Choice of basis for Fujikawa method to derive chiral anomaly I am studying the Fujikawa method of determining the chiral anomalies in a $U(1)$ theory. As we know the basis vectors selected are the eigenstates of the Dirac operator. One of the reasons given is that the eigenstates diagonalize the action which is needed for determining an exact quantity such as Ward-Takahashi identities. Anyone care to explain? I am referring to  Path Integrals and Quantum Anomalies by Kazuo Fujikawa and Hiroshi Suzuki.
 A: Alright, I will try to answer why we need Dirac eigenstates in this procedure, but I am not sure if it is anything more that the tautology that the Fujikawa method is precisely defined by using the Dirac eigenstates.
Let me briefly recap what the idea is: (All of this is for a Euclidean theory.)
We consider the infinitesimal local transformaton
$$\psi(x) \mapsto (1 + \mathrm{i}\alpha(x)\gamma_5)\psi(x) \equiv \psi'(x)$$
which, as it is a gauge transformation, should not change the value of the path integral:
$$ Z = \int \mathcal{D}\psi\mathcal{D}\bar\psi \mathrm{e}^{\int\bar\psi \mathrm{i}\not{D}\psi\mathrm{d}^4x} \overset{!}{=} \int\mathcal{D}\psi'\mathcal{D}\bar\psi'\mathrm{e}^{\int \bar\psi\not{D}\psi\mathrm{d}^4x + \int\alpha\partial_\mu j_5^\mu\mathrm{d}^4x}$$
To derive the anomaly term, we must examine what the change of the measure $\mathcal{D}\psi \mapsto \mathcal{D}\psi'$ is. In general, we can say that it is, by the usual transformation formulae for Grassmannian integrations, $\det(M)^{-1}$ for the operator acting as $\psi' = M\psi$. 
Now, we must recall that the functional measure $\mathcal{D}\psi$ is only defined in the limit of some (UV) regularized theory, it does not exist on its own.. So to actually calculate the correct $\det(M)$, we must obtain it as the limit $\lim_{\Lambda \rightarrow \infty}\det(M_\Lambda)$ of some $M_\Lambda$ and some UV cutoff scale (not necessarily hard) $\Lambda$. UV regulators suppress high momenta. And what are the eigenstates of the Dirac operator? ... Right, they are the momentum modes! So, the most natural regularisation of our theory is to exponentially suppress states with high Dirac eigenvalues as per
$$ \widetilde{\psi}_n = \mathrm{e}^{-\frac{\not{D}^2}{2\Lambda}}\psi_n$$
where the $\psi_n$ are the unregularised Dirac eigenstates. And by choosing the regulator that way, the only consistent way to define the measure is to see it as the integral over the coefficients of these modes, i.e.
$$\mathcal{D}\psi = \lim_{\Lambda \rightarrow \infty}\prod_n \mathrm{d}\widetilde{a}_n$$
where $\psi = \sum_n \widetilde{a}_n\widetilde{\psi}_n$. (The sum over the $n$ is actually an integral if we use no IR cutoff, but it doesn't matter here, you could as well use one, but it clutters notation a bit and contributes nothing insightful.)
Now, the rest of the Fujikawa anomaly method follows through as hopefully described in your book. I hope this is at least an approximate answer to your question.
A: The choice of regulator has nothing to do with the choice of basis. The usual choice of regulator is
$$
f(-\not D^2/\Lambda^2)
$$
which is universal, and makes no reference to any specific basis.
There is a choice of basis that makes the regulator particularly simple to work with, namely a basis of eigenfunctions of $\not D$. In such a basis, the regulator can be expressed as
$$
f(-\lambda^2/\Lambda^2)
$$
with $\lambda$ the eigenvalues of $\not D$. This basis simplifies the evaluation of $f$, but there is nothing that forces you to use this basis. You can choose any basis you want -- that is precisely the point of bases -- but if you do not make a smart choice, you will have to work much harder.
To give a trivial example, the partition function of a QM system is
$$
Z=\text{tr}(\exp(-\beta H))
$$
This can be evaluated in any basis you want, but the energy eigenbasis is particularly convenient
$$
Z=\sum_E e^{-\beta E}
$$
You can also compute $Z$ in a position basis if you want. But the calculation might become more cumbersome than in the energy basis.

Another important fact, which is perhaps more tangential to the question but seems important to me, is that you are not really evaluating the anomaly of the axial symmetry. What you are evaluating is the mixed anomaly between the axial symmetry and the gauge group.
As such, what exactly is anomalous is a matter of conventions. A mixed anomaly in $G_1\times G_2$ can be moved around -- there are counterterms that shift the anomaly from being a pure $G_1$ anomaly to a pure $G_2$ anomaly, and anything in between. Of course, in your example $G_2$ is the gauge group, and therefore we are forced to move the anomaly to $G_1$, as a gauge anomaly is forbidden.
This freedom to redefine mixed anomalies is reflected, in this particular case, in the freedom to choose different regulators; some regulators might make the axial symmetry non-anomalous, consistent with the fact that counterterms might make it look like a pure gauge anomaly instead. But this is forbidden: gauge groups shall not be anomalous. If a regulator preserves the axial symmetry but breaks the gauge group instead, that regulator is a bad one, and cannot be used.
