Let $L$ be a lattice (finitely generated free abelian group), and $M\subseteq L$ be a subgroup of finite index (which is again a lattice).
Consider the dual lattices
$$L^\vee={\rm Hom}(L,\mathbb Z)\quad\text{and}\quad M^\vee={\rm Hom}(M,\mathbb Z).$$
Then naturally $L^\vee\subseteq M^\vee$.
We have perfect pairings
$$L^\vee\times L\to{\mathbb Z}, \qquad M^\vee\times M\to {\mathbb Z}.$$
Question. How can one construct a natural perfect pairing of finite abelian groups
$$M^\vee/L^\vee\times L/M\to {\mathbb Q}/{\mathbb Z}\ ?$$
EDIT: This elementary result is certainly stated and proved somewhere. I would be happy to have a reference.
Best Answer
You can define, for $f\in M^\lor$ and $x\in L$: $$\langle f,x\rangle = \frac{1}{d}f(dx)\in \mathbb{Q}$$ where $d=[L:M]$ is the index of $M$ in $L$. (The point being that $dx\in M$.)
Then this goes to the quotient to define $\langle f,x\rangle\in \mathbb{Q}/\mathbb{Z}$ for $f\in M^\lor/L^\lor$ and $x\in L/M$ since $\langle f,x\rangle\in \mathbb{Z}$ when $f\in L^\lor$ or $x\in M$.