# When is pulling back along an algebra morphism right adjoint to "scalar extension"?

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 $$- \otimes_A B \dashv f^*$$?

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)
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}