Why can i replace a gauge field by the current it couples to in the calculation of a greens function?

i am reading about Anomalies in QFT at the moment and there is a question that appeared on the way.

Often i find it that people are calculating the time ordered expectation value of some fields (in QED for example) and are replacing the field $A_\mu$ by the current $j_\mu = \bar{\psi}\gamma_\mu\psi$. For examle in Peskin Schröder sec. 9.6 on p. 311 he derives a the Ward identities in QED and states that the object $\langle T j^\mu(x)\psi(y)\overline{\psi}(z)\rangle$ corresponds to the electron-photon vertex. But should the vertex not correspond to $\langle T A^\mu(x)\psi(y)\overline{\psi}(z)\rangle$ whith the EM potential $A^\mu$ ?

The same for the calculation of the triangle anomaly. Often you find that people are calculating something like $\langle j_1^\mu(x)j_2^\nu(y)j_3^\rho(z)\rangle$ where the j's are (axial or vector) currents of fermions coupled to some vector fields. This gives than the diagram with three external vector bosons and a circulating fermion.

It is clear to me that $\langle T j^\mu \ldots \rangle$ and $\langle T A^\mu \ldots \rangle$ give me the same diagrams when i start to calculate, i am just wondering if i can prove in any way (from a generating functional or so) that these greens fuctions are really the same. (I use Greens fuction in the sense of time ordered vacuum expectation value of fields.)

Hopefully somebody can clarify this a little.

-