Solution
= Solution
The required condition is the <common-refinement condition for nonempty-sieve coverage>: for every pair $f:V\to U$, $g:W\to U$, there are arrows $h:T\to V$, $k:T\to W$ with
$$
\boxed{fh=gk.}
$$
Necessity follows by pulling back the nonempty sieve generated by $f$ along $g$: a member $k$ of the <pullback> sieve supplies such an $h$. Conversely, this condition makes the <pullback> of every nonempty <sieve on a category> nonempty. The maximal sieve is nonempty, and the transitivity axiom holds: if a sieve $S$ is locally covering along every member of a nonempty covering sieve $R$, choose $f\in R$ and then $g\in f^*S$; their composite is in $S$. Hence the nonempty sieves form a <Grothendieck topology>, called the <atomic topology>.
For all functions between nonempty finite sets, the two maps from a singleton to different points of a two-point set have no common refinement. Every potential domain remains nonempty, so the two constant composites cannot agree. The condition fails.
For surjections it holds: $V\times_UW$ is nonempty and both projections are surjective. Work from now on in a small skeleton $\mathcal D$ of nonempty finite sets and surjections. Every morphism $f:P\twoheadrightarrow U$ is a <regular epimorphism>, with <kernel pair> $P\times_UP\rightrightarrows P$, and is the coequalizer of that pair in $\mathcal D$.
A matching family in a <representable> $yV$ on the sieve generated by $f$ is determined by a surjection $t:P\to V$ equalizing that <kernel pair>. It factors uniquely through a function $U\to V$, which is surjective because $t$ is. This gives the unique amalgamation. A general nonempty covering sieve contains such an $f$; after amalgamating there, common refinements with any other member force agreement on the entire sieve. Therefore \b[every <representable> is a sheaf], so this atomic site is <subcanonical>.
For any sheaf $F$, every restriction $F(f)$ is injective: equality after a covering arrow forces equality by the separated part of the sheaf condition. We shall also use descent along any surjection $q:n\twoheadrightarrow k$:
$$
F(k)\longrightarrow F(n)\rightrightarrows F(n\times_kn)
$$
is an <equalizer> of sets. These are the <descent identities for the atomic finite-surjection site>.
Consider primitive $x\in F(m)$, $y\in F(n)$ with a common restriction along $\alpha:P\twoheadrightarrow m$, $\beta:P\twoheadrightarrow n$. Suppose $a,b\in P$ have $\alpha(a)=\alpha(b)$ but $\beta(a)\ne\beta(b)$. Let $q:n\twoheadrightarrow n-1$ identify just the two points $\beta(a),\beta(b)$. Define the finite nonempty set
$$
T=\{(r,s)\in P^2:\alpha(r)=\alpha(s),\ q\beta(r)=q\beta(s)\}.
$$
Both projections $t_1,t_2:T\twoheadrightarrow P$ are surjective, since $T$ contains every diagonal pair. There is also a surjection
$$
h:T\twoheadrightarrow n\times_{n-1}n,\qquad(r,s)\longmapsto(\beta(r),\beta(s)).
$$
Indeed the target consists of diagonal pairs, which are reached because $\beta$ is surjective, and the two off-diagonal pairs corresponding to $\beta(a),\beta(b)$, reached by $(a,b)$ and $(b,a)$.
Since $\alpha t_1=\alpha t_2$, the common-restriction equality gives $F(\beta t_1)y=F(\beta t_2)y$. If $\pi_1,\pi_2$ are the target kernel-pair projections, this is
$$
F(h)F(\pi_1)y=F(h)F(\pi_2)y.
$$
Injectivity of $F(h)$ gives the kernel-pair matching condition on $y$. Descent along $q$ then writes $y=F(q)y'$, contradicting primitivity. Thus $\ker\alpha\subseteq\ker\beta$; interchange the roles to obtain equality. This is the <primitive-element kernel rigidity lemma>.
Equal kernels produce a unique bijection $\gamma:m\to n$ with $\beta=\gamma\alpha$. Now $F(\alpha)x=F(\alpha)F(\gamma)y$, and injectivity implies
$$
\boxed{x=F(\gamma)y.}
$$
In particular equivalent primitive elements have the same cardinality and differ only by transport along a bijection.
Every element $z\in F(p)$ descends to a primitive one: whenever it is not primitive, descend along a surjection reducing the cardinality by one; this process terminates at or before cardinality one. Kernel rigidity shows that all primitive ancestors of $z$ lie in one equivalence class. Let $F_C(P)$ be the elements with primitive-ancestor class $C$. Restriction along a surjection preserves this class, so each $F_C$ is a subfunctor and
$$
F(P)=\coprod_CF_C(P)
$$
pointwise. Each $F_C$ is a sheaf. A matching family glues in $F$, and one member along a nonempty covering arrow already determines the primitive class of the glued element; it must be $C$.
Choose a representative primitive $x\in F(m)$ of $C$. The Yoneda map $ym\to F$ named by $x$ has image exactly $F_C$. It reaches all descendants of $x$, and every equivalent primitive ancestor is its transport along a bijection. It is therefore pointwise surjective onto $F_C$ and is epic as a map of sheaves. We obtain the <primitive decomposition of an atomic finite-surjection sheaf>
$$
\boxed{F\cong\coprod_CF_C,\qquad ym\twoheadrightarrow F_C.}
$$
This includes the empty <coproduct> for an empty sheaf.
Each nonempty $F_C$ is an <atom in a topos>. If a sheaf <subobject> $S\hookrightarrow F_C$ has an element $F(\alpha)x$ at some $P$, membership descends along the covering surjection $\alpha:P\to m$, so $x$ belongs to $S$. All its restrictions then belong to $S$, giving $S=F_C$. Thus every <subobject> of any $F$ selects entire components of this <coproduct>, and its complementary selection is again a sheaf <subobject>. Its characteristic map sends selected components to $\top$ and all others to $\bot$.
The constant two-element presheaf is a sheaf: a matching family on a nonempty sieve has the same value on all its arrows, since any two have a common refinement. The value extends uniquely. It therefore supplies these characteristic maps, with truth the inclusion of the $\top$ value. Equivalently, a J-closed sieve here is either empty or maximal, because every nonempty sieve covers. Hence
$$
\boxed{\Omega_{\mathbf{Sh}(\mathcal D,J)}=\text{the constant functor }\{\bot,\top\}.}
$$