A Map on the Whole Product
A homotopy compares two maps by placing them at the two ends of a parameter interval. The essential point is that the parameter and the input vary together: continuity is required on the whole product, not merely at each fixed parameter. We use the product topology to make this requirement precise and to check it in examples.
Let \(I=[0,1]\) have the subspace topology inherited from \(\mathbb{R}\) with its standard topology. For topological spaces \((X,\mathcal{T})\) and \((Y,\mathcal{U})\), give \(X\times I\) the product topology. Its basic open sets are \(U\times V\), where \(U\) is open in \(X\) and \(V\) is open in \(I\).
The endpoint conditions can also be written \(H(\,\cdot\,,0)=f\) and \(H(\,\cdot\,,1)=g\). For \(t\in I\), the map \(x\mapsto H(x,t)\) is the time-\(t\) slice. The definition asks for one continuous map on \(X\times I\); checking slices alone is not sufficient.
Two Continuity Facts for Product Topologies
The product topology is useful here because its basic open sets let us test maps into a product one coordinate at a time.
Proof. Suppose first that \(F\) is continuous. The projection \(\pi_1:Y\times W\to Y\) is continuous: for every open \(U\subseteq Y\), its preimage is \(U\times W\), which is open in the product topology. Likewise, \(\pi_2:Y\times W\to W\) is continuous because the preimage of an open \(V\subseteq W\) is \(Y\times V\). Since \(h=\pi_1\circ F\) and \(k=\pi_2\circ F\), both are continuous. Here the composition of continuous maps is continuous because the preimage of an open set under a composition is the successive preimage under the two maps.
Conversely, suppose \(h\) and \(k\) are continuous. If \(U\subseteq Y\) and \(V\subseteq W\) are open, then \(F^{-1}(U\times V)=h^{-1}(U)\cap k^{-1}(V)\), which is open in \(Z\). Every open set in the product is a union of basic open rectangles \(U\times V\). Its preimage under \(F\) is therefore a union of open sets, hence is open. Thus \(F\) is continuous. \(\square\)
Proof. Fix \(t\in I\) and define \(i_t:X\to X\times I\) by \(i_t(x)=(x,t)\). This inclusion is continuous. Indeed, for a basic open set \(U\times V\), its preimage under \(i_t\) is \(U\) if \(t\in V\), and is \(\varnothing\) if \(t\notin V\); both sets are open in \(X\). Since every open set in \(X\times I\) is a union of basic open sets, all its preimages under \(i_t\) are open. Now \(H_t=H\circ i_t\), so \(H_t\) is continuous as a composition of continuous maps. Taking \(t=0\) and \(t=1\) proves the endpoint claim. \(\square\)
This theorem also gives a useful necessary condition: a discontinuous map cannot be an endpoint of a homotopy. To establish a homotopy, however, continuity of the endpoints alone does not settle whether a proposed \(H\) is continuous on the product.
Worked Homotopies
Worked Example: A Homotopy on the Real Line
Give \(\mathbb{R}\) its standard topology and consider \(H:\mathbb{R}\times I\to\mathbb{R}\) defined by \(H(x,t)=x^2+t\). Verify that it is a homotopy and identify its endpoint maps.
The first coordinate projection \((x,t)\mapsto x\) and second coordinate projection \((x,t)\mapsto t\) are continuous by the product topology. Squaring is continuous on \(\mathbb{R}\), and addition is continuous on \(\mathbb{R}\times\mathbb{R}\). Therefore the map \((x,t)\mapsto x^2+t\) is continuous.
At the first endpoint, substitution gives \(H(x,0)=x^2+0=x^2\), so \(f(x)=x^2\). At the second endpoint, \(H(x,1)=x^2+1\), so \(g(x)=x^2+1\). Thus \(H\) is a homotopy from \(f\) to \(g\). As a check, setting \(x=2\) gives \(H(2,0)=2^2+0=4=f(2)\) and \(H(2,1)=2^2+1=5=g(2)\).
Worked Example: Rotating the Circle
Give \(S^1=\{z\in\mathbb{C}:|z|=1\}\) the subspace topology inherited from \(\mathbb{C}\cong\mathbb{R}^2\). Define \(H:S^1\times I\to S^1\) by \(H(z,t)=e^{\pi i t}z\). Verify continuity and find the endpoint maps.
Write \(z=x+iy\), where \(x^2+y^2=1\). Using \(e^{\pi i t}=\cos(\pi t)+i\sin(\pi t)\), the two real coordinates of the proposed output are \((x\cos(\pi t)-y\sin(\pi t),\,x\sin(\pi t)+y\cos(\pi t))\). Each coordinate is continuous in \((z,t)\): the coordinate functions \(x,y\) are continuous on the subspace \(S^1\), and sine, cosine, addition, and multiplication are continuous on their standard real domains. By Theorem 1, the resulting map to \(\mathbb{R}^2\) is continuous.
For every \((x,y)\) on the circle, the squared norm of the output is \((x\cos(\pi t)-y\sin(\pi t))^2+(x\sin(\pi t)+y\cos(\pi t))^2\). On expanding, the mixed terms cancel, and the result is \((x^2+y^2)(\cos^2(\pi t)+\sin^2(\pi t))=1\cdot1=1\). Thus the output lies in \(S^1\). A continuous map into \(\mathbb{R}^2\) whose image lies in \(S^1\) is continuous as a map into the subspace \(S^1\): the preimage of a relatively open set \(O\cap S^1\), with \(O\) open in \(\mathbb{R}^2\), is the preimage of \(O\). So \(H\) is continuous with the stated target.
Since \(\cos(0)=1\) and \(\sin(0)=0\), \(H(z,0)=z\). Since \(\cos(\pi)=-1\) and \(\sin(\pi)=0\), \(H(z,1)=-z\). Hence this homotopy joins the identity map of the circle to the map \(z\mapsto -z\). The norm calculation also checks the endpoint outputs: if \(z=1\), they are \(1\) and \(-1\), both of norm \(1\).
Worked Example: A Finite Discrete Domain
Let \(X=\{a,b\}\) have the discrete topology, in which every subset is open, and give \(\mathbb{R}\) its standard topology. Define \(H:X\times I\to\mathbb{R}\) by \(H(a,t)=\cos(\pi t)\) and \(H(b,t)=2\). Verify continuity and the endpoint maps.
The sets \(\{a\}\times I\) and \(\{b\}\times I\) are open in \(X\times I\), since \(\{a\}\) and \(\{b\}\) are open in \(X\), and \(I\) is open in itself. On the first of these pieces, the formula is the continuous function \(t\mapsto\cos(\pi t)\); on the second, it is the constant function with value \(2\).
For any open \(O\subseteq\mathbb{R}\), the preimage \(H^{-1}(O)\) is the union of its preimages inside these two pieces. The first preimage is \(\{a\}\times\{t\in I:\cos(\pi t)\in O\}\), which is open in the product because the time preimage is open in \(I\). The second is either \(\{b\}\times I\), if \(2\in O\), or \(\varnothing\), if \(2\notin O\); both are open. Their union is open, so \(H\) is continuous.
At \(t=0\), \(H(a,0)=\cos(0)=1\) and \(H(b,0)=2\), giving \(f(a)=1,\ f(b)=2\). At \(t=1\), \(H(a,1)=\cos(\pi)=-1\) and \(H(b,1)=2\), giving \(g(a)=-1,\ g(b)=2\). Thus \(H\) is a homotopy from \(f\) to \(g\). The unchanged value at \(b\) and the computed endpoint values at \(a\) check both endpoint conditions.
Common Pitfall: Checking Each Slice Is Not Enough
A frequent error is to verify that \(x\mapsto H(x,t)\) is continuous for every fixed \(t\), and then conclude that \(H\) is continuous. Separate continuity of all slices does not imply continuity on the product. The following formula makes the gap explicit. On \(\mathbb{R}\times I\), set \(H(0,0)=0\), and at every other point set \(H(x,t)=\frac{xt}{x^2+t^2}\).
For each fixed \(t=0\), this is the constant-zero function of \(x\). For each fixed \(t>0\), the denominator \(x^2+t^2\) is positive for all \(x\), so the formula is continuous in \(x\). For each fixed \(x=0\), it is constant zero as a function of \(t\). For each fixed \(x\ne0\), the denominator is positive for all \(t\in I\), so it is continuous in \(t\). Thus every fixed-time slice and every fixed-\(x\) slice is continuous.
Nevertheless, \(H\) is not continuous at \((0,0)\). For each \(n\in\mathbb{N}\), take \(x=1/n\) and \(t=1/n\). Then \(H(1/n,1/n)=\frac{(1/n)(1/n)}{(1/n)^2+(1/n)^2}=\frac{1/n^2}{2/n^2}=\frac12\). These inputs converge to \((0,0)\), but their output values do not converge to \(H(0,0)=0\). The endpoint maps are continuous—\(H(x,0)=0\) and \(H(x,1)=x/(x^2+1)\)—but this does not make \(H\) a homotopy. The correction is to verify joint continuity on \(X\times I\), as the definition requires.
Why the Product Formulation Matters
The product domain records the two independent inputs to a deformation: a point of \(X\) and a time in \(I\). Its basic open rectangles support direct continuity checks, while the coordinate criterion handles maps whose outputs have several coordinates. The slice theorem gives a quick necessary check on proposed endpoint maps; the pitfall shows why that check cannot replace joint continuity.
Check Your Understanding
For each question, use the product topology and the endpoint conditions explicitly.
- State the definition of a homotopy from \(f:X\to Y\) to \(g:X\to Y\). Which two values of the time parameter specify the endpoint maps?
- Explain why the coordinate criterion tests continuity of a map into \(Y\times W\).
- For \(H:\mathbb{R}\times I\to\mathbb{R}\) given by \(H(x,t)=x+t\), identify both endpoint maps and justify continuity.
- Why does continuity of every time slice not suffice to prove that a map \(X\times I\to Y\) is continuous?
- If \(H:X\times I\to Y\) is continuous, prove that \(x\mapsto H(x,t)\) is continuous for a fixed \(t\in I\).