I'm looking at an exercise that reads:
Problem. Let $A = \begin{bmatrix}a & b\\c & d\end{bmatrix} \in GL(2, \mathbb{C})$ and $\Lambda = \langle z \mapsto z + \omega_1, z \mapsto z + \omega_2\rangle$ with $\omega_1, \omega_2$ linearly independent over $\mathbb{R}$. Prove that if $\Lambda = \langle z \mapsto z + \omega_1', z \mapsto z + \omega_2'\rangle$ where $$ \begin{bmatrix}\omega_1'\\\omega_2'\end{bmatrix} = A\begin{bmatrix}\omega_1\\\omega_2\end{bmatrix} $$ then $A \in GL(2, \mathbb{Z})$ with $\det(A) = \pm1$.
The solutions state that clearly $A \in GL(2, \mathbb{Z})$ otherwise $A$ doesn't take lattice points to lattice points, which I don't follow—being a linear map, wouldn't $A$ take lattice points to lattice points anyway?
Suppose $A$ is not an element of $GL(2, \mathbb{Z})$. There are 2 cases here, firstly when $A$ does not have integer entries and secondly when $A$ does not have determinant $\pm 1$.
In the first case we know that $\omega_1' = a\omega_1 + b\omega_2$ and $\omega_2' = c\omega_1 + d\omega_2$. But if one of $a, b, c, d$ is not an integer then the corresponding $\omega_i'$ is not contained in $\Lambda$ (since $\Lambda$ consists of the sums $n\omega_1 + m\omega_2$ with $n, m \in \mathbb{Z}$).
In the second case the map is not surjective on lattice points.