Let $C$ be a small category. Let's say a presheaf $P\colon C^{\mathrm{op}}\to \mathsf{Set}$ has "full support" if $P(X)\neq \varnothing$ for all objects $X$. We could also say that $P$ is "inhabited everywhere".
Recall that an Ind-object on $C$ is a presheaf on $C$ which is a filtered colimit of representable presheaves. Equivalently, a presheaf is an Ind-object if and only if its category of elements is filtered (see Proposition 4.8 here).
Question: Is there a nice characterization of those categories $C$ which admit an Ind-object with full support?
One characterization that comes from unwinding the definition is that there is a filtered diagram $D$ in $C$ such that every object in $C$ has an arrow to some object in $D$. But I'm wondering if there is some more "concrete" condition, i.e. one that doesn't involve quantifying over arbitrary filtered diagrams.
An obvious sufficient condition is that $C$ itself is filtered, since then the colimit of all representable presheaves by all arrows between them is an Ind-object with full support.
An obvious necessary condition is that $C$ is directed, in the sense that for all objects $X$ and $Y$ in $C$, there exists an object $Z$ in $C$ and arrows $X\to Z$ and $Y\to Z$. This follows from directedness of the category of elements of a presheaf of full support.
But neither of these conditions answer the question fully: I know examples of non-filtered $C$ which admit Ind-objects of full support, and of $C$ which are directed but admit no Ind-objects of full support.
Thanks to Sridhar Ramesh for providing the following example of a directed category with no Ind-object with full support.
Let $S$ be an uncountable set, and consider the free category $\newcommand{\C}{\mathscr{C}}\C$ generated by the directed graph $ (\mathcal{P}^{\mathrm{fin}}(S),\subset)$. So the objects of $\C$ are finite subsets of $S$, an arrow $A\to B$ is a finite chain $(A_0,\dots,A_n)$ with $n\geq 0$, $A_0 = A$, $A_n = B$, and $A_i\subset A_{i+1}$ for all $i$ (note that $\subset$ means proper subset), and composition is concatenation of chains.
$\C$ is directed, since for any objects $A$ and $B$, we have arrows $(A,A\cup B)$ and $(B,A\cup B)$.
Observe that if we have a commutative square in $\C$: $$\require{AMScd} \begin{CD} A @>>> B\\ @VVV @VVV \\ C @>>> D \end{CD}$$ then the composite arrow $A\to D$ is a chain containing both $B$ and $C$, so either $B\subseteq C$ or $C\subseteq B$.
Now suppose that $\newcommand{\D}{\mathscr{D}}\D\colon F\to \C$ is a filtered diagram, i.e. a functor from a filtered category $F$ to $\C$.
Let $x\in F$, and write ${\uparrow}(x)$ for the set of objects in $F$ which admit an arrow from $x$. Given $y,z\in {\uparrow}(x)$, since $F$ is filtered, there is a commutative square in $F$: $$\require{AMScd} \begin{CD} x @>>> y\\ @VVV @VVV \\ z @>>> w \end{CD}$$ It follows that $\D(y)$ and $\D(z)$ are comparable under $\subseteq$. So $\D({\uparrow}(x))$ is linearly ordered by $\subseteq$. Now distinct elements of $\D({\uparrow}(x))$ must have different finite cardinalities, so $\D({\uparrow}(x))$ is countable, and $\bigcup \D({\uparrow}(x))$ is a countable subset of $S$. If $s\in S\setminus \bigcup \D({\uparrow}(x))$, then $\{s\}$ does not admit an arrow into any object in $\D({\uparrow}(x))$. Since $F$ is filtered, it follows that $\{s\}$ does not admit an arrow into any object in $\D(F)$. Thus the Ind-object corresponding to $\D$ does not have full support.
A similar argument shows that the free category generated by the directed graph $(\omega_1,<)$ is directed but admits no Ind-object with full support.