= 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 double
$$
H(A)=(G(A)\times\{0,1\})/\sim,
$$
where exactly the two copies of each element of $\operatorname{im}\alpha_A$ 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 $\operatorname{im}\alpha_A$ would be distinct, every element must lie in that image. Hence \b[$\alpha$ 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.
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