Help finding mistake in proof involving the quotient map. Consider the plane $\mathbb{R}^2=\mathbb{R}\times\mathbb{R}$ with the product topology which has basis consisting of all open squares of the form
$$\tag{1}
\left]a,b\right[ \times \left]c,d\right[ \subseteq\mathbb{R}\times\mathbb{R},\quad a<b,\ c<d.
$$
Define the quotient space (with the quotient topology)
$$\tag{2}
\mathbb{R}^2/\mathbb{Z}:=\mathbb{R}^2/\mathord{\sim}
$$
where the equivalence relation is given by
$$\tag{3}
((x_1,x_2)\sim(y_1,y_2))\iff(x_1-y_1\in\mathbb{Z}\text{ and } x_2-y_2=0).
$$
I want to show that
$$\tag{4}
\pi^{-1}\bigg(\pi\Big(\left]a,b\right[ \times \left]c,d\right[\Big)\bigg)=\bigcup_{n\in\mathbb{Z}}\Big(\left]a+n,b+n\right[\times\left]c,d\right[\Big).
$$
where $\pi:\mathbb{R}^2\rightarrow \mathbb{R}^2/\mathbb{Z}$ is the quotient map. If we consider a element $(x_1,x_2)\in\mathbb{R}^2$ then we can see from eq. $(3)$ that the effect of the quotient map is
$$\tag{5}
\pi\left(\left(x_{1}, x_{2}\right)\right)=\left\{\left(y_{1}, y_{2}\right) \mid x_{1}-y_{2} \in \mathbb{Z} \text { and } x_{2}-y_{2}=0\right\}=\left\{\left(x_{1}+n, x_{2}\right) \mid n \in \mathbb{Z}\right\}.
$$
Is the inverse of this
$$
\tag{6}
\pi^{-1}(\left\{\left(x_{1}+n, x_{2}\right) \mid n \in \mathbb{Z}\right\})=\bigcup_{n\in\mathbb{Z}}\{\left(x_{1}+n, x_{2}\right)\}?
$$
So that the inverse of all elements in $\pi(\left]a,b\right[ \times \left]c,d\right[)$ becomes the right side of eq. $(4)$. If this is correct (I assume it is not), then it appears that $\pi^{-1}$ has no effect, since $\left\{\left(x_{1}+n, x_{2}\right) \mid n \in \mathbb{Z}\right\}$ and $\bigcup_{n\in\mathbb{Z}}\{\left(x_{1}+n, x_{2}\right)\}$ both are sets with elements of the type $(x_1+n,x_2)$ with $n\in\mathbb{Z}$. So it seems I am making a mistake, but I need help identifying and correcting it.
 A: What you've written in that last equation is in fact correct. (I haven't checked the details of the rest of your proof since it's not necessary to answer your question.) More generally, let $S$ be any set, let $\sim$ be an equivalence relation on $S$, and consider the quotient map $\pi\colon S \to S/\mathord{\sim}$. For any $x \in S$, we have $\pi(x) = \{y \in S \mid x \sim y\}$, and
$$\begin{aligned}
\pi^{-1}(\pi(x)) &= \pi^{-1}(\{y \in S \mid x \sim y\}) \\
&= \{y \in S \mid \pi(x) = \pi(y) \} \\
&= \{y \in S \mid x \sim y\} = \pi(x).
\end{aligned}$$
That is, $\pi(x)$ and $\pi^{-1}(\pi(x))$ are literally the same set. Remember, the elements of the quotient set $S/\mathord{\sim}$ are the equivalence classes of $\sim$, which are subsets of $S$.
However, the preimage function $\pi^{-1}$ in general sends subsets of the codomain to subsets of the domain, and $\pi^{-1}(s)$ is really shorthand for $\pi^{-1}(\{s\})$ (where $s$ is an element of the codomain). (Not to be confused with the inverse function of a bijective function, which uses the same notation but operates on elements rather than subsets.) The above coincidence of sets only works when you're taking the preimage of a single element, not the preimage of a bigger subset. In general, if $T \subseteq S$, then $\pi(T) = \{\pi(x) \mid x \in T\}$, and so
$$\begin{aligned}
\pi^{-1}(\pi(T)) &= \pi^{-1}(\{\pi(x) \mid x \in T\}) \\
&= \{y \in S \mid \exists x \in T : \pi(x) = \pi(y) \} \\
&= \bigcup_{x \in T} \{y \in S \mid \pi(x) = \pi(y) \} = \bigcup_{x \in T} \pi(x).
\end{aligned}$$
In other words, on the level of sets, the preimage of a quotient map takes the union of the cosets.
