Monomorphism into separated presheaf Let $X, Y : \mathbb{C}^\text{op} \to \mathbf{Set}$, where $\mathbb{C}$ is a small category, and assume the natural transformation $\alpha : X \to Y$ is a monomorphism. If $Y$ is separated with respect to a coverage on $\mathbb{C}$, then is $X$ necessarily separated (w.r.t. the same coverage)?
 A: The natural transformation $\alpha$ is a monomorphism precisely when all of its components are injections. That way we may see $X$ as just a subpresheaf of $Y$. That is, for each object $C$ in $\mathbb{C}$ we have $X(C) \subseteq Y(C)$ and $\alpha_C$ is just this inclusion.
I do not know what definition of separated you precisely use, but since you are talking about a "coverage" I assume it is the one that links to a Grothendieck topology. Just to be clear, let me recall the definition here:

A presheaf $X$ is separated (with respect to a fixed Grothendieck topology) if the following holds: for every object $C$ in $\mathbb{C}$ and all $x, y \in X(C)$, if the sieve $\{f: C' \to C \mid X(f)(x) = X(f)(y)\}$ is covering, then $x = y$.

Now assume $Y$ is separated. Let $x, y \in X(C)$ such that $S = \{f: C' \to C \mid X(f)(x) = X(f)(y)\}$ is covering. Then because $X(C) \subseteq Y(C)$ we have that $x,y \in Y(C)$. Since $S$ is covering for $C$ and $Y$ is separated we conclude that $x = y$, and so we see that indeed $X$ is separated.
A: Another way of defining sheaves that I find elegant is the following: $Y$ is a sheaf (with respect to a given covering) if and only if for every covering sieve $S$ on $U$ represented as a subfunctor of $\mathsf{Hom}(-,U)$ we have $y\mapsto(f\mapsto Y(f)(y)):Y(U)\to\mathsf{Nat}(S,Y)$ is an isomorphism. Here, $\mathsf{Nat}(S,Y)$ is the set of natural transformations from $S$ to $Y$. $Y$ is a separated presheaf if this is only a monomorphism.
Given that $Y$ is separated and $\alpha$ is a mono, then so is $x\mapsto(f\mapsto Y(f)(\alpha_U(x))):X(U)\to\mathsf{Nat}(S,Y)$ for every $U$ and $S$ covering $U$ because it's a composition of monomorphisms. Naturality of $\alpha$ states that $Y(f)\circ\alpha_U=\alpha_V\circ X(f)$, in other words $$x\mapsto(f\mapsto Y(f)(\alpha_U(x))) = x\mapsto(f\mapsto \alpha(X(f)(x)))$$
This means $x\mapsto(f\mapsto X(f)(x))$ is a mono since it factors through a mono via post-composition by a mono.
