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For an object $c$ of a site $C$ with terminal object $*$ there is a functor \begin{equation} f:\operatorname{Sh}(C)/c\to \operatorname{Sh}(C)/*=\operatorname{Sh}(C) \end{equation} from the slice topos into the whole topos given by composiong $F\to c$ with $c\to *$ where I abuse the notation for the Yoneda embedding.

What is an example of such a situation such that $f$ does not preserve monomorphisms?

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@ShawnHenry: No, this functor is a left adjoint, and cannot be a right adjoint in general because it does not preserve products. –  Zhen Lin Oct 25 '12 at 22:47
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There is no such situation, and the reason is purely category-theoretic.

Let $\mathcal{C}$ be any category. I claim that the monomorphisms in the slice category $(\mathcal{C} \downarrow Z)$ are exactly the same as the monomorphisms in $\mathcal{C}$. It's not hard to see that if a morphism in $(\mathcal{C} \downarrow Z)$ is a monomorphism in $\mathcal{C}$ then it must also be a monomorphism in $(\mathcal{C} \downarrow Z)$ because composition in $(\mathcal{C} \downarrow Z)$ is inherited from $\mathcal{C}$. On the other hand, suppose we have a monomorphism $f : A \to B$ in $(\mathcal{C} \downarrow Z)$, and two arrows $g, h : X \to A$ in $\mathcal{C}$ such that $f \circ g = f \circ h$. Then, if we denote by $a : A \to Z$ and $b : B \to Z$ the structural morphisms, we must have $$a \circ g = b \circ f \circ g = b \circ f \circ h = a \circ h$$ and so we can think of $g, h : X \to A$ as morphisms in $(\mathcal{C} \downarrow Z)$ as well, and thus $f \circ g = f \circ h$ implies $g = h$, as required to be a monomorphism in $\mathcal{C}$.

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