 Solution

For a pair $(L,t)$ choose an oriented basis $({\omega }_1,{\omega }_2)$ with $\mathrm{Im}({\omega }_1/{\omega }_2)>0$ and $t={\omega }_2/N$ modulo $L$. Such a basis exists because an exact-order point gives a primitive vector modulo $N$, which can be completed to a determinant-one basis. Define the marked-lattice model of a modular form byChanging to a basis with the same marked point uses a matrix with $c\equiv 0$, $d\equiv 1$ and therefore $a\equiv 1(\mathrm{mod}N)$, exactly ${\Gamma }_1(N)$. The weight-$k$ transformation of $f$ cancels the factor from ${\omega }_2^{-k}$, proving independence of the basis.

The resulting function has homogeneity $F_f(uL,ut)=u^{-k}{F}_f(L,t)$ for $u\in {\mathbb{C}}^{\times }$. Conversely evaluating at $(\mathbb{Z}\tau +\mathbb{Z},1/N)$ recovers $f$. Holomorphy in $\tau$ and holomorphy in the local parameters of degenerating lattices at every modular cusp characterize the functions arising from modular forms, rather than arbitrary homogeneous lattice functions.