Shrinking a Space onto a Subspace
A retraction sends a space onto a subspace while leaving every point of that subspace fixed. A deformation retraction adds a further requirement: the retraction must be reachable from the identity map by a continuous homotopy. This records how the whole space moves onto the subspace, not merely where its points end up. A strong deformation retraction imposes an additional constraint: points already in the subspace must stay fixed throughout the motion.
Let \(X\) be a topological space and \(A\subseteq X\) have the subspace topology. Write \(i:A\to X\) for the inclusion, \(i(a)=a\). A homotopy from \(f:X\to X\) to \(g:X\to X\) is a continuous map \(H:X\times I\to X\), where \(I=[0,1]\), such that \(H(x,0)=f(x)\) and \(H(x,1)=g(x)\) for each \(x\in X\). Continuity here is for the product topology on \(X\times I\).
The endpoint condition in either definition is \(H(x,0)=x\) and \(H(x,1)=i(r(x))\). Since \(r(a)=a\), the endpoint at \(t=1\) fixes \(A\); the extra condition for a strong deformation retraction requires that \(A\) remain fixed at every intermediate time. Thus every strong deformation retraction is a deformation retraction, but a particular homotopy witnessing a deformation retraction need not be strong.
The Homotopy-Equivalence Consequence
A continuous map \(f:X\to Y\) is a homotopy equivalence if there is a continuous map \(g:Y\to X\) such that \(g\circ f\) is homotopic to \(\operatorname{id}_X\) and \(f\circ g\) is homotopic to \(\operatorname{id}_Y\). Such a \(g\) is called a homotopy inverse. The homotopies need not be equalities of maps.
Proof. The maps \(i\) and \(r\) are continuous: \(i\) is the subspace inclusion, and continuity of \(r\) is part of the definition of retraction. For every \(a\in A\), \(r(i(a))=r(a)=a\), so \(r\circ i=\operatorname{id}_A\). In the other order, \(i\circ r:X\to X\) is homotopic to \(\operatorname{id}_X\) by the homotopy in the definition of deformation retraction. These are exactly the two conditions for \(r\) to be a homotopy inverse to \(i\). A strong deformation retraction includes such a homotopy, so the same argument applies. \(\square\)
This conclusion separates the roles of the maps and the homotopy. The retraction gives an exact identity on \(A\), while the deformation homotopy supplies the possibly non-exact identity relation on \(X\). Neither condition alone can be silently substituted for the other.
Worked Example: The Interval Deformation-Retracts onto an Endpoint
Give a strong deformation retraction of \(X=[0,1]\), with its subspace topology from \(\mathbb{R}\), onto \(A=\{0\}\), also with its subspace topology.
Define \(r:X\to A\) by \(r(x)=0\). This is continuous because it is constant, and \(r(0)=0\), so it is a retraction. Define \(H:X\times I\to X\) by \(H(x,t)=(1-t)x\). The formula uses continuous real operations, and its values lie in \(X\): when \(x,t\in[0,1]\), both factors are nonnegative and \((1-t)x\le 1\).
At \(t=0\), \(H(x,0)=(1-0)x=x\). At \(t=1\), \(H(x,1)=(1-1)x=0=i(r(x))\). For the only point \(a=0\) of \(A\), \(H(0,t)=(1-t)0=0\) for every \(t\). Thus this is a strong deformation retraction. By the theorem, the inclusion of \(\{0\}\) into \([0,1]\) is a homotopy equivalence with \(r\) as homotopy inverse.
Check: Substituting \(x=1\) gives \(H(1,0)=1\) and \(H(1,1)=0\), as required; substituting \(x=0\) gives \(H(0,t)=0\) throughout.
What the “Strong” Condition Adds
The distinction concerns the motion at intermediate times, not the endpoint retraction. A homotopy may move points of \(A\) away from their original positions and return them by time \(1\). The following example verifies that possibility while keeping the entire homotopy inside the space.
Worked Example: A Deformation Homotopy That Is Not Strong
Again let \(X=[0,1]\) and \(A=\{0\}\), with the topologies just specified. Keep the retraction \(r(x)=0\), but define a different map \(K:X\times I\to X\) by \(K(x,t)=(1-t)x+\frac12(1-x)\sin(\pi t)\).
For fixed \(t\), this expression is affine in \(x\). Its values at the endpoints of the \(x\)-interval are \(K(0,t)=\frac12\sin(\pi t)\) and \(K(1,t)=1-t\). Both belong to \([0,1]\), since \(0\le\sin(\pi t)\le1\) and \(0\le1-t\le1\). Every value of an affine function on \([0,1]\) lies between its endpoint values, so \(K(x,t)\in[0,1]\) for all \(x,t\). The formula is continuous.
At \(t=0\), \(K(x,0)=x+\frac12(1-x)\sin(0)=x\). At \(t=1\), \(K(x,1)=0+\frac12(1-x)\sin(\pi)=0=i(r(x))\). It therefore witnesses a deformation retraction. But at the point \(0\in A\) and time \(t=\frac12\), \(K(0,\frac12)=\frac12\sin(\frac{\pi}{2})=\frac12\), which is not \(0\). This particular homotopy is not strong. The pair \((X,A)\) still admits the strong homotopy in the preceding example; the point is that a deformation homotopy need not itself fix \(A\) throughout.
Check: The endpoint values of \(K(0,t)\) are \(K(0,0)=0\) and \(K(0,1)=0\), while its intermediate value at \(t=\frac12\) is \(\frac12\). This verifies both the return to the retraction at time \(1\) and the failure to stay fixed along the way.
A Retraction Alone Is Not Enough
The homotopy requirement is substantive. A continuous retraction supplies a candidate endpoint map, but there may be no continuous way to move the identity to that map. A discrete space gives a direct verification of this failure.
Worked Example: A Retraction That Is Not a Deformation Retraction
Let \(X=\{0,1\}\) have the discrete topology, in which every subset is open, and let \(A=\{0\}\) have its subspace topology. The map \(r:X\to A\) defined by \(r(0)=r(1)=0\) is continuous: the preimage of either open subset of \(A\) is either \(\varnothing\) or \(X\). It fixes \(A\), so it is a retraction.
Suppose, for contradiction, that there were a homotopy \(H:X\times I\to X\) from \(\operatorname{id}_X\) to \(i\circ r\). Fix \(x=1\). The map \(h:I\to X\) given by \(h(t)=H(1,t)\) is continuous, since it is the composition of \(H\) with the continuous map \(t\mapsto(1,t)\). A continuous map from the connected interval \(I\) to a discrete space must be constant: if it took both values, the preimages of \(\{0\}\) and \(\{1\}\) would be disjoint, nonempty open sets partitioning \(I\), contradicting connectedness of the interval.
But the proposed homotopy requires \(h(0)=1\) and \(h(1)=0\), so \(h\) is not constant. This contradiction proves that the retraction is not a deformation retraction. In particular, the existence of \(r\) alone does not establish that \(i\) is a homotopy equivalence.
Check: The identity sends \(1\) to \(1\), while \(i\circ r\) sends \(1\) to \(0\). A homotopy would have to connect these distinct endpoints by a path in the discrete space, and the continuity argument shows that no such path exists.
Common Pitfall: Checking Only the Endpoints
A common error is to verify \(H(x,0)=x\) and \(H(x,1)=i(r(x))\), then conclude that \(H\) is a deformation homotopy. These endpoint equations do not establish that \(H\) is continuous or that its values lie in \(X\). In the discrete example, the tempting formula \(H(x,t)=(1-t)x\) has the right endpoint values for \(x=1\), but at \(t=\frac12\) it gives \(H(1,\frac12)=\frac12\), which is not an element of \(X=\{0,1\}\). It is not even a map \(X\times I\to X\), so it cannot be a homotopy in that space. For a strong deformation retraction, one must also check \(H(a,t)=a\) for every \(a\in A\) and every \(t\in I\), not just at \(t=0\) and \(t=1\).
Deformation retracts matter because they turn a geometric description of a space being continuously reduced to a subspace into an algebraic consequence: the inclusion is a homotopy equivalence. That consequence is weaker than being a homeomorphism. For instance, the interval and the one-point space \(\{0\}\) are homotopy equivalent by the first example, but they are not homeomorphic: a homeomorphism must be a bijection, and no map from a two-point-or-larger set such as \([0,1]\) to a singleton is bijective.
Check Your Understanding
For each question, distinguish the endpoint retraction from the homotopy that is meant to realize it.
- State the equations a homotopy \(H:X\times I\to X\) must satisfy to witness that \(A\) is a deformation retract of \(X\).
- What additional equation defines a strong deformation retraction?
- If \(A\) is a deformation retract of \(X\), identify maps that are homotopy inverses and state which composite is exactly the identity.
- Why does the constant retraction from the discrete two-point space onto \(\{0\}\) fail to give a deformation retraction?
- Can a homotopy witnessing a deformation retraction move points of \(A\) at intermediate times? Explain using the interval example.