= Solution
First prove the hinted statement. Let $U\subseteq y_n$ be the domain of a partial presheaf map $\phi:U\to y_{n+1}$. If $U$ is nonempty, take a section $f:m\to n$ in $U(m)$ and write $g=\phi_m(f):m\to n+1$. 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 gives
$$
gh=\phi_{m+1}(fh)=\phi_{m+1}(fk)=gk,
$$
a contradiction. Therefore \b[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. Hence
$$
\boxed{\operatorname{Hom}_{\operatorname{Sh}(C,J)}(A_n,A_{n+1})=\varnothing.}
$$
The stronger intermediate statement also says $A_{n+1}(n)=\varnothing$, by the Yoneda correspondence. It will be useful for the last clause.
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