When is $N\otimes_A B \to N$ an isomorphism? Let $A, B$ be commutative (unital) rings and $f\colon A \to B$ an $A$-algebra. There then exists a canonical functor $f_*\colon \mathbf{Mod}_B \to \mathbf{Mod}_A$ such that, for every morphism of $B$-module $g\colon N \to N'$, it associates the morphism of $A$-modules $f_*(g)\colon f_*(N) \to f_*(N')$, where, if $\rho_N\colon B\times N\to N$ (resp. $\rho_{N'}\colon B\times N' \to N'$) is the action of $B$ on $N$ (resp. on $N'$), $f_*(N)$ (resp. $f_*(N')$) has the induced $A$-module structure

$\rho_N\circ(f, \text{id}_N)\colon A\times N \to B\times N \to N\quad \text{(resp. }\rho_{N'}\circ(f, \text{id}_{N'})\text{)}$

and $f_*(g)$ is the same morphism of abelian group canonically $A$-linear with these induced structure of $A$-modules on $f_*(N)$ and $f_*(N')$.
The functor $f_*$ is faithful and has both left and right adjoints, resp. $f^*$ and $f^!$, such that, for every morphism $h\colon M\to M'$ of $A$-modules,

$f^*(g)= \text{id}_B \otimes h\colon B\otimes_A M \to B\otimes_A M'$

(the base change) and

$f^! = h\circ -\colon \text{Hom}_A(B, M) \to \text{Hom}_A(B, M'): u\mapsto h\circ u.$



Question. When is $f_*$ also full as functor, i.e., for every $B$-modules $N, N'$, when is $f_*\bigl(\text{Hom}_B(N, N')\bigr) \cong \text{Hom}_A\bigl(f_*(N), f_*(N')\bigr)$ as $A$-modules?

Equivalently, we can ask when the counit

$(\epsilon_f)_N\colon f^*(f_*(N)) = B\otimes_A f_*(N) \to N: b\otimes n \mapsto bn$

of the adjunction $(f^*\dashv f_*)$ is an isomorphism of $B$-modules for every $B$-module $N$.
Remark. For every multiplicative subset $S$ of $A$, it holds for the canonical $A$-algebra $i^S_A\colon A \to S^{-1}A$, but I cannot say much more that this.
Any reference is welcome.
 A: A basic fact about category theory states that a morphism $f \colon A \to B$ is an epimorphism in a category $\mathcal C$ if and only if the pushout of it with itself (i.e., the cobase change of $f$ along itself) exists and
$\require{AMScd}$
\begin{CD}
    A @>f>> B\\
    @V f V V @VV \text{id}_B V\\
    B @>>\text{id}_B> B
    \end{CD}
is a cocartesian square.
In the case in which $\mathcal C = \mathbf{CRing}$, the pushout square of a morphism $f$ along itself is given by
\begin{CD}
    A @>f>> B\\
    @V f V V @VV p_2 V\\
    B @>>p_1> B\otimes_A B
    \end{CD}
where $p_1$ sends $b\mapsto b\otimes 1_B$ and $p_2$ maps $b \mapsto 1_B \otimes b$. If $g, g'\colon B \to C$ are two morphisms such that $gf = g'f$, then the universal morphism $B\otimes_A B \to C$ maps $b\otimes b'$ to $g(b)g'(b')$.
Therefore $f$ is a epimorphism in $\mathbf{CRing}$ if and only if the canonical morphism $\phi\colon B\otimes_A B \to B$ sending $b\otimes b' \mapsto bb'$ is an isomorphism.
The functor $N\otimes_B -$ preserves and reflects isomorphisms, so we get that $N\otimes_B \phi$ from $N\otimes_B (B\otimes_A B)$ to $N\otimes_B B$ is an isomorphism if and only if $\phi$ is so.
But, by the associativity of tensor product in $\mathbf{CRing}$, we have that $N\otimes_B (B\otimes_A B)$ is canonically isomorphic to $(N\otimes_B B)\otimes_A B$ (the mapping being the obvious one).
Further, there is a canonical isomorphism $N\otimes_B B \to N$ given by $n\otimes b\mapsto bn$.
Finally, we can conclude that the morphism $N\otimes_B \phi$ is isomorphic to the counit $(\epsilon_f)_N$ and hence the latter is an isomorphism if and only if the former is so, and this holds if and only if $f$ is an epimorphism.
