The square with opposite edges identified is homeomorphic to the product $(\mathbb{R}\mathrm{/}\mathbb{Z})\times (\mathbb{R}\mathrm{/}\mathbb{Z})$

$Q$ is continuous, its components $q\circ {\pi }_0$ and $q\circ {\pi }_1$ being composites of continuous maps, and surjective, since every $(a,b)\in T\times T$ is $(q(u),q(v))$ for some $u,v$ by surjectivity of $q$.

For $U,V\subseteq \mathbb{R}$: $Q[U\times V]=q[U]\times q[V]$. Indeed $Q(u,v)=(q(u),q(v))$ gives the inclusion $\subseteq$, and conversely $(q(u),q(v))$ with $u\in U$, $v\in V$ is $Q(u,v)$.

$Q(x,y)=Q(x^{\mathrm{\prime }},y^{\mathrm{\prime }})$ exactly when $x-x^{\mathrm{\prime }}\in \mathbb{Z}$ and $y-y^{\mathrm{\prime }}\in \mathbb{Z}$, by [A1] applied in each coordinate.

$E=Q\upharpoonright S$ is continuous by [L3], and surjective: given $(a,b)\in T\times T$ write $a=q(u)$, $b=q(v)$; then $u-\lfloor u\rfloor$ and $v-\lfloor v\rfloor$ lie in $[0,1)$ by [A2] and $E(u-\lfloor u\rfloor ,v-\lfloor v\rfloor )=(a,b)$ by [A1].

Fix integers $m,n$ and put $R:=[m-1,m+1]\times [n-1,n+1]$. For $\alpha ,\beta \in \{0,1\}$ let $R_{\alpha \beta }:=[m-1+\alpha ,m+\alpha ]\times [n-1+\beta ,n+\beta ]$ and define $g_{\alpha \beta }:R_{\alpha \beta }\to S\mathrm{/}\text{\hspace{0.17em}⁣}\sim$ by $g_{\alpha \beta }(x,y):=P(x-(m-1+\alpha ),y-(n-1+\beta ))$. Each $g_{\alpha \beta }$ is continuous, being $P$ composed with a translation of each coordinate into $[0,1]$, which is continuous by [L2], [L3] and [L5].

For $(s,t),(s^{\mathrm{\prime }},t^{\mathrm{\prime }})\in S$: $E(s,t)=E(s^{\mathrm{\prime }},t^{\mathrm{\prime }})$ exactly when $(s,t)\sim (s^{\mathrm{\prime }},t^{\mathrm{\prime }})$, by step 1.3 and the definition of $\sim$. So the fibres of $E$ are exactly the classes of $\sim$.

$F$ is constant on the fibres of $Q$: if $Q(x,y)=Q(x^{\mathrm{\prime }},y^{\mathrm{\prime }})$ then $x^{\mathrm{\prime }}=x+m$ and $y^{\mathrm{\prime }}=y+n$ for integers $m,n$ by step 1.3, and then $x^{\mathrm{\prime }}-\lfloor x^{\mathrm{\prime }}\rfloor =x-\lfloor x\rfloor$ and $y^{\mathrm{\prime }}-\lfloor y^{\mathrm{\prime }}\rfloor =y-\lfloor y\rfloor$ by [A2].

$Q$ is an open map: by [L1] and [L2] the boxes $U\times V$ with $U,V$ open in $\mathbb{R}$ form a basis of ${\mathbb{R}}^2$, their images are the boxes $q[U]\times q[V]$ by step 1.2, which are open in $T\times T$ by [A1] and [L1], and the image of a union is the union of the images. With step 1.1 and [L4] this makes $Q$ an open quotient map, which is claim 1.

The four maps of step 1.5 agree on the overlaps of the $R_{\alpha \beta }$, which are contained in the lines $x=m$ and $y=n$. On $x=m$ the two candidate values differ only in that the first coordinate of the argument of $P$ is $1$ in one and $0$ in the other, and $(1,u)\sim (0,u)$; on $y=n$ the same holds in the second coordinate, and at $(m,n)$ all four values are $P$ of the four corners of $S$, which are all $\sim$-equivalent.

By steps 1.5 and 2.4 and the finite closed cover $\{R_{\alpha \beta }\}$ of $R$, [L3] gives a continuous $g:R\to S\mathrm{/}\text{\hspace{0.17em}⁣}\sim$ restricting to each $g_{\alpha \beta }$; and $g=F$ on $R$, since for $x\in [m-1,m)$ one has $\lfloor x\rfloor =m-1$, for $x\in [m,m+1)$ one has $\lfloor x\rfloor =m$, and at $x=m+1$ the value $P(\dots ,0,\dots )$ agrees with $P(\dots ,1,\dots )$ by $\sim$, the same three cases applying to $y$.

$F$ is continuous: the open sets $(m-1,m+1)\times (n-1,n+1)$, $m,n\in \mathbb{Z}$, cover ${\mathbb{R}}^2$ by [L1] and [L2], and on each of them $F$ is the restriction of the continuous $g$ of step 3.1, hence continuous by [L3]; the open cover clause of [L3] then gives continuity of $F$.

By step 2.1 and [L4] applied to the quotient map $P$ and the continuous map $E$ of step 1.4, there is exactly one continuous $\bar{E}:S\mathrm{/}\text{\hspace{0.17em}⁣}\sim \to T\times T$ with $\bar{E}\circ P=E$; by step 2.2 and [L4] applied to the quotient map $Q$ of step 2.3 and the continuous $F$ of step 4.1, there is exactly one continuous $\bar{F}:T\times T\to S\mathrm{/}\text{\hspace{0.17em}⁣}\sim$ with $\bar{F}\circ Q=F$.

$\bar{F}\circ \bar{E}=\mathrm{i}\mathrm{d}$: for $(s,t)\in S$ one has $\bar{F}(\bar{E}(P(s,t)))=\bar{F}(Q(s,t))=F(s,t)=P(s-\lfloor s\rfloor ,t-\lfloor t\rfloor )$, which equals $P(s,t)$ in every case, since $\lfloor u\rfloor =0$ for $u\in [0,1)$ and $1-\lfloor 1\rfloor =0$ with $(0,\cdot )\sim (1,\cdot )$ and $(\cdot ,0)\sim (\cdot ,1)$; and $P$ is surjective.

$\bar{E}\circ \bar{F}=\mathrm{i}\mathrm{d}$: for $(x,y)\in {\mathbb{R}}^2$ one has $\bar{E}(\bar{F}(Q(x,y)))=\bar{E}(F(x,y))=E(x-\lfloor x\rfloor ,y-\lfloor y\rfloor )=Q(x-\lfloor x\rfloor ,y-\lfloor y\rfloor )=Q(x,y)$ by [A1] and [A2]; and $Q$ is surjective by step 1.1.

By steps 6.1 and 6.2 the continuous maps $\bar{E}$ and $\bar{F}$ are mutually inverse, so $\bar{E}$ is a homeomorphism, which is claim 2; with step 2.3 both claims are proved.