Retractions: Maps That Fix a Subspace
A continuous map from a space into one of its subspaces need not preserve any of the points already in that subspace. A retraction is a particular kind of map that does preserve them all. This pointwise condition is central: it distinguishes a retraction from an arbitrary map with the same codomain, and it gives a useful way to recognize retractions among maps from a space to itself.
Let \((X,\mathcal{T})\) be a topological space and let \(A\subseteq X\) have the subspace topology. The inclusion map \(i:A\to X\) is defined by \(i(a)=a\). A map \(r:X\to A\) is continuous when preimages of open sets in \(A\) are open in \(X\). When discussing fixed points, we regard \(r\) also as the map \(i\circ r:X\to X\), which sends \(x\) to the same point \(r(x)\), now viewed as a point of \(X\).
The equation \(r\circ i=\operatorname{id}_A\) says that starting at a point of \(A\), including it into \(X\), and applying \(r\) returns the same point. This is stronger than merely requiring \(r(X)\subseteq A\). Indeed, a map \(X\to A\) has its values in \(A\) by its codomain, but it may move points of \(A\) to other points.
The Fixed-Point Property
Proof. If \(a\in A\), the retraction condition gives \(R(a)=i(r(a))=i(a)=a\), so every point of \(A\) is fixed. Conversely, if \(R(x)=x\), then \(x=R(x)=i(r(x))\). Since \(r(x)\in A\), the point \(i(r(x))\) belongs to \(A\), and therefore \(x\in A\). These two inclusions show that the fixed-point set is exactly \(A\). \(\square\)
Thus the phrase “a retraction fixes its target” can be made precise in either of two ways: the map \(r:X\to A\) restricts to the identity on \(A\), and the associated self-map \(R:X\to X\) has precisely \(A\) as its set of fixed points. The second statement also rules out a possible ambiguity: a point outside \(A\) cannot be fixed by \(R\), because every value of \(R\) lies in \(A\).
Worked Example: Clamping the Real Line to an Interval
Give a retraction from \(\mathbb{R}\), with its standard topology, onto \(A=[0,1]\), with the subspace topology.
Define \(r:\mathbb{R}\to[0,1]\) by \(r(x)=\min(1,\max(0,x))\). For \(x\le 0\), \(\max(0,x)=0\), so \(r(x)=0\). For \(0\le x\le 1\), \(\max(0,x)=x\), so \(r(x)=x\). For \(x\ge 1\), \(\max(0,x)=x\) and \(\min(1,x)=1\), so \(r(x)=1\). The formulas agree at the endpoints: \(r(0)=0\) and \(r(1)=1\).
For continuity, \(\max(0,x)=(x+|x|)/2\) is continuous, and for any real \(u\), \(\min(1,u)=(1+u-|1-u|)/2\) is continuous. Hence their composition \(r\) is continuous. If \(a\in[0,1]\), then \(\max(0,a)=a\) and \(\min(1,a)=a\), so \(r(a)=a\). Thus \(r\) is a retraction. For example, \(r(-2)=\min(1,\max(0,-2))=\min(1,0)=0\), while \(r(3/4)=\min(1,\max(0,3/4))=\min(1,3/4)=3/4\). The latter computation checks that a point already in the target is unchanged.
Check: The image is contained in \([0,1]\) by the formula, and every \(a\in[0,1]\) is the image of itself because \(r(a)=a\). Therefore the image is exactly \([0,1]\), as required for this retraction.
There is a useful algebraic consequence. A map from \(X\) to itself is called idempotent if applying it twice has the same effect as applying it once. Retractions, viewed as self-maps, are idempotent because after one application the point is in \(A\), where the map acts as the identity.
Proof. First suppose \(r\) is a retraction. For any \(x\in X\), the point \(R(x)=i(r(x))\) lies in \(A\). Since \(r\) fixes \(A\), applying \(R\) to this point leaves it unchanged: \(R(R(x))=R(x)\). This holds for every \(x\), so \(R\circ R=R\).
Conversely, suppose \(R:X\to X\) is continuous and \(R(R(x))=R(x)\) for every \(x\). Set \(A=R(X)\). The same function, with codomain restricted to \(A\), is continuous: if \(O=A\cap U\) is open in \(A\), where \(U\) is open in \(X\), then \(R^{-1}(O)=R^{-1}(U)\), since all values of \(R\) lie in \(A\); this preimage is open. Now take any \(a\in A\). There is \(x\in X\) with \(a=R(x)\). Then \(R(a)=R(R(x))=R(x)=a\). Thus the restricted map fixes every point of \(A\) and is a retraction onto \(A\). \(\square\)
Examples and Non-Examples
Worked Example: Projection of a Square onto Its Bottom Edge
Let \(X=[0,1]\times[0,1]\) with the subspace topology from \(\mathbb{R}^2\), and let \(A=[0,1]\times\{0\}\) with its subspace topology. Define \(r:X\to A\) by \(r(x,y)=(x,0)\).
The values lie in \(A\), since \(0\le x\le1\). To see continuity, an open set of \(A\) has the form \(A\cap U\) for some open set \(U\subseteq\mathbb{R}^2\). Its preimage is \(\{(x,y)\in X:(x,0)\in U\}\), which is open in \(X\): the map \((x,y)\mapsto(x,0)\) is continuous, as its two coordinate functions \(x\mapsto x\) and \((x,y)\mapsto0\) are continuous. Finally, if \((x,0)\in A\), then \(r(x,0)=(x,0)\). Therefore \(r\) is a retraction.
For a direct calculation, \(r(3/4,2/5)=(3/4,0)\), while \(r(3/4,0)=(3/4,0)\). The first point is sent to the bottom edge; the second, already on that edge, is fixed. Every point \((x,0)\) of \(A\) is fixed by the same calculation, not just these two test points.
Check: Applying the map twice gives \(r(r(x,y))=r(x,0)=(x,0)=r(x,y)\). This agrees with the idempotence proposition.
Worked Example: A Retraction in a Finite Discrete Space
Let \(X=\{a,b,c\}\) have the discrete topology, in which every subset is open, and let \(A=\{a,c\}\) have its subspace topology. Define \(r:X\to A\) by \(r(a)=a\), \(r(b)=a\), and \(r(c)=c\).
Every map from a discrete space is continuous: for any open subset \(O\subseteq A\), the preimage \(r^{-1}(O)\) is a subset of \(X\), and every subset of \(X\) is open. The map fixes \(a\) and \(c\), which are all the points of \(A\). Thus it is a retraction. Its values are \(r(a)=a\), \(r(b)=a\), and \(r(c)=c\), so its image is exactly \(\{a,c\}\).
For contrast, define \(g:X\to A\) by \(g(a)=c\), \(g(b)=a\), and \(g(c)=c\). This map is also continuous, for the same discrete-topology reason, and its image is still \(A\). But \(g(a)=c\ne a\), so \(g\) is not a retraction onto \(A\). This example shows that continuity and having image \(A\) do not replace the requirement to fix every point of \(A\).
Worked Example: A Continuous Bijection That Is Not a Retraction
Take \(X=A=[0,1]\), with the standard topology, and define \(f:X\to A\) by \(f(x)=1-x\). This map is continuous, and it is a bijection: applying it twice gives \(f(f(x))=1-(1-x)=x\). Nevertheless, it is not a retraction onto \(A\), because a retraction onto the whole space \(A=X\) must fix every point. At \(x=1/4\), the calculation is \(f(1/4)=1-1/4=3/4\ne1/4\).
The point \(1/2\) is fixed, since \(f(1/2)=1-1/2=1/2\), but a retraction onto \([0,1]\) must fix \(1/4\) and every other point as well. This example distinguishes being continuous and bijective from being a retraction. In the particular case \(A=X\), the identity map is the only retraction, because the defining condition already requires \(r(x)=x\) for every \(x\in X\).
Check: The formula gives \(f(f(1/4))=f(3/4)=1-3/4=1/4\), confirming that \(f\) is idempotent after two applications, not after one. In fact \(f\circ f=\operatorname{id}_X\), whereas \(f\ne\operatorname{id}_X\), so it does not satisfy the idempotence condition for a retraction onto its image.
A Subspace That Cannot Be a Retract
Not every subspace admits a retraction from the space containing it. The intermediate value theorem gives a simple obstruction. The interval \([0,1]\) cannot retract onto its two endpoints, even though a map that sends every point to one of the endpoints can certainly be defined. The obstacle is continuity together with the requirement to fix both endpoints.
Worked Example: No Retraction from the Interval onto Its Endpoints
Give \([0,1]\) and \(A=\{0,1\}\) their subspace topologies from \(\mathbb{R}\). Suppose, for contradiction, that \(r:[0,1]\to A\) is a retraction. Then \(r(0)=0\) and \(r(1)=1\). The inclusion \(A\to\mathbb{R}\) is continuous, so regarding \(r\) as a real-valued function gives a continuous function on \([0,1]\) with endpoint values \(0\) and \(1\).
By the intermediate value theorem, this real-valued function takes the value \(1/2\) at some point of \([0,1]\). But every value of \(r\) must belong to \(A=\{0,1\}\), and \(1/2\notin\{0,1\}\). This contradiction proves that no such retraction exists. The use of the intermediate value theorem is essential here: it uses continuity on the whole interval, not merely the endpoint conditions.
For comparison, the constant map \(c:[0,1]\to\{0,1\}\) given by \(c(x)=0\) is continuous, since a constant map has open preimages of open sets. But \(c(1)=0\ne1\), so it is not a retraction. Defining a map into the proposed target is possible; fixing every point of that target continuously is the part that fails.
What to Check When Testing a Retraction
For a proposed map \(r:X\to A\), check the conditions separately. First, confirm that \(A\) is being given the subspace topology and that \(r\) is continuous as a map into \(A\). Second, take an arbitrary \(a\in A\) and verify \(r(a)=a\); checking only selected points is not enough. If instead a self-map \(R:X\to X\) is given, a useful equivalent test is to verify that it is continuous, idempotent, and then take its image as the target.
A common mistake is to infer “retraction” from the fact that the formula has values in \(A\), or even from the stronger fact that its image is all of \(A\). The reflection \(f(x)=1-x\) on \([0,1]\) disproves both shortcuts: it is continuous and onto \([0,1]\), but \(f(1/4)=3/4\), so it does not fix the target. The correct test is the defining identity on every point of the target.
Retractions also identify a subspace through the fixed points of a continuous self-map: for a retraction, the fixed-point set is exactly its target. This gives a practical way to check a formula and a conceptual way to recognize the role of the subspace. It does not say that every subspace is a retract; the endpoint example shows that continuity can prevent one from existing.
Check Your Understanding
For each question, distinguish the continuity condition from the requirement that the target be fixed pointwise.
- Verify that \(r:\mathbb{R}\to\{0\}\), defined by \(r(x)=0\), is a retraction when \(\{0\}\) has the subspace topology.
- Let \(R:X\to X\) be continuous and satisfy \(R\circ R=R\). Explain why every point of \(R(X)\) is fixed by \(R\).
- On \(X=[0,1]\times[0,1]\), what is the fixed-point set of \(R(x,y)=(x,0)\)?
- Why is the continuous map \(g:\{a,b,c\}\to\{a,c\}\) defined by \(g(a)=c\), \(g(b)=a\), and \(g(c)=c\) not a retraction onto \(\{a,c\}\)? Assume the discrete topology.
- Use the intermediate value theorem to explain why a retraction from \([0,1]\) onto \(\{0,1\}\) cannot exist.