Keeping a Subspace Fixed During a Homotopy
A homotopy deforms one continuous map into another through a continuous family of maps. Sometimes the deformation must leave certain points untouched at every intermediate time, not merely return them to their original positions at the beginning and end. This is the meaning of a homotopy relative to a subspace. The requirement is pointwise in both the chosen subspace and the time parameter.
Let \(X\) and \(Y\) be topological spaces, let \(A\subseteq X\) have the subspace topology, and let \(f,g:X\to Y\) be continuous. Write \(I=[0,1]\) with its standard topology. A homotopy from \(f\) to \(g\) 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 \(x\in X\). The product \(X\times I\) is given the product topology.
The condition requires the same value at every time: for each fixed \(a\in A\), the function \(t\mapsto H(a,t)\) is constant with value \(f(a)\). It is stronger than asking only that \(H(a,0)=H(a,1)\). A homotopy can start and finish at the same value on \(A\) but move those points in between.
Checking the Relative Condition
Worked Example: An Interval Fixed at Both Ends
Give a homotopy relative to \(A=\{-1,1\}\) between \(f,g:I\to\mathbb{R}\), where \(I=[-1,1]\) has its subspace topology from \(\mathbb{R}\), \(\mathbb{R}\) has its standard topology, and \(f(x)=x\), \(g(x)=x^3\).
Both maps are continuous because they are polynomial functions restricted to \(I\). Their values agree at the two points of \(A\): \(f(-1)=-1=g(-1)\) and \(f(1)=1=g(1)\). Define \[ H(x,t)=(1-t)x+tx^3. \] This is continuous on \(I\times I\), since it is a polynomial in the coordinates \(x\) and \(t\). At the initial time, \(H(x,0)=(1-0)x+0x^3=x=f(x)\). At the final time, \(H(x,1)=(1-1)x+1x^3=x^3=g(x)\).
Now check every point of the subspace at every time. If \(a=-1\), then \[ H(-1,t)=(1-t)(-1)+t(-1)^3=-(1-t)-t=-1=f(-1). \] If \(a=1\), then \[ H(1,t)=(1-t)(1)+t(1)^3=1-t+t=1=f(1). \] Thus \(H(a,t)=f(a)\) for both \(a\in A\) and all \(t\in I\), as required.
Check: at \(x=1/2\) and \(t=1/2\), the formula gives \[ H(1/2,1/2)=(1/2)(1/2)+(1/2)(1/8)=1/4+1/16=5/16. \] This is an intermediate value, not an endpoint value. At the fixed point \(x=1\), the same time gives \(H(1,1/2)=(1/2)(1)+(1/2)(1)=1\), confirming the relative condition there.
This example uses linear interpolation of the two maps. When the target is a convex subset of a real vector space, the straight-line formula \(H(x,t)=(1-t)f(x)+tg(x)\) stays in the target; if \(f\) and \(g\) agree on \(A\), it also fixes \(A\). The formula is not automatically available for an arbitrary target space, since addition and scalar multiplication may not be defined there.
Worked Example: A Disk Fixed on Its Boundary
Let \(D^2=\{x\in\mathbb{R}^2:\|x\|\leq 1\}\), with its subspace topology from Euclidean \(\mathbb{R}^2\), and let \(A=S^1=\{x\in\mathbb{R}^2:\|x\|=1\}\). The target is also \(D^2\). Construct a homotopy relative to \(S^1\) from the identity map \(f(x)=x\) to \(g(x)=(2-\|x\|)x\).
First check that \(g\) takes values in \(D^2\). Put \(r=\|x\|\), so \(0\leq r\leq 1\). Since \(2-r\geq 0\), \(\|g(x)\|=r(2-r)\), and \[ 1-r(2-r)=1-2r+r^2=(1-r)^2\geq 0. \] Thus \(0\leq\|g(x)\|\leq 1\). The map is continuous because the norm and scalar multiplication are continuous. On the boundary \(r=1\), \(g(x)=(2-1)x=x=f(x)\).
Define \[ H(x,t)=(1-t)x+tg(x)=\bigl(1+t(1-\|x\|)\bigr)x. \] The formula is continuous. For \(r=\|x\|\), its output has norm \(r(1+t(1-r))\). Since \(0\leq t\leq1\), \[ r(1+t(1-r))=(1-t)r+t\,r(2-r)\leq (1-t)\cdot1+t\cdot1=1. \] So \(H\) takes values in \(D^2\) throughout the homotopy. Substituting \(t=0\) gives \(H(x,0)=x=f(x)\); substituting \(t=1\) gives \(H(x,1)=(2-\|x\|)x=g(x)\). Finally, for \(a\in S^1\), \(\|a\|=1\), so \[ H(a,t)=\bigl(1+t(1-1)\bigr)a=a=f(a) \] for every \(t\). This verifies the relative condition on the entire boundary.
Check at the centre: \(H(0,t)=\bigl(1+t(1-0)\bigr)0=0\), so the formula is defined and remains in the disk there as well. The boundary calculation and this check are consistent with the fact that the formula uses the norm, which is defined at every point of \(D^2\).
Endpoint Agreement Is Not Enough
Worked Example: A Homotopy That Moves the Subspace
Let \(X=[0,1]\) with its standard subspace topology, \(Y=\mathbb{R}\) with its standard topology, and \(A=\{1/2\}\). Take \(f=g:X\to\mathbb{R}\) to be the constant zero map. The endpoint maps agree on \(A\). Consider \[ H(x,t)=4t(1-t), \] which is continuous and independent of \(x\). At both endpoint times, \(H(x,0)=0=f(x)\) and \(H(x,1)=0=g(x)\), so it is a homotopy from \(f\) to \(g\).
It is not a homotopy relative to \(A\): at \(a=1/2\) and \(t=1/2\), \[ H(1/2,1/2)=4(1/2)(1-1/2)=1, \] whereas \(f(1/2)=0\). The point of \(A\) moves during this homotopy. There is, however, a different homotopy relative to \(A\), namely \(K(x,t)=0\) for all \(x,t\). This illustrates the distinction: the condition concerns the chosen homotopy at every time, not just its endpoint maps.
Common pitfall. Endpoint agreement on \(A\) is necessary, but it does not make every homotopy between \(f\) and \(g\) relative to \(A\). In the example, \(f|_A=g|_A\), yet the displayed \(H\) fails the pointwise test. To check a proposed relative homotopy, substitute an arbitrary \(a\in A\) into the formula and verify the resulting equality for every \(t\in I\); checking only \(t=0\) and \(t=1\) is insufficient.
There is also a necessary condition that can be checked before searching for a homotopy: the endpoint maps must agree on \(A\). For example, if \(A=\{a\}\) and \(f(a)\ne g(a)\), then a relative homotopy would require \(H(a,0)=f(a)\) and \(H(a,1)=g(a)\), while also requiring \(H(a,t)=f(a)\) for every \(t\). Taking \(t=1\) would force \(g(a)=f(a)\), a contradiction.
Relative Homotopy as an Equivalence Relation
For a fixed subset \(A\subseteq X\) and a fixed continuous map \(r:A\to Y\), consider the set of continuous maps \(f:X\to Y\) satisfying \(f|_A=r\). Relative homotopy is an equivalence relation on this set: each such map is related to itself, the relation is symmetric, and it is transitive.
For the transitive step, we use the finite closed pasting lemma: if a space is covered by finitely many closed subsets and continuous maps on those subsets agree on overlaps, the map they define on the whole space is continuous. To see why, for any closed set \(F\) in the target, the preimage of \(F\) is the union of the preimages within the finitely many closed pieces. Each such preimage is closed in its piece, hence closed in the whole space because the piece is closed. Their finite union is closed. Taking complements gives continuity by the closed-set criterion.
Proof. For reflexivity, the map \(H(x,t)=f(x)\) is continuous, has both endpoint maps equal to \(f\), and satisfies \(H(a,t)=f(a)=r(a)\) for every \(a\in A\) and \(t\in I\).
For symmetry, suppose \(H\) is a homotopy relative to \(A\) from \(f\) to \(g\). Define \(K(x,t)=H(x,1-t)\). The map \((x,t)\mapsto(x,1-t)\) is continuous, so \(K\) is continuous. Its endpoint maps are \(K(x,0)=H(x,1)=g(x)\) and \(K(x,1)=H(x,0)=f(x)\). For \(a\in A\), \(K(a,t)=H(a,1-t)=r(a)\). Thus \(K\) is a homotopy relative to \(A\) from \(g\) to \(f\).
For transitivity, suppose \(H\) is relative to \(A\) from \(f\) to \(g\), and \(J\) is relative to \(A\) from \(g\) to \(h\). Define \[ L(x,t)= \begin{cases} H(x,2t),&0\leq t\leq 1/2,\\ J(x,2t-1),&1/2\leq t\leq 1. \end{cases} \] At \(t=1/2\), the first formula gives \(H(x,1)=g(x)\), and the second gives \(J(x,0)=g(x)\), so the formulas agree. The two time slabs \(X\times[0,1/2]\) and \(X\times[1/2,1]\) are closed in \(X\times I\). Each formula is continuous on its slab, and the formulas agree on the overlap; the finite closed pasting lemma therefore gives continuity of \(L\). Its endpoint maps are \(L(x,0)=f(x)\) and \(L(x,1)=h(x)\). If \(a\in A\), each applicable formula gives \(r(a)\), so \(L(a,t)=r(a)\) throughout. Hence \(L\) is a homotopy relative to \(A\) from \(f\) to \(h\). \(\square\)
Two useful consequences follow directly from the definition. First, if \(H\) is relative to \(A\) and \(B\subseteq A\), then \(H\) is relative to \(B\), because every \(b\in B\) is also in \(A\). Second, if \(u:Y\to Z\) is continuous, then \(u\circ H:X\times I\to Z\) is relative to \(A\): for \(a\in A\), \((u\circ H)(a,t)=u(f(a))\), independent of \(t\). These observations allow a fixed-subspace condition to be retained under restriction to a smaller subspace or under continuous postcomposition.
What to Verify in Practice
When checking a proposed homotopy relative to \(A\), organize the verification into three separate checks:
- Confirm that the formula defines a continuous map \(H:X\times I\to Y\) with values in the stated target.
- Substitute \(t=0\) and \(t=1\) to recover the required endpoint maps.
- For an arbitrary \(a\in A\), compute \(H(a,t)\) and show that it equals \(f(a)\) for every \(t\in I\).
The third check is the defining feature. It is not enough to know that the two endpoint maps agree on \(A\), and it is not enough to test a few selected times. A relative homotopy gives a continuous deformation of the whole map while preserving all specified values on the subspace at every instant.
Check Your Understanding
For each question, distinguish the endpoint condition from the condition imposed at every intermediate time.
- For \(f(x)=x\) and \(g(x)=x^3\) on \([-1,1]\), why does the straight-line homotopy fix \(\{-1,1\}\), and what is its value at \(x=1/2,\ t=1/2\)?
- In the disk example, verify directly from the formula that \(H(a,t)=a\) whenever \(\|a\|=1\).
- For \(H(x,t)=4t(1-t)\) and \(A=\{1/2\}\), calculate \(H(1/2,1/2)\). Does endpoint agreement make this homotopy relative to \(A\)?
- Suppose a homotopy \(H\) is relative to \(A\). Explain why it is relative to every subset \(B\subseteq A\).
- Describe the piecewise formula used to concatenate a homotopy from \(f\) to \(g\) with one from \(g\) to \(h\), and identify why the formulas agree at the joining time.