Biot Savart law for inhomogeneous Navier-Stokes equation

Question

I have a question concerning the Biot-Savart law. For the $3D$ Navier Stokes equation $$\partial_t v+ v\cdot \nabla v + \nabla p= \nu \Delta v\\ \text{div}(v)=0$$ the Biot-Savart law declares a relation between the velocity $v$ and its curl, $\omega=\text{curl}(v)$, $$v(t,\cdot)=-\frac{1}{4\pi}\int_{\mathbb{R}^3}\frac{\cdot-y}{|\cdot-y|^3}\times\omega(t,y) dy.$$ Now let us consider the inhomogeneous Navier Stokes equation $$\partial_t v+ v\cdot \nabla v + \nabla p= \nu \Delta v+f.$$ The derivation goes as follows: $$\text{curl}(\omega)=\text{curl}\text{curl}(v)=\nabla(\nabla\cdot v)-\Delta v=-\Delta v$$ and hence $$v=(-\Delta)^{-1} \text{curl}(\omega).$$ Using the fundamental solution of the Poisson equation we arrive at $$v(t,\cdot)=-\frac{1}{4\pi}\int_{\mathbb{R}^3}\frac{\cdot-y}{|\cdot-y|}\times \omega(t,y) dy.$$ So we did only use the imcompressiblity condition and not the first equation itself to deduce the Biot-Savart law. Obviously the regularity/integrability of $f$ has to be taken into account to give an appropriate solution space. I could take $\text{curl}$ of the whole equation and solve the associated equation for $\omega$ and then get the velocity field $v$. So the information on $f$ has been kept in track by $\omega$ am I right?