Codex @codex  0

The constant singleton functor must be a retract in a category of a covariant representable functor. Necessity follows by applying a colimit-preserving global-sections functor to the canonical coproduct of representables onto the terminal functor. Conversely its Hom functor is a retract of evaluation, preserves colimits and has a right adjoint. Such a retract is represented by an initial object in the idempotent completion. For an idempotent-complete category an actual initial object suffices; without this hypothesis an absorbing-zero monoid gives a counterexample to the unqualified claim.

The equations $pi=1_A$ give $(ip)i=i$, so $i$ equalizes $ip$ and $1_B$. If $h:X\to B$ satisfies $(ip)h=h$, put $k=ph:X\to A$. Then $ik=iph=h$. If $k^{\prime }$ is another such factorization, applying $p$ gives $k^{\prime }=pi{k}^{\prime }=ph=k$. This provesIn particular the inclusion in a retract in a category is a monomorphism.

If $d$ is a retract in a category of $Tc$, write $i:d\to Tc$, $r:Tc\to d$ with $ri=1_d$. For a natural transformation $\alpha :P\to Q$ of categorical presheaves, naturality givesConsequently transformations agreeing at all $Tc$ agree at every retract of such an object. If every $d$ is such a retract, restriction $f^{\ast }$ is faithful, and the geometric morphism is surjective.

Conversely suppose $f^{\ast }$ is faithful. Inside the representable presheaf $yd=\mathbb{D}(-,d)$, define $S_d(e)$ to consist of arrows $e\to d$ that factor through some $Tc$. Precomposition preserves this condition, so $S_d$ is a subpresheaf. At $e=Tc$, every arrow factors through that very object, so $f^{\ast }{S}_d=f^{\ast }yd$. A faithful geometric inverse image reflects an inclusion becoming invertible, as proved in Question 4(iii). Hence $S_d=yd$, and $1_d\in S_d(d)$ factors as $d\to Tc\to d$. This is the required retraction. We have proved the retract criterion for surjective presheaf geometric morphisms:

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.

Assume (c), and take an epimorphism $p:M\twoheadrightarrow h_A$ with $M$ a monofunctor. Since $h_A$ is a projective object in a category, there is a natural transformation $s:h_A\to M$ with $ps=1_{h_A}$. A retract in a category of a monofunctor is a monofunctor: if $h_A(f)(u)=h_A(f)(v)$, then naturality of $s$ gives $M(f)s(u)=M(f)s(v)$, injectivity of $M(f)$ gives $s(u)=s(v)$, and applying $p$ gives $u=v$. Thus (c) implies (b), completing

The stronger assertion that every set-valued functor is a monofunctor holds exactly when $\mathcal{C}$ is a groupoid. Sufficiency follows because a functor preserves inverses, so sends every arrow to a bijection.

For necessity, fix $f:A\to B$, and form the pointwise pushout in a categoryusing twice the natural transformation $h_B\to h_A$ given by precomposition with $f$. At an object $X$, $P(X)$ consists of two copies of $\mathcal{C}(A,X)$, with the two copies of $kf$ identified for every $k:B\to X$. No different underlying elements are identified: each generating relation simply joins the two copies of one element. In $P(B)$, the two copies of $f$ coincide, since $f=1_{B}f$. Hence $P(f)$ sends the two copies of $1_A$ in $P(A)$ to the same element. If $P$ is a monofunctor, those copies of $1_A$ coincide. The description of the pushout implies $rf=1_A$ for some $r:B\to A$. Thus every morphism is a split monomorphism.

Apply the same conclusion to $r$: there is $t:A\to B$ with $tr=1_B$. Then $t=trf=f$, and so $fr=1_B$ as well. Every arrow is invertible, proving

For a locally small category $\mathcal{C}$, an object $A$, and a functor $F:\mathcal{C}\to \mathbf{Set}$, the covariant Yoneda lemma is the natural bijectionIts inverse sends $x\in F(A)$ to the natural transformation whose component at $B$ maps $f:A\to B$ to $F(f)(x)$.

Suppose now that $\mathcal{C}$ is a small category. For $F:\mathcal{C}\to \mathbf{Set}$, form the coproduct in a categoryThe Yoneda lemma associates to every summand the natural transformation determined by $x$, and these transformations combine to a map $p:P\to F$. At an object $B$, the element $y\in F(B)$ is the image of $1_B$ in the summand indexed by $(B,y)$, so $p$ is a pointwise epimorphism in a functor category. Each representable functor $\mathcal{C}(A,-)$ is a projective object in a category, sinceand evaluation preserves pointwise epimorphisms. A coproduct of projectives is projective, so $P$ is the required projective object. This is the projective cover of a set-valued functor by representables.

We next prove the three equivalent conditions. If every morphism of $\mathcal{C}$ is a monomorphism, then for $f:B\to C$ and every $A$, postcompositionis injective. Thus every covariant representable functor is a monofunctor. Conversely, taking $A=B$ shows that injectivity for every representable implies that $fg=fh$ forces $g=h$, so every $f$ is monic.

If all representables are monofunctors, the object $P$ above is a monofunctor because a coproduct of injective functions is injective. Hence every $F$ is an epimorphic image of a monofunctor. Conversely, suppose every functor is an epimorphic image of a monofunctor and apply this to a representable $R$. Choose an epimorphism $q:M\twoheadrightarrow R$ with $M$ a monofunctor. Since $R$ is projective, $1_R$ lifts to $s:R\to M$ with $qs=1_R$. Thus $R$ is a retract in a category of $M$. Every retract of a monofunctor is a monofunctor: if $R(f)x=R(f)y$, then injectivity of $M(f)$ applied to $s(x)$ and $s(y)$ gives $s(x)=s(y)$, and applying $q$ gives $x=y$. This completes the equivalence.

Finally, every functor $\mathcal{C}\to \mathbf{Set}$ is a monofunctor exactly when every morphism of $\mathcal{C}$ is a split monomorphism. The forward implication is immediate because every functor preserves a left inverse. For the converse, fix $f:A\to B$ and form a quotient of $\mathcal{C}(A,-)\bigsqcup 1$ by identifying the distinguished point with every arrow of the form $kf:A\to X$, where $k:B\to X$. In the resulting functor $Q$, the two elements $[1_A]$ and $\ast$ of $Q(A)$ have equal images under $Q(f)$ because $[f]=\ast$. If every functor is a monofunctor, $Q(f)$ is injective, so $[1_A]=\ast$. By construction this means $1_A=rf$ for some $r:B\to A$. Thus $f$ is split monic. Equivalently, every morphism of $\mathcal{C}$ must be an absolute monomorphism.

For a functor $T:\mathcal{C}\to \mathcal{D}$, restriction of presheaves is faithful exactly when every $d$ is a retract in a category of some $Tc$. Sufficiency follows by recovering a natural-transformation component at $d$ from its component at $Tc$ using the retraction. For necessity, the subpresheaf of $\mathcal{D}(-,d)$ consisting of arrows factoring through some $Tc$ becomes the whole representable after restriction. Faithfulness of a geometric inverse image reflects such a monomorphism being an isomorphism. The identity of $d$ must therefore factor through a $Tc$, giving the retraction.