Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 22 / 7 / ii

We prove Booleanity directly from the dense-cover description. Let $B$ be a subsheaf of a sheaf $F$. Its pseudocomplement $N$ consists of sections $x\in F(c)$ such that no restriction along any arrow into $c$ belongs to $B$. This is a subpresheaf. It is also a sheaf: if a section of $F$ restricts into $N$ on a dense covering sieve on a category, any hypothetical restriction into $B$ has a further restriction in that cover; it would then belong to both $B$ and $N$, impossible. Thus the unique amalgamation in $F$ of an $N$-matching family lies in $N$.

For any $x\in F(c)$, the sieve on a category of arrows on which its restriction lies in $B$ or $N$ is dense. Given an arrow $u:d\to c$, either some further restriction belongs to $B$, supplying the required member of that sieve on a category, or none does, in which case the restriction along $u$ itself belongs to $N$. Therefore $B\cup N$ is locally all of $F$, so their join in the sheaf topos is $F$. Their intersection is empty. Every subsheaf consequently has a complement, proving the topos is Boolean.

For two-valuedness, let $U\hookrightarrow 1$ be a nonempty subterminal sheaf. At some stage $m$, its value is the singleton. For every $k\ge m$ there is an arrow $k\to m$, so restriction makes $U(k)$ the singleton as well. At any stage $n$, the sieve on a category of arrows into $n$ whose domains are at least $m$ is dense: after an arrow with domain $l$, choose a further domain $k\ge \max (l,m)$ and use the existence of arrows downwards. On this covering sieve on a category the unique terminal section has a matching family in $U$, so the sheaf condition gives $U(n)=1$. Hence $U=1$. Since the empty sieve on a category never covers, the empty presheaf is an initial sheaf distinct from $1$. Thusand the topos is two-valued, in the distinct global sense of a two-valued topos.