= Solution
For the <quotient map> $\pi$, a subset $U$ of the quotient is open exactly when $\pi^{-1}(U)$ is open in $X$. This definition immediately makes $\pi$ continuous. If $g$ is continuous then $g\circ\pi$ is continuous. Conversely, <continuity> of the composition gives, for every open $V\subseteq Z$, that $\pi^{-1}(g^{-1}(V))$ is open. By the defining criterion $g^{-1}(V)$ is open, so $g$ is continuous.
A map $h$ constant on each <equivalence class> defines a unique map $\bar h([x])=h(x)$. It is well-defined by the assumed class constancy, and its composition with $\pi$ is $h$. The preceding criterion proves that $\bar h$ is continuous.
For the integer <lattice> quotient, define
$$
F(x,y)=(e^{2\pi ix},e^{2\pi iy}).
$$
This continuous map is onto the product of the two unit circles, and its fibres are exactly the <lattice> equivalence classes. It therefore induces a <continuous bijection> $\bar F:\mathbb R^2/\mathbb Z^2\to\mathbb T^2$. Every class has a representative in $[0,1]^2$, so the quotient is the continuous image of that <compact> square and is <compact>. The target is <Hausdorff> as a subspace of $\mathbb C^2$. A <continuous bijection> from a <compact> space to a <Hausdorff> space is a <homeomorphism>, proving
$$
\boxed{\mathbb R^2/\mathbb Z^2\cong\mathbb T^2.}
$$
Finally, if $U\subseteq\mathbb R^2$ is open, then $\pi^{-1}(\pi(U))=\bigcup_{m\in\mathbb Z^2}(U+m)$ is open, being a union of translates. The quotient criterion makes $\pi(U)$ open. Thus \b[the <quotient map> is an <open map>].
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