Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 20 / 4

Write $\hat{\mathcal{C}}=[{\mathcal{C}}^{\mathrm{op}},\mathbf{Set}]$ and similarly for $\mathcal{D}$. The geometric morphism induced by a functor hasPrecomposition preserves all pointwise limits and colimits, so in particular it preserves finite limits. The Right Kan extension exists because the categories are small and sets have all small limits; its universal property gives $f^{\ast }⊣f_{\ast }$. Thus these functors define a geometric morphism $\hat{\mathcal{C}}\to \hat{\mathcal{D}}$.

There is also $f_{!}={\mathrm{Lan}}_{F^{\mathrm{op}}}$, a left Kan extension, with $f_{!}⊣f^{\ast }$. The Yoneda lemma identifies $f_{!}(yC)\cong y(FC)$, since for every $P$,This is the representable calculation used in the next parts.

A representable functor $yC$ is an indecomposable projective object. Given an epimorphism ${\coprod }_i{B}_i\twoheadrightarrow yC$, evaluate at $C$. Epimorphisms and coproducts in a presheaf category are pointwise, so $1_C\in yC(C)$ is the image of some element of a particular $B_i(C)$. By the Yoneda lemma, that element defines $s:yC\to B_i$, and its composite into $yC$ corresponds to $1_C$, hence is the identity. The selected component is split epic. More generally, evaluation sends any epimorphism to a surjection, so a map from $yC$ lifts through any epimorphism; this also proves its ordinary projectivity.

Conversely, every presheaf $A$ has the canonical epimorphismwhose component is the natural transformation named by $x$. It is pointwise surjective, since an element at $D$ is reached from its own summand $(D,x)$ at $1_D$. If $A$ is indecomposable projective, one component $r:yC\to A$ has a section $s:A\to yC$. The endomorphism $sr$ of $yC$ is idempotent and therefore corresponds to an idempotent morphism $e:C\to C$.

If idempotents split in $\mathcal{C}$, choose $C\stackrel{p}{\to }D\stackrel{i}{\to }C$ with $ip=e$, $pi=1_D$. Then $A\cong yD$: the mutually inverse maps are $y(p)s:A\to yD$ and $ry(i):yD\to A$. ThusWithout that hypothesis the argument still proves that every such object is a retract of a representable. The initial presheaf is not indecomposable projective, since its identity is the empty-coproduct epimorphism and has no component to select.

The key fact is that the extra left adjoint $f_{!}$ sends each representable $yC$ to an indecomposable projective object. Let ${\coprod }_i{B}_i\twoheadrightarrow f_{!}yC$ be epic. The inverse image $f^{\ast }$ preserves epimorphisms and coproducts, because it is a left adjoint between toposes. Apply it and lift the unit $\eta :yC\to f^{\ast }{f}_{!}yC$ through the resulting epimorphism, using projectivity of $yC$. A map from $yC$ to a coproduct selects one component, by evaluation at $C$ and the Yoneda lemma. Thus for some $i$ we obtain $t:yC\to f^{\ast }{B}_i$ with $f^{\ast }(B_i\to f_{!}yC)t=\eta$.

Transpose $t$ across $f_{!}⊣f^{\ast }$ to $\bar{t}:f_{!}yC\to B_i$. The displayed equality says that its composite back to $f_{!}yC$ is the identity. This proves the required indecomposable-projective property.

Since idempotents split in $\mathcal{D}$, part (ii) supplies objects $FC\in \mathcal{D}$ and isomorphisms $f_{!}yC\cong y(FC)$. Full faithfulness of the Yoneda embedding transports the action of $f_{!}$ on representable arrows to a functor $F:\mathcal{C}\to \mathcal{D}$. For $P\in \hat{\mathcal{D}}$,These identifications are natural in both $C$ and $P$. HenceIts right adjoint is consequently the right Kan extension from part (i), uniquely up to natural isomorphism. Thus the entire geometric morphism is induced by $F$.

The canonical geometric morphism $\hat{\mathcal{C}}\to \mathbf{Set}$ has inverse image the constant-presheaf functor $\Delta$ and direct image the global sections functorIf the presheaf topos is a local topos, $\Gamma$ is also the inverse image of a geometric morphism $g:\mathbf{Set}\to \hat{\mathcal{C}}$. This morphism has an extra left adjoint $\Delta$. Apply part (iii) with source category $1$ and target category $\mathcal{C}$, using its idempotent-splitting hypothesis. Then $g$ is induced by a functor $1\to \mathcal{C}$, choosing an object $C_0$, and $\Gamma$ is naturally evaluation at $C_0$.

Since evaluation at $C_0$ is $\mathrm{Hom}(y{C}_0,-)$, this says $\mathrm{Hom}(1,-)\cong \mathrm{Hom}(y{C}_0,-)$. Uniqueness of representing objects gives $y{C}_0\cong 1$. Thus $\mathcal{C}(C,C_0)$ is a singleton for every $C$: $C_0$ is terminal.

Conversely, if $C_0$ is terminal, $y{C}_0=1$ and $\Gamma (P)=P(C_0)$. Evaluation at that object preserves finite limits and has a right Kan extension as right adjoint, so is an inverse image functor. Therefore, under the permitted idempotent-completeness assumption,