Example of a functor from the slice topos into the whole topos which does not preserve monomorphisms 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?
 A: 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 $\mathcal{C}$.
