----- > [!proposition] Proposition. ([[the product of covering maps is a covering map]]) > If $p : E \to B$ and $p': E' \to B'$ are [[covering space|covering maps]], then $p \times p' : E \times E' \to B \times B'$ > is a [[covering space|covering map]]. > [!proof]- Proof. ([[the product of covering maps is a covering map]]) > Let $(b,b') \in B \times B'$. Obtain [[neighborhood]]s $U \ni b$, $U' \ni b'$ s.t. $p ^{-1}(U)$ and $p' ^{-1}(U')$ each admit partitions $\{ V_{\alpha} \}$, $\{ V_{\beta}' \}$ into slices respectively. Then $(p \times p')^{-1}(U \times U')$ equals the union of all the sets $V_{\alpha} \times V_{\beta}'$. These are disjoint open sets of $E \times E'$, and each is mapped [[homeomorphism|homeomorphically]] onto $U \times U'$ by $p$. > [!basicexample] > Consider the [[torus]] $\mathbb{T}=\mathbb{S}^{1}\times \mathbb{S}^{1}$. The product map $p \times p: \mathbb{R} \times \mathbb{R} \to \mathbb{S}^{1} \times \mathbb{S}^{1}$ is a [[covering space|covering]] of $\mathbb{T}$ by the plane $\mathbb{R}^{2}$, where $p$ denotes a[[covering space#^706eb3|covering map of the circle]]. Each of the unit squares $[n,n+1] \times [m, m+1]$ gets wrapped by $p \times p$ entirely around the torus. > ![[CleanShot 2024-03-30 at [email protected]]] \ Note that in this picture we visualized the torus not as the product $\mathbb{S}^{1}\times \mathbb{S}^{1}$, which lives in $\mathbb{R}^{4}$, but via the usual parameterization in $\mathbb{R}^{3}$. In [[torus]] we check that the two spaces are [[homeomorphism]] [[TODO]]. ----- #### ---- #### References > [!backlink] > ```dataview TABLE rows.file.link as "Further Reading" FROM [[]] FLATTEN file.tags GROUP BY file.tags as Tag > [!frontlink] > ```dataview TABLE rows.file.link as "Further Reading" FROM outgoing([[]]) FLATTEN file.tags GROUP BY file.tags as Tag