Does ring epimorphism induce reversed fully faithful functors in the module categories? Let $R$ and $S$ be commutative rings, and $\varphi : R \to S$ be an epimorphism in the category of rings.
Is the induced functor $F : S\text{-Mod} \to R\text{-Mod}$ fully faithful?
This is the case when $R = \Bbb Z$ and $S = \Bbb Q$, but I wonder if this holds in general.
Also, what if I omit the requirement that $R$ and $S$ be commutative?
 A: It's not difficult to see that without any assumptions on $\varphi$, the restriction of scalars $F$ is faithful, since it doesn't change the morphisms as maps at all.
So it's really about showing that $F$ is full under the assumption that $\varphi$ is an epimorphism. I'll assume that all rings are commutative, since pushouts in the category of not-necessarily commutative rings are too complicated for this argument to work.
We start with a lemma

If $\varphi: R \to S$ is an epimorphism, then the maps $i_1:S \to S \otimes_R S, s \mapsto s \otimes 1$ and $i_2: S \to S \otimes_R S, s \mapsto 1 \otimes s$ are equal to each other and isomorphisms.

This is actually just the special case of the following statement about epimorphisms and pushouts in any category:

If $\mathcal C$ is a category, then $f:A\to B$ in $\mathcal C$ is an epimorphism iff this diagram is a pushout square: $$\require{AMScd} \begin{CD}
A @>{f}>>B
\\ @V{f}VV @VV{\operatorname{id}_B}V\\
B @>>{\operatorname{id}_B}> B
\end{CD}$$

Proof: This is really just a restatement of the definitions. Let's only write down the proof of the direction we actually need. We check the defining property of the pushout. Suppose that $X$ is an object in $C$ and we have morphisms $g_1, g_2:B \to C$ such that $g_1 \circ f = g_2 \circ f$, i.e. a cocone over the diagram 
$$\require{AMScd} \begin{CD}
A @>{f}>>B
\\ @V{f}VV\\
B 
\end{CD}$$
Then by definition of $f$ being an epimorphism, we get that $g_1=g_2$, so there's a unique morphism $B \to C$, given by $g_1=g_2$ which makes the necessary diagrams commute. Thus $B$ is the push-out as claimed.
Let $M$ and $N$ be $S$-modules, then $F(M)$ and $F(N)$ are $R$-modules via restriction of scalars. Let $f:F(M) \to F(N)$ be a $R$-linear map.
Our goal is that if we identify $M "=" F(M)$ and $N"="F(N)$, $f:M \to N$ is $S$-linear.
Let $m \in M$, then consider the map $S \times S \to N, (s,s') \mapsto sf(s'm)$. By the universal property of the tensor product (using that $f$ is $R$-linear), this induces a map $h: S \otimes_R S \to N, s \otimes s' \mapsto sf(s'm)$. But since the maps $S \mapsto S \otimes_R S, s \mapsto s \otimes 1$ and $S \mapsto S \otimes_R S, s \mapsto 1 \otimes s$ are equal, we get that $f(sm)=h(1 \otimes s)=h(s \otimes 1)=sf(m)$. This shows that $f$ is $S$-linear.
