Note that $ker(f)$ is a sublattice of the integer lattice $\mathbb{Z}^r$. And conversely any such integer sublattice will give you such a homomorphism $f$. The index of this sublattice is exactly what you call $h$ (since the homomorphism is surjective). If $C$ is a symmetric convex body in $\mathbb{R}^r$ and $L$ a full rank lattice then the number of lattice points in $t C$ (the dilation of $C$ by a real factor of $t$) is asymptotic to $t^r/D$, t^r vol(C)/D$, where$D$is the volume of the fundamental domain of$L$. In your case$D=h$and$C$is the unit hypercube. Getting a good remainder is usually a much harder job. In general the remainder is bounded by something proportional to the area of$C$. When$C$has a smooth boundary (which alas, the hypercube doesn't) you can get better estimates (but usually have to work quite hard). When you're working in a high dimension a large fraction of the volume is close to the vertices -- making things much more difficult. In particular you might look at the work of Martin Huxley. For example his book "Lattice Points, Area and Exponential Sums" or a number of his papers on this subject. 1 Note that$ker(f)$is a sublattice of the integer lattice $\mathbb{Z}^r$. And conversely any such integer sublattice will give you such a homomorphism$f$. The index of this sublattice is exactly what you call$h$(since the homomorphism is surjective). If$C$is a symmetric convex body in $\mathbb{R}^r$ and$L$a full rank lattice then the number of lattice points in$t C$(the dilation of$C$by a real factor of$t$) is asymptotic to$t^r/D$, where$D$is the volume of the fundamental domain of$L$. In your case$D=h$and$C\$ is the unit hypercube. Getting a good remainder is usually a much harder job.