Consider the model

$$

\mathcal{L} = \bar{\psi}\big[i\gamma^{\mu}(\partial_{\mu}-i\gamma_{5}A_{\mu} - iV_{\mu}) -me^{2i\gamma_{5}\phi}\big] \psi+\frac{m_{Z}^{2}}{2}(A_{\mu}+\partial_{\mu}\phi)^{2} \quad (1)

$$

Here $A_{\mu}$ is an axial-vector gauge boson, $V_{\mu}$ is a vector-like gauge boson, $\phi$ is the Goldstone boson associated with the Higgs mechanism which generates the masses $m_{A}$ and $\psi$. The Lagrangian is classically invariant under the axial transformations

$$

\psi \to e^{i\gamma_{5}\alpha(x)}\psi, \quad A_{\mu} \to A_{\mu}+\partial_{\mu}\alpha, \quad \phi \to \phi - \alpha$$

Since $m_{A}$ and $m$ are non-zero, the physical particle content is $\{\psi, Z_{\mu}\equiv A_{\mu}+\partial_{\mu}\phi, V_{\mu}\}$. If I perform the redefinition

$$

\psi \equiv e^{-i\gamma_{5} \phi }\kappa, \quad (2)

$$

where $\kappa$ is an axial gauge invariant fermion, then classically, the Lagrangian $(1)$ will transform to the form

$$

\mathcal{L} = \bar{\kappa}\big[ i\gamma^{\mu}(\partial_{\mu} - i\gamma_{5}Z_{\mu}-iV_{\mu}) - m\big]\kappa + \frac{m_{Z}^{2}}{2}Z_{\mu}^{2} \quad (3)

$$

Naively, now the axial symmetry is completely rendered out, since all the quantities in (3) are gauge invariant by definition. However, actually the model is anomalous, and on the quantum level an additional gauge variant term will arise.

My question is the following. Am I right that the redefinition $(2)$ induce the anomalous term to the Lagrangian as well as if we consider it as the axial transformation, inducing the well known anomalous term $\frac{\phi}{4 \pi}V_{\mu\nu}\tilde{V}^{\mu\nu}$?