= Solution
{c}
An <operator equation in a path integral> $\mathcal O(x)=0$ means that insertion of $\mathcal O(x)$ into every <time-ordered correlation function> vanishes when $x$ is separated from the other insertion points. At coincident points, <Ward identity contact terms> can remain; an <operator equation> does not assert that these distributions vanish.
Let $X$ be a product of local insertions. Make the local phase change of integration variables from part <1/b>. The <functional measure> has unit <functional Jacobian>: the opposite phases of $\psi$ and $\bar\psi$ cancel in a vector-symmetry-preserving <regularization in quantum field theory>. There is no vector <gauge anomaly>. Invariance under a change of integration variables gives
$$
0=\langle\delta_\alpha X\rangle
+i\int d^4x\,\alpha(x)\langle T\{\partial_\mu j^\mu(x)X\}\rangle.
$$
If $X=\prod_r\mathcal O_r(x_r)$ and $\delta_\alpha\mathcal O_r=iq_r\alpha(x_r)\mathcal O_r$, comparison of the coefficients of the arbitrary $\alpha(x)$ yields the <Ward-Takahashi identity>
$$
\boxed{\langle T\{\partial_\mu j^\mu(x)X\}\rangle
=-\sum_rq_r\delta^{(4)}(x-x_r)\langle TX\rangle.}
$$
Consequently $\partial_\mu j^\mu=0$ is an <operator equation> at separated points. This is stronger than just $\langle\partial_\mu j^\mu\rangle=0$. For example, $X=\psi(y)\bar\psi(z)$ has $q_y=1$, $q_z=-1$, so its two <Ward identity contact terms> are $[-\delta^{(4)}(x-y)+\delta^{(4)}(x-z)]\langle T\psi(y)\bar\psi(z)\rangle$; dropping them would give a false identity. Equivalently, the <Dirac equation> and the <adjoint Dirac equation> give $\partial_\mu(\bar\psi\gamma^\mu\psi)=0$, with the same coincident-point qualification in <time-ordered correlation functions>.
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