= Solution
We first justify the relevant <pointwise epimorphism in a functor category> criterion. A componentwise <surjective function> gives an <epimorphism>, because equality of two composites can be cancelled at each component. Conversely, for a <natural transformation> $t:X\to Y$, form the pointwise <pushout> $P=Y\amalg_XY$. Restrictions of $Y$ preserve the identifications imposed by $t$, so these pointwise <sets> and restrictions form a <categorical presheaf>. The two inclusions $j_1,j_2:Y\to P$ satisfy $j_1t=j_2t$. If $t$ is epic, $j_1=j_2$. At an object $U$, an element outside the image of $t_U$ would retain two distinct copies in $P(U)$, contradicting this equality. Thus every $t_U$ is surjective.
Apply this criterion to $H_\bullet(f)$ at $B$. Surjectivity of $\mathcal C(B,A)\to\mathcal C(B,B)$ supplies $g:B\to A$ with $fg=1_B$. Conversely, if $fg=1_B$, then $H_\bullet(f)H_\bullet(g)=1_{H_B}$, making $H_\bullet(f)$ a <split epimorphism> and hence an <epimorphism>. Therefore
$$
\boxed{H_\bullet(f)\text{ epic}\iff f\text{ split epic}.}
$$
Back to article page
