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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.

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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 $Y \otimes_\mathcal{A} Y$.

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Very nice, thank you very much for this answer. –  user49789 Nov 17 '12 at 22:35
    
(In one of the last lines, it should be "k(1,0)=(2,0)~(1,1)~...", I think?) –  user49789 Nov 17 '12 at 22:37
    
Yes, that's right. –  Zhen Lin Nov 17 '12 at 22:41
    
Actually, what happens if the codomain $Y$ of the monomorphism $X \hookrightarrow Y$ is representable, and $X$ is a colimit of subobjects $X_i \subset X$ in $\mathcal{A}$? Is it then possible to deduce that we still have a monomorphism after applying $F$? –  user49789 Nov 17 '12 at 22:43
    
Sorry, I meant: "..., and $X$ is a colimit of subobjects $X_i \subset Y$ in $\mathcal{A}$?" –  user49789 Nov 17 '12 at 22:47
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