Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 20 / 1 / iv / 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 $P=Y{⨿}_{X}Y$. 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 $j_{1}t=j_{2}t$. 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