Does the Yoneda extension of a functor preserving monomorphisms also preserve monomorphisms? I have a quick question ...
(Notation: $\mbox{Set}_A$ shall denote the category of presheaves on the category $A$.)
Suppose given two categories $A$ and $B$, and a functor $F \colon \, \mbox{Set}_A \rightarrow \mbox{Set}_B$ having the property that 1.) $F$ commutes with colimits, and 2.) $F_{|\,A}$, the restriction of $F$ to the presheaves that are in the image of the Yoneda embedding, preserves monomorphisms. Does $F$ then also preserve monomorphisms? Does it do so if the codomain of the monomorphism at hand is representable?
Thanks.
 A: First things first: if $F$ fails to preserve monomorphisms, then by composing with an appropriate evaluation functor $[\mathcal{B}^\textrm{op}, \textbf{Set}] \to \textbf{Set}$ we obtain a functor $[\mathcal{A}^\textrm{op}, \textbf{Set}] \to \textbf{Set}$ that fails to preserve monomorphisms. In other words, we may assume without loss of generality that $\mathcal{B}$ is the terminal category. 
To construct a counterexample, we choose a simple $\mathcal{A}$: let $\mathcal{A}$ be a category with one object $*$ and three morphisms: $\textrm{id}, f, f^2$, with $f^3 = f^2$. Notice that the only monomorphism in $\mathcal{A}$ is $\textrm{id}$, so any functor $\mathcal{A} \to \textbf{Set}$ whatsoever vacuously preserves monomorphisms.
The objects of $[\mathcal{A}^\textrm{op}, \textbf{Set}]$ are sets $X$ equipped with a distinguished endomorphism $f_X : X \to X$ such that $f_X^3 = f_X^2$. Given any two objects $X$ and $Y$ in $[\mathcal{A}^\textrm{op}, \textbf{Set}]$ we may form a "tensor product" $X \otimes_\mathcal{A} Y$, which is the set of pairs $(x, y)$ modulo the smallest equivalence relation $\sim$ such that $(f_X (x), y) \sim (x, f_Y (y))$. As usual, there is a tensor–hom adjunction
$$(-) \otimes_\mathcal{A} Y \dashv \textbf{Set}(Y, -) : \textbf{Set} \to \textbf[\mathcal{A}^\textrm{op}, \textbf{Set}]$$
and in fact all cocontinuous functors $[\mathcal{A}^\textrm{op}, \textbf{Set}] \to \textbf{Set}$ are of the form $(-) \otimes_\mathcal{A} Y$ for a suitable $Y$. 
In commutative algebra, it is well-known that tensoring need not preserve monomorphisms. The same is true here. Consider the following objects of $[\mathcal{A}^\textrm{op}, \textbf{Set}]$:
\begin{align}
X & = \{ 0, 1, 2 \} && f_X (0) = 1, f_X (1) = 2, f_X (2) = 2 \\
Y & = \{ 0, 1 \} && f_Y (0) = 1, f_Y (1) = 1
\end{align}
Clearly, the map $y \mapsto y + 1$ defines a monomorphism $h : Y \to X$. I claim that $k = h \otimes_\mathcal{A} Y : Y \otimes_\mathcal{A} Y \to X \otimes_\mathcal{A} Y$ is not injective. Indeed, we have
$$k(0, 0) = (1, 0)$$
$$k(1, 0) = (2, 0) \sim (1, 1) \sim (0, 1) \sim (1, 0)$$
and $(0, 0) \nsim (1, 0)$ in $Y \otimes_\mathcal{A} Y$.
