Let \(p_i:(\widetilde X_i,\widetilde x_i)\to(X_i,x_i)\) be covering spaces for \(i=1,2\).
Show that the product space \(\widetilde X_1\times\widetilde X_2\) together with the map \(p:\widetilde X_1\times\widetilde X_2\to X_1\times X_2\) defined by \(p(y,z):=(p_1(y),p_2(z))\) is also a covering space.
Find the universal covering of the space \(S^1\times D^2\times S^1\).
Solution
1. Let \((x_1,x_2)\in X_1\times X_2\).Choose evenly covered open neighborhoods \(U_i\) of \(x_i\) for \(p_i\) such that\[ p_i^{-1}(U_i)=\coprod_{\alpha\in A_i}V_{i,\alpha}, \]where every restriction\[ p_i|_{V_{i,\alpha}}:V_{i,\alpha}\longrightarrow U_i \]is a homeomorphism.
This is exactly the covering-space property of \(p_i\) at \(x_i\), for \(i=1,2\).
2. The product neighborhood \(U_1\times U_2\) is evenly covered by\[ p:\widetilde X_1\times\widetilde X_2\longrightarrow X_1\times X_2, \qquad p(y,z)=(p_1(y),p_2(z)). \]
We have \[ \begin{aligned} p^{-1}(U_1\times U_2) &=p_1^{-1}(U_1)\times p_2^{-1}(U_2)\\ &= \left(\coprod_{\alpha\in A_1}V_{1,\alpha}\right) \times \left(\coprod_{\beta\in A_2}V_{2,\beta}\right)\\ &= \coprod_{(\alpha,\beta)\in A_1\times A_2} \bigl(V_{1,\alpha}\times V_{2,\beta}\bigr). \end{aligned} \] Each product \(V_{1,\alpha}\times V_{2,\beta}\) is open in \(\widetilde X_1\times\widetilde X_2\), and the products are pairwise disjoint. On such a product sheet, \(p\) restricts to \[ (p_1|_{V_{1,\alpha}})\times(p_2|_{V_{2,\beta}}): V_{1,\alpha}\times V_{2,\beta} \longrightarrow U_1\times U_2. \] This is a homeomorphism, with inverse the product of the two inverse homeomorphisms. Thus \(U_1\times U_2\) is evenly covered.
3. Hence \(p\) is a covering map, proving part (a).
Every point \((x_1,x_2)\) has the evenly covered neighborhood produced in <1>1–<1>2.
4. Define\[ P:\mathbb R\times D^2\times\mathbb R \longrightarrow S^1\times D^2\times S^1 \]by\[ P(s,z,t)=\bigl(e^{2\pi i s},z,e^{2\pi i t}\bigr). \]Then \(P\) is a covering map.
The exponential map \[ q:\mathbb R\longrightarrow S^1, \qquad q(s)=e^{2\pi i s}, \] is the standard covering of the circle, while \[ \operatorname{id}_{D^2}:D^2\longrightarrow D^2 \] is a one-sheeted covering. Applying part (a) twice to \[ q\times\operatorname{id}_{D^2}\times q \] gives exactly the map \(P\).
5. The covering in 4 is the universal covering of \(S^1\times D^2\times S^1\).
The space \(D^2\) is contractible, as is \(\mathbb R\). Therefore their product \[ \mathbb R\times D^2\times\mathbb R \] is contractible, for example by the straight-line homotopy to \((0,0,0)\). In particular it is path-connected and simply connected. A connected covering with simply connected total space is a universal covering. Hence \(P\) is the universal covering requested in part (b).