 Solution

In the Category of sets, an epimorphism is exactly a surjective function. A surjective function is right-cancellable. If $q:X\to Y$ misses $y\in Y$, the constant-zero function and the function that is one at $y$ and zero elsewhere are distinct maps $Y\to \{0,1\}$ with equal composites with $q$.

If every ${\alpha }_A$ is surjective, equality $\beta \alpha =\gamma \alpha$ for natural transformations $\beta ,\gamma :G\Rightarrow H$ gives ${\beta }_A={\gamma }_A$ for every $A$. Thus $\alpha$ is an epimorphism in the functor category.

For the converse, construct the pointwise amalgamated doublewhere exactly the two copies of each element of $\mathrm{im}{\alpha }_A$ are identified; different elements of $G(A)$ remain different. Define $H(u)[y,i]=[G(u)y,i]$. Naturality of $\alpha$ implies that $G(u)$ takes its image into the image at the target, so this formula is well-defined and gives a functor. The maps $j_{i,A}(y)=[y,i]$ form natural transformations $j_0,j_1:G\Rightarrow H$, with $j_0\alpha =j_1\alpha$.

If $\alpha$ is an epimorphism, $j_0=j_1$. Since the two copies of an element outside $\mathrm{im}{\alpha }_A$ would be distinct, every element must lie in that image. Hence $\alpha$ is epic if and only if every component is epic. This proves the pointwise epimorphism in a functor category criterion directly, including the needed existence and naturality of the separating functor.