= Solution
Suppose every component $\alpha_A$ is a <monomorphism> in the <Category of sets>, hence an <injective function>. If $\beta,\gamma:H\to F$ are <natural transformations> with $\alpha\beta=\alpha\gamma$, then $\alpha_A\beta_A=\alpha_A\gamma_A$ at every object. Componentwise cancellation gives $\beta_A=\gamma_A$, so $\beta=\gamma$. Thus $\alpha$ is a <monomorphism> in the <functor category>.
Conversely, suppose $\alpha$ is a <monomorphism> and $x,y\in F(A)$ satisfy $\alpha_A(x)=\alpha_A(y)$. The <Yoneda lemma> supplies <natural transformations> $\theta^x,\theta^y:Y(A)\to F$. Its <naturality> in $F$ identifies $\alpha\theta^x$ and $\alpha\theta^y$ with the equal elements $\alpha_A(x)$ and $\alpha_A(y)$. Cancellation by $\alpha$ gives $\theta^x=\theta^y$; evaluating at $1_A$ gives $x=y$. Therefore \b[a <natural transformation> is a <monomorphism> exactly when all its components are <injective functions>]. This is the <pointwise monomorphism in a set-valued functor category> criterion.
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