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

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.