 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 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.