Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 22 / 5 / Solution

Write $t:1\hookrightarrow \Omega$ for truth. Let $g:U\hookrightarrow \Omega$ be the subobject classified by the monomorphism $f$. Its classifier pullback gives $fg=t{!}_U$. The left side is monic, so ${!}_U:U\to 1$ is monic as well: $U$ is subterminal.

Let $i:V\hookrightarrow U$ be classified by $g$. The pullback square for $i$ says that $V\to 1$ is also the pullback of $g$ along $t$. Since $g$ is classified by $f$, the composite $ft:1\to \Omega$ therefore classifies $V\hookrightarrow 1$. Pulling this subobject back along ${!}_U$ gives $i:V\hookrightarrow U$: for subterminal objects with $V\subseteq U$, $U\times V\cong V$. Uniqueness of characteristic maps now gives $ft{!}_U=g$. Combining with $fg=t{!}_U$ proves

To deduce the full identity, let $u:1\to \Omega$ classify $U\hookrightarrow 1$. The pullback of $g$ along $f$ is $t{!}_U:U\hookrightarrow \Omega$. Indeed $ft{!}_U=g$, and if $fa=gb$, monicity of $f$ forces $a=t{!}_{U}b$, giving the universal property. The classifier of that pullback is $f^2$. But the subobject $t{!}_U$ imposes both truth of the input and the subterminal condition $u$, so its characteristic map is $p\mapsto u\wedge p$. The meet operation here is defined using only finite limits and the classifier. Thus $f^2=u\wedge (-)$. In particular $f^{2}t=u=f^{2}u$, by the unit and idempotence laws for meet. Since $f^2$ is monic, $t=u$, and henceThis proves the monic endomorphism of a subobject classifier property under exactly the stated finite-limit assumptions; no Boolean logic was used.

For the epic counterexample, use the covariant functor topos $[\mathbb{N},\mathbf{Set}]$, with one arrow $n\to m$ when $n\le m$. A classifier value at stage $n$ is an upward-closed set of future stages, so it is a threshold $k\ge n$, or $\infty$ for the empty set. The transition from $n$ to $n+1$ sends $k$ to $\max (k,n+1)$ and fixes $\infty$. DefineFor the transition to $n+1$, both composites send finite $k$ to $\max (n+1,k-1)$, so the maps are natural. Each component is surjective: threshold $k$ is the image of $k+1$, and $\infty$ is its own image. Epimorphisms in a functor topos are objectwise surjective, so $F$ is epic. But $F_n(n)=F_n(n+1)=n$, making it noninjective and not invertible. Thus the classifier has an epic endomorphism which is not an isomorphism, the shift epimorphism of the natural-number presheaf classifier.