Interpolating Between Maps
A homotopy changes one continuous map into another through a continuously varying family of maps. When the target is a convex subset of a real vector space, the most direct construction is linear interpolation: at each point of the domain, move along the straight segment between the two map values. Convexity is exactly what guarantees that this segment stays inside the target.
Write \(I=[0,1]\) with its standard topology. A homotopy from \(f:X\to Y\) to \(g:X\to Y\) is a continuous map \(H:X\times I\to Y\), where \(X\times I\) has the product topology, such that \(H(x,0)=f(x)\) and \(H(x,1)=g(x)\) for every \(x\in X\). The maps \(f\) and \(g\) are then called homotopic.
The topology on \(\mathbb{R}^n\) is the standard Euclidean topology, and a subset \(C\) is given the subspace topology. Thus a map into \(C\) is continuous when it is continuous as a map into that subspace. The formula for interpolation is evaluated in the ambient vector space \(\mathbb{R}^n\); convexity will ensure its values actually lie in \(C\).
The Straight-Line Homotopy Theorem
Proof. Regard \(f\) and \(g\) as continuous maps into \(\mathbb{R}^n\), using the continuous inclusion of the subspace \(C\) into \(\mathbb{R}^n\). The projection maps \(\pi_X:X\times I\to X\) and \(\pi_I:X\times I\to I\) are continuous by the definition of the product topology. Consequently, \((x,t)\mapsto f(\pi_X(x,t))\), \((x,t)\mapsto g(\pi_X(x,t))\), and \((x,t)\mapsto\pi_I(x,t)\) are continuous. Addition and scalar multiplication in \(\mathbb{R}^n\) are continuous, so the displayed formula defines a continuous map \(H:X\times I\to\mathbb{R}^n\). For each \((x,t)\), both \(f(x)\) and \(g(x)\) belong to \(C\), and \(t\in I\). Convexity therefore gives \((1-t)f(x)+tg(x)\in C\). Hence \(H\) is continuous as a map into the subspace \(C\). Finally, substituting \(t=0\) gives \(H(x,0)=f(x)\), and substituting \(t=1\) gives \(H(x,1)=g(x)\). Thus \(H\) is a homotopy from \(f\) to \(g\). \(\square\)
This theorem gives a complete answer for a common question about maps into a convex target: any two continuous maps with the same domain and that target are homotopic. There is no further condition on the domain. If \(X\) is empty, the unique maps and the unique homotopy are empty functions, so the conclusion still holds. If \(C\) is empty and maps \(X\to C\) exist, then \(X\) must be empty, which is already covered.
Worked Example: Two Maps into an Interval
Take \(X=I\) and \(C=I\), both with their standard topologies. Define \(f(x)=x^2\) and \(g(x)=x\). These maps are continuous, and their values lie in \(I\) because \(0\le x\le1\) implies \(0\le x^2\le1\). The straight-line homotopy is
For \(x,t\in I\), the coefficients \(1-t\) and \(t\) are nonnegative and sum to \(1\). Since \(x^2,x\in[0,1]\), their weighted average lies in \([0,1]\). The formula is continuous because it is a polynomial in \(x\) and \(t\).
For a numerical check, substitute \(x=\frac12\) and \(t=\frac13\). First \(x^2=\frac14\), \(1-t=\frac23\), and \(tx=\frac13\cdot\frac12=\frac16\). Therefore \(H(\frac12,\frac13)=\frac23\cdot\frac14+\frac16=\frac16+\frac16=\frac13\). At the endpoints of the parameter, \(H(x,0)=x^2=f(x)\) and \(H(x,1)=x=g(x)\), as required. The computed value \(\frac13\) is in the target interval.
Worked Example: Maps into a Triangle
Let \(X=\{a,b,c\}\) have the discrete topology, and let the target be the convex triangle \(C=\{(u,v)\in\mathbb{R}^2:u\ge0,\ v\ge0,\ u+v\le1\}\), with the subspace topology. Define maps by \(f(a)=(1,0)\), \(f(b)=(0,1)\), \(f(c)=(0,0)\), and \(g(a)=(0,0)\), \(g(b)=(\frac12,\frac12)\), \(g(c)=(0,1)\). Every listed point belongs to \(C\): its coordinates are nonnegative and their sum is at most \(1\). Every map from a discrete space is continuous, since the preimage of any open set is a subset of a discrete space and hence open.
At a general parameter \(t\in I\), interpolation gives \(H(a,t)=(1-t,0)\), \(H(b,t)=(\frac{t}{2},1-\frac{t}{2})\), and \(H(c,t)=(0,t)\). For \(a\), the coordinates are nonnegative and sum to \(1-t\le1\). For \(b\), they are nonnegative because \(0\le t\le1\), and their sum is \(\frac{t}{2}+1-\frac{t}{2}=1\). For \(c\), they are nonnegative and sum to \(t\le1\). Thus all three values remain in the triangle at every parameter.
At \(t=\frac12\), the three values are respectively \((\frac12,0)\), \((\frac14,\frac34)\), and \((0,\frac12)\). Their coordinate sums are \(\frac12\), \(1\), and \(\frac12\), respectively, so each satisfies the triangle condition. At \(t=0\) these formulas give \(f(a),f(b),f(c)\); at \(t=1\) they give \(g(a),g(b),g(c)\). This checks both the target condition and the endpoints directly.
Convex Targets in the Plane
Worked Example: Two Maps from the Circle into the Disk
Give \(S^1=\{(x,y)\in\mathbb{R}^2:x^2+y^2=1\}\) the subspace topology from \(\mathbb{R}^2\), and give \(D^2=\{(u,v)\in\mathbb{R}^2:u^2+v^2\le1\}\) its subspace topology. The disk is convex: if \(p,q\in D^2\) and \(t\in I\), the triangle inequality gives \(\|(1-t)p+tq\|\le(1-t)\|p\|+t\|q\|\le(1-t)+t=1\). The nonnegative coefficients and their sum \(1\) justify the first and second inequalities.
Define \(f:S^1\to D^2\) by \(f(x,y)=(\frac{x}{2},\frac{y}{2})\), and let \(g(x,y)=(\frac14,0)\). These maps are continuous coordinate formulas. For \(f\), \(\|f(x,y)\|=\frac12\), since \(x^2+y^2=1\); for \(g\), \(\|g(x,y)\|=\frac14\). Thus both have values in \(D^2\). Their straight-line homotopy is $$ H((x,y),t)=\left((1-t)\frac{x}{2}+\frac{t}{4},\ (1-t)\frac{y}{2}\right). $$
To verify the target condition throughout this homotopy, use \(\|(x,y)\|=1\) and the triangle inequality: \[ \|H((x,y),t)\| \le (1-t)\left\|\left(\frac{x}{2},\frac{y}{2}\right)\right\|+t\left\|\left(\frac14,0\right)\right\| =\frac{1-t}{2}+\frac{t}{4} =\frac12-\frac{t}{4} \le\frac12\le1. \] The first equality uses the norms \(\frac12\) and \(\frac14\); the last inequalities hold because \(0\le t\le1\). Therefore \(H((x,y),t)\in D^2\). At \(t=0\) it equals \(f(x,y)\), and at \(t=1\) it equals \(g(x,y)\).
For an additional pointwise check, take \((x,y)=(0,1)\) and \(t=\frac12\). The first coordinate is \((\frac12)\cdot0+(\frac12)\cdot\frac14=\frac18\), and the second is \((\frac12)\cdot\frac12=\frac14\). The squared norm is \((\frac18)^2+(\frac14)^2=\frac1{64}+\frac4{64}=\frac5{64}\), so the norm is \(\frac{\sqrt5}{8}\le\frac12\), since \(5\le16\). This agrees with the general disk bound.
Where the Construction Can Fail
The hypothesis that the target is convex matters for this particular formula. For example, give \(S^1\) its subspace topology and consider the two constant maps from a one-point space into \(S^1\), with values \(p=(1,0)\) and \(q=(-1,0)\). Linear interpolation at parameter \(t\) gives \((1-2t,0)\). At \(t=\frac12\) this is \((0,0)\), whose squared norm is \(0^2+0^2=0\), not \(1\). Thus the formula does not define a homotopy into \(S^1\).
This calculation shows only that the straight-line construction fails for this pair of values in this target. It does not show that the two maps are not homotopic: another path in the circle can join the points. Convexity is a sufficient condition that makes the interpolation work for every pair of maps; it is not a necessary condition for two particular maps to be homotopic.
A related error is to assume that a straight-line formula stays in the target merely because its endpoints do. The calculation above corrects that assumption: both endpoints have norm \(1\), but their midpoint has norm \(0\). Always check that the intermediate values belong to the specified target, or invoke convexity when its hypotheses have been verified.
What the Theorem Gives
For a fixed convex subset \(C\subseteq\mathbb{R}^n\), the result can be stated as a practical test: given any topological space \(X\) and any continuous maps \(f,g:X\to C\), they are homotopic. The proof constructs the homotopy and verifies continuity, target membership, and both endpoint conditions. The maps need not agree at any point, and the construction does not assert that they are equal. It asserts that their values can be connected by a continuous family while remaining in the common target.
| Question | What to check |
|---|---|
| Is the formula a straight-line homotopy? | Use \(H(x,t)=(1-t)f(x)+tg(x)\). |
| Are its values in the target? | Verify that the target is convex and that \(f(x),g(x)\) lie in it. |
| Is it continuous? | Use continuity of \(f,g\), the product projections, and vector addition and scalar multiplication. |
| Does it join the required maps? | Substitute \(t=0\) and \(t=1\) into the formula. |
Check Your Understanding
For each question, identify the target condition and verify any proposed interpolation at its endpoints.
- For maps \(f,g:X\to\mathbb{R}^n\), write the straight-line homotopy and explain why it always stays in the target.
- Let \(C=[-1,2]\) and let \(f,g:X\to C\) be continuous. Why does the straight-line construction take values in \(C\)?
- For the triangle in the worked example, compute \(H(b,\frac13)\) from the displayed general formula and check its coordinates against the triangle inequalities.
- Why does the midpoint of \((1,0)\) and \((-1,0)\) show that their straight-line interpolation is not a homotopy into \(S^1\)? What does it not prove?
- State the continuity facts used to prove that the straight-line formula defines a continuous map on \(X\times I\).