Prove that this group action is continuous from $S\times\Bbb{R}^2\to\Bbb{R}^2$

Question

Let me give some description about the notations used-
S denotes the collection of all transformations on $\Bbb{R}^2$ by integer coordinates.
i.e. $S=\{t_v|t_v(x)=x+v,\forall x\in\Bbb{R}^2; \forall v\in\Bbb{Z}^2\}$.
Notice that, it is a subgroup of collection of all isometries(distance preserving map) on $\Bbb{R}^2$ under function composition.It is also easy to observe that $S\cong\Bbb{Z}^2$ (by this isomorphism $\phi(t_v)=v\in\Bbb{Z}^2$)
Now, this group $S$ is given discrete metric $d$. Again take $\Bbb{R}^2$ with usual euclidean metric $d_u$.
Define a group action $\sigma:S\times\Bbb{R}^2\to\Bbb{R}^2$ by $\sigma(t_v,x)=t_v(x)=x+v$. Here, we have taken product metric $d'$ on $S\times\Bbb{R}^2$ i.e. $d'=\max\{d, d_u\}$.
Prove that $\sigma$ is continuous map.
There is a small hint on the book which says- prove for any open set $U$ in $\Bbb{R}^2$, you will get "deck" of open sets in $S\times\Bbb{R}^2$ under $\sigma$.
First of all I don't know what "deck" means here. Secondly, I can't prove why $\sigma^-1(U)$ is open.
Can anybody solve this problem? Thanks for assistance in advance.

Answer 1

We can explicitly write $\sigma^{-1}(U)$ as the union of the sets $\{t_v\}\times(U-v)$ for all $v\in\Bbb Z^2 $, probably this was meant by the 'deck'.
Now, this is easily seen to be open: let $(t_v, x)\in \sigma^{-1}(U)$, then $x\in U-v$, that is, $x+v\in U$, so by the translation being diffeomorphic, $U-v$ is an open neighborhood of $x$, and then $\{t_v\} \times (U-v)$ will be an open neighborhood of $(t_v, x)$ within $\sigma^{-1}(U)$.