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

A dense sieve, also called a persistently nonempty sieve, is one whose every pullback sieve on a category is nonempty: a member factoring through $f$ has the form $fh$, corresponding to a member $h$ of $f^{\ast }R$. The maximal sieve on a category clearly has this property. If $R$ is persistently nonempty, then $u^{\ast }R$ is too: for any $v$ into the domain of $u$, apply density to $uv$ to find an extension $uvh\in R$, giving $vh\in u^{\ast }R$.

For transitivity, suppose $R$ is dense on $n$ and $S$ is a sieve on a category such that $r^{\ast }S$ is dense for each $r\in R$. Given $f:m\to n$, choose $h$ with $fh\in R$. Density of $(fh{)}^{\ast }S$, tested on its identity arrow, supplies $k$ with $fhk\in S$. Thus $S$ is dense. These are the maximality, pullback and transitivity axioms of a Grothendieck topology. Therefore the specified sieves on a category form the coverage $J$, namely the double-negation topology. The empty sieve on a category is never covering, because its identity pullback is empty.

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.

Let $y_n=C(-,n)$. Suppose two sections $f,g\in y_n(m)$ have equal restrictions on a covering sieve on a category $R$ on $m$. Such a sieve on a category is nonempty, so choose $h:l\to m$ in it. Then $fh=gh$. Applying hypothesis (b) with the two right-hand factors both $h$ gives $f=g$, since $f,g$ have the same source and target. Hence every representable is separated for $J$.

In the associated-sheaf construction, sections of $F^{+}$ are matching families modulo agreement after covering refinement. The map $F\to F^{+}$ identifies two original sections only if they agree on a covering sieve on a category. Thus it is monic when $F$ is separated. The plus construction is itself separated, and applying it a second time gives the associated sheaf; its second unit is monic as well. Applying this to $y_n$ proves the unit is a monomorphismThe same construction shows the unit image is dense: a section of $F^{++}$ is locally represented by sections of $F^{+}$, which are in turn locally represented by sections of $F$, and transitivity combines the two covering refinements.

First prove the hinted statement. Let $U\subseteq y_n$ be the domain of a partial presheaf map $\varphi :U\to y_{n+1}$. If $U$ is nonempty, take a section $f:m\to n$ in $U(m)$ and write $g={\varphi }_m(f):m\to n+1$. Hypothesis (c) supplies $h,k:m+1\to m$ with $fh=fk$ but $gh\ne gk$. Because $U$ is a subpresheaf, both restrictions belong to $U(m+1)$; naturality givesa contradiction. Therefore every such partial map has empty domain.

Now suppose a presheaf map $y_n\to A_{n+1}$ existed. Pull back the monomorphism $y_{n+1}\hookrightarrow A_{n+1}$ along it. Its unit image is dense by the preceding plus-construction argument, and density is stable under pullback, so the resulting domain $U\subseteq y_n$ is dense. It gives exactly a partial map $U\to y_{n+1}$ of the prohibited kind. Such a domain cannot be empty: at the section $1_n$ of $y_n(n)$, density would otherwise require the empty sieve on a category to cover. Thus there is no map $y_n\to A_{n+1}$.

Any sheaf map $A_n\to A_{n+1}$, composed with $y_n\hookrightarrow A_n$, would be such a presheaf map. HenceThe stronger intermediate statement also says $A_{n+1}(n)=\varnothing$, by the Yoneda correspondence. It will be useful for the last clause.

The sheaf $A_n$ is noninitial: it contains $y_n$, whose value at $n$ includes $1_n$. In a two-valued topos its support, the image of $A_n\to 1$, can therefore only be $1$. An image equal to $0$ would force $A_n=0$, since the initial object of a topos is strict. Consequently every $A_n\to 1$ is an epimorphism.

Products of sheaves on a small site exist and are computed as products of the underlying presheaves, since a product of unique matching-family amalgamations again uniquely amalgamates. For the countable product $P={\prod }_{n\in \mathbb{N}}{A}_n$, its value at stage $m$ is ${\prod }_n{A}_n(m)$. The factor with $n=m+1$ is empty by the stronger statement proved in the preceding part. Thus $P(m)=\varnothing$ at every stage $m$, with the unique empty restriction maps. It is the empty presheaf, already an initial sheaf because no empty sieve on a category covers. HenceThis proves the final PDF-only request and exhibits the separating tower site with vanishing product. It is a genuine failure of countable choice in this Boolean two-valued sheaf topos; it does not contradict the classical external construction of the site.