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For the covariant category in this part, put $\mathcal{E}=[\mathcal{C},\mathbf{Set}]$ and $r_c=\mathcal{C}(c,-)$. Its global-sections functor is $\Gamma =\mathrm{Hom}(1,-)$, the limit of the diagram. If $\Gamma$ is also an inverse image, it is a left adjoint and preserves all small colimits.

Consider the canonical pointwise-surjective mapAt $d$, the summand $r_d(d)$ contains $1_d$, so $q$ is an epimorphism. In a presheaf topos epimorphisms are regular, and a colimit-preserving functor preserves their coequalizer presentations. Thus $\Gamma q$ is surjective. Since $\Gamma$ preserves coproducts, some summand has a global element, yielding a section $s:1\to r_c$ of its unique map $r_c\to 1$. The constant singleton functor is a retract of a covariant representable.

Conversely, if $1$ is such a retract in a category, $\mathrm{Hom}(1,-)$ is a retract of $\mathrm{Hom}(r_c,-)$, which is evaluation at $c$ by the Yoneda lemma. Evaluation preserves colimits pointwise. The colimit comparison for a retract functor is a retract of the evaluation comparison; a retract of an isomorphism is an isomorphism, so $\Gamma$ preserves all small colimits. It also preserves finite limits because it is representable. A right adjoint can be constructed byFor $h:d\to d^{\prime }$, precomposition gives $r_{d^{\prime }}\to r_d$, and applying $\Gamma$ and then $\mathrm{Set}(-,S)$ gives $R(S)(d)\to R(S)(d^{\prime })$. The presentation of every functor as a colimit of representables gives $\mathrm{Hom}(F,R(S))\cong \mathrm{Set}(\Gamma F,S)$. Hence $\Gamma ⊣R$, so $\Gamma$ is an inverse image and the topos is local.

The retract condition has a precise interpretation as the initial object criterion for a covariant local topos. Write the section as a natural family $u_d:c\to d$. Naturality says $h{u}_d=u_{d^{\prime }}$ for every $h:d\to d^{\prime }$. Put $e=u_c$. Then $e^2=e$ and every $f:c\to d$ satisfies $fe=u_d$. In the idempotent completion of $\mathcal{C}$, the object $(c,e)$ is initial: an arrow from it to $(d,k)$ must satisfy $v=kve$, and the identities force $v=u_d$; this arrow exists because $k{u}_d=u_d$ and $u_{d}e=u_d$.

Conversely an initial object $(c,e)$ in the idempotent completion gives that natural family and the retract. ThereforeIf $\mathcal{C}$ is already idempotent-complete, this is equivalent to an initial object in $\mathcal{C}$ itself, and $\Gamma$ is evaluation there. The word is initial, not terminal, because this question uses covariant functors.

One cannot omit idempotent completion for a general small category. Take the one-object category of the monoid $\{1,0\}$ with absorbing zero. It has no initial object because its endomorphism set has two elements, but the natural family $u=0$ satisfies $m0=0$ for every $m$. Its covariant functor category is local. Concretely, a functor is a set with an idempotent operator, and its global sections are the fixed points of that operator.

For a topological space $X$, there is an equally explicit criterion. If $\mathbf{Sh}(X)$ is local, apply $\Gamma$ to the epimorphism ${\coprod }_i{U}_i\to 1$ associated with any open cover $X={\bigcup }_i{U}_i$, regarding the opens as subterminal sheaves. Since $\Gamma U_i$ is a singleton exactly when $U_i=X$, and otherwise empty, coproduct and epimorphism preservation force some $U_i=X$.

Thus every open cover of $X$ contains $X$ itself. Conversely this open cover criterion for a local sheaf topos condition is equivalent to existence of a point $x$ whose only open neighborhood is $X$: the union of all proper open sets cannot be all of $X$, so choose $x$ outside it. Such a point plainly forces any cover to contain $X$. Empty spaces fail the condition.

At this point, the stalk functor $F\mapsto F_x$ is simply $F(X)=\Gamma F$, since there is only one open neighborhood. Its right adjoint is the skyscraper sheaf of setsA morphism $F\to R(S)$ is determined exactly by its map $F(X)\to S$, because all proper opens omit $x$. Hence $\Gamma ⊣R$, and $\Gamma$ preserves finite limits, proving locality. ConsequentlyIn a $T_0$ space this distinguished point is unique and closed. In a $T_1$ space the condition forces $X$ to be a singleton. Indistinguishable points in a non-$T_0$ space can all have this property; uniqueness of an actual point is not part of the general criterion.