The bosonization map relates the fermionic current $\bar{\psi}\gamma\psi$ to the bosonic current $\partial\phi$, and also the components of $\psi$ to $e^{i\sqrt{\pi}\left(\phi\pm\bar\phi\right)}$. Here I'm using $\bar{\phi}$ to denote the dual field $\partial_\mu \phi = \epsilon_{\mu\nu}\partial^\nu\bar\phi$.
Now I was under the impression that global $U(1)$ vector symmetry $\psi \rightarrow e^{i\theta}\psi$ is associated with shift symmetry $\phi\rightarrow \phi+\frac{1}{\sqrt{\pi}}\theta$, and similarly for axial symmetry and shifts in $\bar{\phi}$.
But we can extend this global symmetry to a gauge symmetry $\theta(x)$. Now the current $\bar{\psi}\gamma\psi$ is gauge invariant, but $\partial\phi\rightarrow \partial\phi+\frac{1}{\sqrt{\pi}}\partial\theta$ is not!
Does gauge symmetry make sense in bosonization?
I'm particularly confused since I've seen use of bosonization of a fermion coupled to a gauge field to explain things like the chiral anomaly and lack of dependence on the vacuum angle. (see e.g. Witten Nucl Phys B149, 285)
For instance suppose we have a gauge invariant Lagrangian like $$\bar{\psi}i(\displaystyle{\not}\partial+i\displaystyle{\not}A)\psi -\frac{1}{4}F^2$$ and after naive bosonization we get something like (forgive me for ignoring the constant factors) $$\frac{1}{2}(\partial{\phi})^2+\partial_\mu\phi A^\mu -\frac{1}{4}F^2$$ but this is missing the $\frac{1}{2}A^2$ term that would be needed for gauge invariance!
