gauge theory - Cutkosky rule for the triangle diagram

Outline - the anomalous vacuum polarization correction

Suppose the abelian anomalous gauge theory (with axial gauge field $A$, vector gauge field $V$ and single massless fermion $\psi$): $$ \tag 1 L = -\frac{1}{4}V_{\mu\nu}V^{\mu\nu} - \frac{1}{4}A_{\mu\nu}A^{\mu\nu} + \bar{\psi}\gamma^{\mu}(i\partial_{\mu} + V_{\mu} + \gamma_{5}gA_{\mu})\psi, $$ and assume the 6th order of perturbation theory vacuum polarization diagram

The corresponding amplitude $M_{A\to A}$ has the form $$ \tag 2 M_{A \to A} = \int \frac{d^{4}q}{(2 \pi)^{4}}\epsilon_{\mu}(p)\Gamma^{\mu\nu\lambda}(p,q,p+q)\Gamma_{\lambda \nu \mu'}(p+q,q,p)\epsilon^{*\mu'}(p) $$ The effective vertex $\Gamma^{\mu\nu\lambda}$ corresponding to the triangle sub-diagram is anomalous and has the pole structure; we have $$ p^{\mu}\Gamma_{\mu\nu\lambda}(p,k_{1},k_{2}) \sim \epsilon_{\nu \lambda \rho \sigma}k_{1}^{\rho}k_{2}^{\sigma}, \quad k_{1}^{\nu}\Gamma_{\mu\nu\lambda}(p,k_{1},k_{2}) \sim \epsilon_{\mu \lambda \rho \sigma}k_{1}^{\rho}k_{2}^{\sigma}, \quad k_{2}^{\lambda}\Gamma_{\mu\nu\lambda}(p,k_{1},k_{2}) \sim \epsilon_{\mu \nu \rho \sigma}k_{1}^{\rho}k_{2}^{\sigma} \Rightarrow $$ $$ \tag 3 \Gamma^{\mu\nu\lambda}(p,k_{1},k_{2}) \sim \frac{p^{\mu}}{p^{2}}\epsilon^{\nu\lambda\rho\sigma}k_{1\rho}k_{2\sigma} + \frac{k_{1}^{\nu}}{k_{1}^{2}}\epsilon^{\mu\lambda \rho \sigma}k_{1\rho}k_{2\sigma} - \frac{k_{2}^{\lambda}}{k_{2}^{2}}\epsilon^{\mu\nu\rho \sigma}k_{1\rho}k_{2\sigma} $$

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The specific problem

In order to check the unitarity of the theory (i.e., to check whether the optical theorem holds) I need to calculate the imaginary part of $(1)$. There is the complication because of the pole structure of the vertex $(3)$. Note that typically vertices are free from poles, so usually for checking the unitarity of the gauge theory we don't need to worry about the imaginary part of vertices. An example is anomaly free non-abelian gauge theory, where the proof of the unitarity is simple (see, for example, section 6.5.4 on p. 237 here); if we have the abelian gauge theory, the checking of the unitarity is even elementary.

My book ("Advanced gauge quantum field theory" by P. van Nieuwenhuizen) "avoids" this problem by introducing the Higgs mechanism for the axial $U(1)$ sector: $$ \Delta L = \frac{1}{2}|\partial_{\mu} - 2igA_{\mu})\varphi|^{2} +\frac{\mu^{2}}{2}|\varphi|^{2} - \lambda |\varphi|^{4} - G (\bar{\psi}_{L}\varphi \psi_{R}+\bar{\psi}_{R}\varphi^{*} \psi_{L}) $$ It generates the mass for the axial gauge field and for the fermion, and then (by manipulating with the $G$ parameter) we can choose the fermion mass to be very large relative to the $A_{\mu}$ mass. Therefore, if the ingoing energy is smaller than the fermion mass, the imaginary part of the triangle diagram vanishes (loop fermions are always off-shell). However, the theory with the lagrangian $L + \Delta L$ is complicated theory whose quantization involves ghosts, and I'm not interested in it. Moreover, it turns out that the massless limit of the modification doesn't exist because of the anomaly, and the theory isn't equivalent to $(1)$.

But specifically in the case of the above diagram (surprisingly) the book states:

In this case these axial vector bosons could even be massless. The cutting relations for such graphs are again based on Ward identities, and if there are anomalies, there are extra terms in these Ward identities which lead to a break down of unitarity.

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My question

It seems that the statement above means that the imaginary part of the triangle diagram vanishes is not relevant independently on the fermion mass (since the masslessness of the axial gauge field leads to the masslessness of the fermion field). But I don't understand why.

Could You help?