Contracting a Product
A contraction moves every point of a space continuously to one fixed point. If a space is a product, it is natural to move each coordinate using a contraction of its factor. The key point is to ensure that the coordinate motions assemble into a continuous map on the product and finish at one fixed product point.
Let \(I=[0,1]\) with its standard topology. A homotopy from \(f:X\to Y\) to \(g:X\to Y\) is a continuous map \(H:X\times I\to Y\) such that \(H(x,0)=f(x)\) and \(H(x,1)=g(x)\) for every \(x\in X\). Products below have the product topology.
Thus a contraction is a homotopy from the identity map \(\operatorname{id}_X\) to the constant map with value \(x_0\). The chosen point \(x_0\) is the endpoint of the contraction; the definition does not require the contraction to fix \(x_0\) at intermediate times.
Why Coordinatewise Formulas Are Continuous
For a product \(P=\prod_{\alpha\in J}X_\alpha\), let \(\pi_\alpha:P\to X_\alpha\) denote projection onto coordinate \(\alpha\). The product topology has a basis made from restrictions on finitely many coordinates. Consequently, a map into \(P\) is continuous exactly when all of its coordinate maps are continuous.
Proof. If \(f\) is continuous, each projection \(\pi_\alpha\) is continuous by the definition of the product topology, so \(\pi_\alpha\circ f\) is continuous. Conversely, suppose every \(\pi_\alpha\circ f\) is continuous. A subbasis for the product topology consists of sets \(\pi_\alpha^{-1}(U)\), where \(U\) is open in \(X_\alpha\). Their preimages under \(f\) are \(f^{-1}(\pi_\alpha^{-1}(U))=(\pi_\alpha\circ f)^{-1}(U)\), which are open in \(Z\). Since the preimage of every subbasis element is open, the preimage of every finite intersection of subbasis elements is open, and then the preimage of every union of such intersections is open. These sets form the product topology, so \(f\) is continuous. \(\square\)
This criterion applies to maps whose domain is itself a product with \(I\), using the product topology on that domain. It is what makes the coordinate formula below a continuous homotopy.
The Product Contraction Theorem
Proof. First suppose \(J\) is nonempty and the stated contractions have been chosen. The point \(x^*=(x_\alpha^*)_{\alpha\in J}\) belongs to \(P\). Define \(H:P\times I\to P\) by the displayed coordinate formula. For each \(\alpha\), its \(\alpha\)-coordinate is \(H_\alpha\circ(\pi_\alpha\times\operatorname{id}_I)\), which is continuous: the projection \(\pi_\alpha\) is continuous, the map \(\pi_\alpha\times\operatorname{id}_I\) is continuous for product topologies, and \(H_\alpha\) is continuous. The coordinate criterion therefore gives continuity of \(H\). For every \(x=(x_\alpha)\in P\), the defining endpoint equations for \(H_\alpha\) give \(H(x,0)=(x_\alpha)=x\) and \(H(x,1)=(x_\alpha^*)=x^*\). Thus \(H\) contracts \(P\). If \(J\) is empty, its product has one point; the constant map on that space is a contraction. \(\square\)
For a finite family, the theorem uses only finitely many choices of contractions. For an arbitrary family, the statement is made with a contraction chosen for each factor; in the usual set-theoretic setting with the axiom of choice, contractibility of every factor supplies such a family of choices.
Finite Products: Simultaneous and Staged Formulas
For a finite product \(X_1\times\cdots\times X_n\), suppose \(H_i\) contracts \(X_i\) to \(x_i^*\). The simultaneous formula is \((x_1,\ldots,x_n,t)\mapsto(H_1(x_1,t),\ldots,H_n(x_n,t))\). Every coordinate begins at its original value and reaches its chosen endpoint at the same time \(t=1\).
There is also a staged formula that contracts one coordinate at a time. During stage \(j\), the earlier coordinates have reached their endpoints, coordinate \(j\) is moving, and the later coordinates have not yet moved. For \(n\geq1\), on the interval \((j-1)/n\leq t\leq j/n\), put \(s=nt-(j-1)\) and set \((K(x,t))_i=x_i^*\) if \(i<j\), \((K(x,t))_j=H_j(x_j,s)\) if \(i=j\), and \((K(x,t))_i=x_i\) if \(i>j\).
Proof that the staged formula is a homotopy. On each closed time interval \([(j-1)/n,j/n]\), every coordinate of \(K\) is continuous: it is constant, a projection, or a contraction composed with a continuous reparametrization. The coordinate criterion proves continuity on that interval. At a shared endpoint \(t=j/n\), stage \(j\) has coordinate \(j\) equal to \(H_j(x_j,1)=x_j^*\), and stage \(j+1\) begins with that same coordinate fixed at \(x_j^*\); all other coordinates agree there as well. The pieces therefore agree on their overlaps. The finite pasting lemma says that continuous maps on a finite closed cover, agreeing on overlaps, combine to a continuous map. Indeed, for a closed set \(F\) in the target, its preimage under the combined map is a finite union of sets closed in the respective closed pieces, hence closed in the whole domain. Thus \(K\) is continuous. At \(t=0\) all coordinates are unchanged, and at \(t=1\) all coordinates equal their chosen endpoints. This proves the staged formula is a contraction. \(\square\)
Worked Example: Contracting the Unit Square
Give \(I\times I\) its product topology, where both factors have the standard topology. Each factor contracts to \(0\) by \(H(x,t)=(1-t)x\). The simultaneous product contraction is \(C((x,y),t)=((1-t)x,(1-t)y)\).
For \((x,y)=(1/2,1/4)\) and \(t=2/3\), the first coordinate is \((1-2/3)(1/2)=(1/3)(1/2)=1/6\), and the second is \((1-2/3)(1/4)=(1/3)(1/4)=1/12\). Hence \(C((1/2,1/4),2/3)=(1/6,1/12)\). Both coordinates remain in \(I\), since \(0\leq 1-t\leq1\) and \(0\leq x,y\leq1\). At \(t=0\), the formula gives \((x,y)\); at \(t=1\), it gives \((0,0)\). Substitution verifies the required endpoints.
Worked Example: A Contraction in Three Coordinates
Give \(\mathbb{R}^3\) its product topology, which agrees with its usual Euclidean topology. Contract the three copies of \(\mathbb{R}\) to \(1,-2,0\), respectively, using the coordinate contractions \(H_i(u,t)=(1-t)u+ta_i\), where \((a_1,a_2,a_3)=(1,-2,0)\). The product formula is \(C((x,y,z),t)=((1-t)x+t,(1-t)y-2t,(1-t)z)\).
At \((2,1,-3)\) and \(t=1/4\), the coordinates are \((3/4)2+1/4=6/4+1/4=7/4\), \((3/4)1-2(1/4)=3/4-1/2=1/4\), and \((3/4)(-3)=-9/4\). Thus the output is \((7/4,1/4,-9/4)\). At \(t=0\) the formula returns \((x,y,z)\); at \(t=1\) it returns \((1,-2,0)\), independent of the input. Each coordinate formula is continuous, so the map is continuous by the product criterion.
Worked Example: A Disk Times an Interval
Let \(D^2=\{(u,v)\in\mathbb{R}^2:u^2+v^2\leq1\}\) have the subspace topology, and give \(I\) its standard topology. The disk contracts to \((0,0)\) by multiplication by \(1-t\), and \(I\) contracts to \(0\) by the same scalar formula. Therefore a contraction on \(D^2\times I\) is \(C(((u,v),z),t)=(((1-t)u,(1-t)v),(1-t)z)\).
At \(((3/5,4/5),1/2)\) and \(t=1/4\), the disk coordinates become \((3/4)(3/5)=9/20\) and \((3/4)(4/5)=12/20=3/5\); the interval coordinate becomes \((3/4)(1/2)=3/8\). The new disk point lies in \(D^2\), since its squared norm is \((9/20)^2+(12/20)^2=81/400+144/400=225/400=9/16\leq1\). The interval coordinate \(3/8\) lies in \(I\). At \(t=0\) the input is unchanged, and at \(t=1\) the output is \(((0,0),0)\). This checks both the formula and that its values remain in the product.
A Common Endpoint Error
Each coordinate must finish at a fixed point that does not depend on the input. For example, a tempting formula on \(I\times I\) is \(G((x,y),t)=((1-t)x,ty)\), on the mistaken grounds that both coordinates are being moved. But at \(t=1\), it gives \(G((x,y),1)=(0,y)\), which still depends on \(y\), so it is not a contraction to a point. At the specific input \((1/2,1/3)\), its endpoint is \((0,1/3)\), whereas at \((1/2,2/3)\) its endpoint is \((0,2/3)\). The corrected formula \(C((x,y),t)=((1-t)x,(1-t)y)\) sends both inputs to \((0,0)\) at \(t=1\). A product contraction requires every factor contraction to use the same terminal time and to reach its chosen fixed endpoint.
The product construction is useful because it turns separate contractions into one contraction without requiring geometric intuition about the whole product. For finite products, simultaneous motion gives a concise formula, while staged motion can make the order of coordinate changes explicit. In either construction, continuity follows from the product topology and continuity in each coordinate.
Check Your Understanding
Use the coordinate formula and verify both endpoint conditions in each proposed construction.
- If \(X\) contracts to \(x_0\) and \(Y\) contracts to \(y_0\), write the simultaneous contraction of \(X\times Y\) and state which topology is used.
- For contractions \(H_1,H_2,H_3\), write the staged time interval and local parameter for the second stage.
- Show directly that \(\mathbb{R}^2\) contracts to \((2,-1)\) using scalar straight-line formulas in its two coordinates.
- For a product indexed by the empty set, what space is obtained, and why is it contractible?
- Explain why the formula \(((1-t)x,ty)\) does not contract \(I\times I\) to a point.