When is pulling back along an algebra morphism right adjoint to “scalar extension”

Question

Let $\mathfrak{M}$ be an arbitrary monoidal category, and let $A, B$ be algebras therein, together with an algebra morphism $f \colon A \to B$.
The algebra morphism always induces a pullback functor, from e.g. right $B$– to right $A$-modules:
\begin{align}
f^* \colon \mathfrak{M}_B \to \mathfrak{M}_A, \quad
(V,\ \rho \colon V \otimes B \to V) \mapsto (V, f^*\rho = \rho \circ V\otimes f) \ .
\end{align}

Assume now that $\mathfrak{M}$ has coequalizers. Then for any $(V, \sigma) \in \mathfrak{M}_A$, we can define an object
\begin{align}
V \otimes_A B =
\operatorname{Coeq}(
V \otimes A \otimes B \xrightarrow{V \otimes (m_B \circ f \otimes B)} V \otimes B,
V \otimes A \otimes B \xrightarrow{\sigma \otimes B} V \otimes B)
\ ,
\end{align}
where by $m_B \colon B \otimes B \to B$ I mean the multiplication of the algebra $B$.

$V \otimes_A B$ is a priori only an object in $\mathfrak{M}$, I would assume, but if tensoring with $B$ preserves (these) coequalizers, then I'm fairly sure that it becomes a $B$-module by simply acting on $B$.
Assuming this works, we have a functor
\begin{align}
– \otimes_A B \colon \mathfrak{M}_A \to \mathfrak{M}_B
\ .
\end{align}
Example:
If $\mathfrak{M} = \textsf{Ab} = \mathbb{Z}\text{-mod}$, then we actually have an adjunction $- \otimes_A B \dashv f^*$.
This really is nothing else than the classical tensor-hom adjunction, since $f^* \cong \operatorname{Hom}_B(_fB_B, -)$.

My questions are:

Q1: Does $M \otimes_A B$ exist as a $B$-module if tensoring with $B$ preserves coequalizers? If not, what must we impose?

Q2: Assuming that we have those two functors, do we always $- \otimes_A B \dashv f^*$?

Answer 1

Let us call an object $B$ in a monoidal category $\mathfrak{M}$ left coflat if the endofunctor $B \otimes - $ preserves coequalizers. (This is standard terminology)

I managed to prove the following theorem, but will not post the proof, since it took me 3.5 pages to write out every detail (if anyone wants to see it, don't hesitate to hit me up). There should be enough detail in the statement such that anyone could prove it.

Theorem. Let $\mathfrak{M}$ be a monoidal category with coequalizers, and let $f \in \operatorname{Alg}_{\mathfrak{M}}(A, B)$ with $B$ (left) coflat. Denote the algebra structure of $B$ by $(B, m^B, u^B)$. Then the pullback functor \begin{align} f^* \colon {}_B \mathfrak{M} \to {}_{A}\mathfrak{M} , \quad f^*(M, \rho) = (M, \rho_f = \rho \circ f \otimes M) , \quad f^*g = g \end{align} is right adjoint to the functor \begin{align} B \otimes_A - \colon {}_{A}\mathfrak{M} \to {}_{B}\mathfrak{M} , \quad (N, \sigma) \mapsto (B \otimes_A N, \triangleright_{\sigma}) , \quad g \mapsto B \otimes_A g \ , \end{align} which sends an $A$-module $(N, \sigma)$ to the coequalizer of $m^B \otimes N$ and $B \otimes \sigma$. The $B$-action $\triangleright_{\sigma}$ and the action of the functor on morphisms $g \colon (N, \sigma) \to (N', \sigma')$ is given by \begin{align} \triangleright_{\sigma} \circ B \otimes \pi^\sigma = \pi^\sigma \circ m^B \otimes N \quad\text{ and }\quad B \otimes_A g \circ \pi^\sigma = \pi^{\sigma'} \circ B \otimes g \ , \end{align} where $\pi^\sigma \colon B \otimes N \to B \otimes_A N$ is the coequalizer morphism. The unit of the adjunction is \begin{align} \eta_{(N, \sigma)} = \pi^\sigma \circ u^B \otimes N \colon (N, \sigma) \to (B \otimes_A N, (\triangleright_\sigma)_f) \ , \end{align} and the counit $\varepsilon_{(M, \rho)} \colon (B \otimes_A M, \triangleright_{\rho_f}) \to (M, \rho)$ is uniquely determined by \begin{align} \varepsilon_{(M, \rho)} \circ \pi^{\rho_f} = \rho \ . \end{align}

In particular, this theorem gives positive answers to both of my questions.