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For a massless Dirac fermion,The local gauge symmetry gives the vector current $J^{\mu }=e\bar{\psi }{\gamma }^{\mu }\psi$. The global chiral transformation $\psi \mapsto e^{i\beta {\gamma }^5}\psi$ gives the classical axial currentUsing the massless Dirac equation and $\{{\gamma }^5,{\gamma }^{\mu }\}=0$ gives ${\partial }_{\mu }{J}_{\mathrm{ax}}^{\mu }=0$ classically.

Quantum mechanically, a gauge-invariant regulator for the fermion measure is not invariant under the axial rotation. In the Fujikawa form, its infinitesimal Jacobian containsThe first nonzero term in the heat kernel expansion is quadratic in $F_{\mu \nu }$. Usingand the Gaussian momentum integral gives the chiral anomalyThe overall sign follows the gamma-matrix, charge, and Levi-Civita conventions stated in the question. The same coefficient is obtained from the one-loop axial-vector-vector triangle diagram.

The matrix ${\gamma }^5=-i{\gamma }^0{\gamma }^1{\gamma }^2{\gamma }^3$ anticommutes with every Dirac matrix. Moving ${\gamma }^{\mu }$ through each power in the exponential series givesUnder the chiral transformationHermiticity of ${\gamma }^5$ and its anticommutation with ${\gamma }^0$ implyConsequentlyso the action is invariant. The corresponding classical axial current is

For the chiral transformation $\psi \mapsto e^{i\alpha {\gamma }^5}\psi$, one also has $\bar{\psi }\mapsto \bar{\psi }{e}^{i\alpha {\gamma }^5}$. The massless kinetic and vector-current terms are invariant because $\{{\gamma }^5,{\gamma }^{\mu }\}=0$. Since ${\gamma }^5$ commutes with $[{\gamma }^{\mu },{\gamma }^{\nu }]$, the Pauli bilinear transforms with $e^{2i\alpha {\gamma }^5}$ and is not invariant. Thus the continuous classical axial symmetry requires $\lambda =0$.

A fermion mass also breaks the symmetry, so in the massive theory one additionally needs $m=0$, which contradicts a genuinely massive fermion. Making $\alpha$ local produces $({\partial }_{\mu }\alpha )\bar{\psi }{\gamma }^{\mu }{\gamma }^5\psi$. No choice of the vector coupling $e$ or Pauli coupling $\lambda$ cancels this term; gauging it requires an axial gauge field, and at the quantum level one must also address the chiral anomaly.