varies is exactly continuity of \(H:X\times I\to X\), which is part of the definition. \(\square\)
The pointwise path requirement is a quick obstruction test: if even one \(x\) cannot be joined to \(i(r(x))\) by a path in \(X\), no such deformation retraction exists. Passing this test does not construct a deformation retraction. One still needs a single continuous map on \(X\times I\) satisfying all three conditions.
Proof. Regard \(r\) as taking values in \(X\) through the inclusion \(i\). Define $$ H(x,t)=(1-t)x+t\,i(r(x)). $$ The identity map, \(i\circ r\), scalar multiplication, and addition are continuous, so \(H\) is continuous. At \(t=0\), \(H(x,0)=x\); at \(t=1\), \(H(x,1)=i(r(x))\). If \(a\in A\), then \(r(a)=a\), and therefore $$ H(a,t)=(1-t)a+ta=a. $$ Since \(x\) and \(i(r(x))\) belong to the convex set \(X\), every point on the segment between them belongs to \(X\). Thus \(H\) takes values in \(X\) and satisfies all the required conditions. \(\square\)
Worked Tests
Worked Example: A Retraction of a Square
Give the square \(X=[-1,1]\times[0,1]\) the Euclidean subspace topology, and let \(A=[-1,1]\times\{0\}\) with its subspace topology. Consider $$ r(x,y)=(x(1-2y),0). $$ We check that this particular map is a deformation-retraction endpoint.
For \((x,y)\in X\), we have \(-1\leq x\leq1\) and \(-1\leq1-2y\leq1\), so \(-1\leq x(1-2y)\leq1\). The second coordinate of \(r(x,y)\) is \(0\), so \(r(x,y)\in A\). If \((x,0)\in A\), then \(r(x,0)=(x,0)\). The formula is continuous, hence \(r\) is a retraction.
Define $$ H((x,y),t)=((1-2ty)x,(1-t)y). $$ Both coordinates are continuous functions of \((x,y,t)\). Since \(0\leq t,y\leq1\), the factor \(1-2ty\) lies in \([-1,1]\); consequently the first coordinate lies in \([-1,1]\). The second lies in \([0,1]\), so \(H\) takes values in \(X\). At \(t=0\), \(H((x,y),0)=(x,y)\). At \(t=1\), \(H((x,y),1)=(x(1-2y),0)=r(x,y)\). If \(y=0\), then \(H((x,0),t)=(x,0)\) for every \(t\). Thus all conditions hold. For instance, \(r(1,1)=(-1,0)\), and the path given by the homotopy is \(H((1,1),t)=(1-2t,1-t)\); it starts at \((1,1)\) and ends at \((-1,0)\).
Check: at \(t=0\), the displayed formula returns \((x,y)\); at \(t=1\), it returns exactly the specified \(r(x,y)\), not merely some map into \(A\). Points of \(A\) remain fixed throughout.
Worked Example: Two Retractions, Two Endpoints
Give \(X=[0,1]\) and \(A=[0,\tfrac12]\) their usual subspace topologies. Define two maps \(q,r:X\to A\) by $$ q(x)= \begin{cases} x,&0\leq x\leq \tfrac12,\\ \tfrac12,&\tfrac12\leq x\leq1, \end{cases} \qquad r(x)= \begin{cases} x,&0\leq x\leq \tfrac12,\\ 1-x,&\tfrac12\leq x\leq1. \end{cases} $$ The formulas agree at \(x=\tfrac12\), so both maps are continuous. Their values lie in \(A\), and both fix every point of \(A\); hence both are retractions.
For the specified retraction \(r\), put $$ H(x,t)=(1-t)x+t\,r(x). $$ Both \(x\) and \(r(x)\) lie in \([0,1]\), so their convex combination lies in \([0,1]\). The formula is continuous because \(r\) is continuous. It starts at \(x\), ends at \(r(x)\), and for \(x\in A\), \(r(x)=x\), so it fixes \(A\). Therefore this is a deformation retraction with endpoint \(r\). The same argument, using \(q\) in place of \(r\), gives a deformation retraction with endpoint \(q\).
It matters which endpoint is specified. The homotopy \(K(x,t)=(1-t)x+tq(x)\) ends at \(q\), not at \(r\). For \(x=\tfrac34\), the second branch gives \(q(\tfrac34)=\tfrac12\) and \(r(\tfrac34)=\tfrac14\). Thus \(K(\tfrac34,1)=\tfrac12\ne\tfrac14=r(\tfrac34)\). This homotopy cannot certify that \(r\) is a deformation-retraction endpoint, even though \(r\) does have its own suitable homotopy.
Worked Example: A Retraction That Cannot Be a Deformation Retraction
Let \(X=\{a,b\}\) have the discrete topology, and give \(A=\{a\}\) its subspace topology. The map \(r:X\to A\) defined by \(r(a)=r(b)=a\) is continuous, takes values in \(A\), and fixes \(a\). It is therefore a retraction.
Suppose there were a deformation retraction with endpoint \(r\). Evaluating it at \(b\) would give a continuous path \(h:I\to X\) with \(h(0)=b\) and \(h(1)=a\). Define \(\phi:X\to\mathbb{R}\) by \(\phi(a)=0\) and \(\phi(b)=1\). This map is continuous because every subset of \(X\) is open. Then \(\phi\circ h:I\to\mathbb{R}\) is continuous and takes the values \(1\) and \(0\) at the endpoints. By the intermediate value theorem it must take the value \(\tfrac12\), but \(\phi\circ h\) takes values only in \(\{0,1\}\). This contradiction proves that no such path, and hence no deformation retraction with endpoint \(r\), exists.
Check: the obstruction is not a failure of the retraction equations; those hold. It is the failure of the necessary path condition for \(b\).
Common Mistakes: Endpoint and Fixed Points
A proposed homotopy should be checked as a map on the whole product \(X\times I\), not only by checking its initial and final slices. For example, on the square \(X=[0,1]\times[0,1]\), let \(A=[0,1]\times\{0\}\) and \(r(x,y)=(x,0)\). The map $$ G((x,y),t)=\bigl(x,(1-t)y+t(1-t)x(1-x)\bigr) $$ is continuous and takes values in \(X\). Indeed, \(0\leq x(1-x)\leq\tfrac14\), so its second coordinate lies between \(0\) and $$ (1-t)+\frac{t(1-t)}4=(1-t)(1+t/4)\leq1. $$ At \(t=0\), \(G((x,y),0)=(x,y)\), and at \(t=1\), \(G((x,y),1)=(x,0)=r(x,y)\). But for \((x,y)=(\tfrac12,0)\) and \(t=\tfrac12\), its second coordinate is $$ \frac12\left(1-\frac12\right)\frac12\left(1-\frac12\right)=\frac1{16}, $$ so it moves a point of \(A\). Thus \(G\) has the right endpoint but is not the required deformation retraction. The vertical homotopy \(((x,y),t)\mapsto(x,(1-t)y)\) does give one for this same \(r\).
In practice, test a proposed construction in this order: verify its values stay in \(X\); check \(H(x,0)=x\); substitute \(t=1\) and compare with the specified \(i\circ r\); then substitute an arbitrary \(a\in A\) and check that \(H(a,t)=a\) for every \(t\). A failure at any one stage is decisive.
Check Your Understanding
For each question, distinguish carefully between the retraction map and a homotopy that might have it as endpoint.
- State the three equations a homotopy must satisfy to be a deformation retraction with endpoint \(r\).
- For a retraction \(r:X\to A\), what path must the homotopy trace for a point \(x\in X\)? What must it do for \(a\in A\)?
- Why does the discrete two-point example rule out a deformation retraction even though the constant map is a retraction?
- In the interval example, evaluate \(q(\tfrac34)\) and \(r(\tfrac34)\). Which endpoint does the homotopy built using \(q\) reach?
- Explain why having the correct initial and final maps is not enough for a deformation retraction.