Consider a 1D spin-$1/2$ chain. After a Jordan-Wigner transformation, the spin-$1/2$ operator $S^z_i$ takes the form
$$ S^z_i = c^\dagger_i c_i - \frac{1}{2} \equiv \rho_i - \frac{1}{2}$$
where $\{ c_i \}$ are a set of fermionic modes obeying $\{ c_i,c_j^\dagger \} = \delta_{ij}$ and $\{ c_i,c_j \} = \{ c_i^\dagger, c_j^\dagger \} = 0$ and $\rho_i = c^\dagger_i c_i$ is the density operator.
Bosonising
The first step to bosonise a theory is to work in the continuum limit and expand the fermionic modes $c_i$ about the two Fermi points $\pm k_F$ as
$$ c_i = e^{ik_F x_i} \psi_R(x_i) + e^{-ik_F x_i} \psi_L(x_i)$$
Substituting this into the expression for $S^z_i$, we would arrive at
$$ S^z_i=\rho_L(x_i) + \rho_R(x_i) + (e^{-2ik_Fx_i} \psi^\dagger_R(x_i) \psi_L(x_i) + \mathrm{h.c.} ) - \frac{1}{2}$$
where $\rho_{L,R} = \psi^\dagger_{L,R} \psi_{L,R}$.
Now in Eq. (6.32) of Quantum Physics in One Dimension by Giamarchi, the author ignored the factor of $1/2$, essentially saying that when we bosonise we can assume $S_i^z = \rho_i$. Why can I just do this? This extra factor of $1/2$ seems to be important if I was looking at something like the interaction in the $XXZ$ model: the interaction coinsists of the terms
$$ S^z_i S^z_{i+1} = (\rho_i - 1/2)(\rho_{i+1} - 1/2) = \rho_i \rho_{i+1} - \frac{1}{2} ( \rho_i + \rho_{i+1}) + \frac{1}{4} $$
However, again in Giamarchi the author has assumed the factor of $1/2$ is not there in the expression for $S^z_i$, so writes in Eq. (6.18)
$$S_i^z S^z_{i+1} = \rho_i \rho_{i+1} $$
This seems to be a huge simplification here as we have thrown away non-trivial density terms.
My question
Why are we allowed to assume that $S_i^z = c^\dagger_i c_i$ when we bosonise this operator?
