Let $\boldsymbol{Ring}$ denote the category of commutative rings with unity (not necessarily different from $0$), and morphisms preserving unity.
I have been wondering if the assignment
$A\in\boldsymbol{Ring}\to A\boldsymbol{mod}$
$(f:A\to B)\to(f^e:=\_\otimes_Af^*B:A\boldsymbol{mod}\to B\boldsymbol{mod})$
is a pseudofunctor from 1-category $A\boldsymbol{mod}$ to the 2-category $\boldsymbol{Cat}$?
(Here $f^*B$ is $B$ viewed as an $A$ module via the ring map $f$, so $f^e$ is the scalar extension functor along the map $f$.)
So far I am stuck at finding , for $f:A\to B,\,g:B\to C$ ring morphisms and $N\in A\boldsymbol{mod}$, a $C\boldsymbol{mod}$ isomorphisms
$\eta_N:g^e(f^eN)=(N\otimes_Af^*B)\otimes_Bg^*C\to (g\circ f)^eN=N\otimes_A(g\circ f)^*C$
natural in $N$. My guess is that, on pure tensors, the map should send
$n\otimes f^*b\otimes g^*c=n\otimes f^*1\otimes g^*g(b)c\to n\otimes(g\circ f)^*g(b)c$, but I don't quite know how to prove it is welldefined.
My question is: can this map be welldefined? Is this assignment a pseudofunctor?
Apperently, it follows from general nonsense: Notice that, for every ring morphism $f$, $f^e$ is left adjoint to the scalar restriction functor $f^*$. Now the assignment from $\boldsymbol{Ring}^{op}\to \boldsymbol{Cat}$, on objects $$A\to A\boldsymbol{mod},$$ on morphisms $$(f^{op}:A\to B)\to(f^*:A\boldsymbol{mod}\to B\boldsymbol{mod})$$ is a pseudofunctor, and thus gives rise to a Grothendieck fibration $p:E\to \boldsymbol{Rings}$. Now, since each $f^*$ has the left adjoint $f^e$, this fibration is also an opfibration (see nlab). So the assignment described in the original post must also be a pseudofunctor. Still it would be interesting to actually see the natural isomorphism worked out.
I guess it always helps to know more category theory :)