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The Yoneda lemma states that for $A\in \mathcal{C}$ and $P:\mathcal{C}\to \mathbf{Set}$ there is a natural bijectionIf $P\twoheadrightarrow Q$ is an epimorphism in the functor category, it is pointwise surjective, so $P(A)\to Q(A)$ is surjective. Yoneda identifies this map withThus every representable functor is a projective object in a category.

The colimit form of the Special adjoint functor theorem says that a colimit-preserving functor from a locally small, cocomplete, well-copowered category with a small generating family into a locally small category has a right adjoint functor. For small $\mathcal{C}$, the category $[\mathcal{C},\mathbf{Set}]$ is locally small and has colimits pointwise. Quotients of $P$ are represented by compatible equivalence relations on the sets $P(C)$, so they form a set; hence the category is well-copowered. The set of representables $\{\mathcal{C}(C,-):C\in \mathcal{C}\}$ generates it by the Yoneda lemma. The theorem therefore gives a right adjoint to every small-colimit-preserving functorIn particular, product with a fixed functor $P$ is computed pointwise, and $-\times P(C)$ preserves colimits in the Category of sets. Hence $-\times P$ preserves all small colimits and has a right adjoint $(-{)}^P$. Thus $[\mathcal{C},\mathbf{Set}]$ is a cartesian closed category.

Now work in $[{\mathcal{C}}^{\mathrm{op}},\mathbf{Set}]$ and write $yA=\mathcal{C}(-,A)$. If $\mathcal{C}$ has binary products, thenThus exponentiation by $yA$ is precomposition with $-\times A$. Precomposition between functor categories has a right adjoint given by Right Kan extension, so $yA$ is a tiny object.

Conversely, suppose $\mathcal{C}$ has a terminal object $1$ and $F$ is tiny. The representable $y1$ is the terminal presheaf, and the exponential adjunction plus Yoneda givesSince $F$ is tiny, $(-{)}^F$ is a left adjoint and preserves all colimits; evaluation at $1$ also preserves pointwise colimits. Therefore the hom functor $[{\mathcal{C}}^{\mathrm{op}},\mathbf{Set}](F,-)$ preserves coproducts and epimorphisms. Preservation of epimorphisms makes $F$ projective, while preservation of coproducts makes it indecomposable.

Solved by gpt-5.6-sol high.

For $F:\mathcal{C}\rightleftarrows \mathcal{D}:G$ with $F⊣G$, the unit and counit of an adjunction areobtained by transposing identity morphisms. They satisfy the triangular identities ${\epsilon }_{F}F\eta =1_F$ and $G\epsilon {\eta }_G=1_G$.

If $G$ is full and faithful, fullness gives a map $u:B\to FGB$ with $G(u)={\eta }_{GB}$; the second triangular identity and faithfulness show that $u$ is inverse to ${\epsilon }_B$. Conversely, if $\epsilon$ is an isomorphism, then for $h:GB\to G{B}^{\prime }$ the unique arrow whose transpose is $h$ iswhich proves that $G$ is full, and the triangular identities prove faithfulness. Hence (i) and (ii) are equivalent. Condition (ii) immediately gives (iii). Conversely, a natural isomorphism $FG\cong 1_{\mathcal{D}}$ combines with the adjunction bijection to show naturally that $\mathcal{D}(B,B^{\prime })\to \mathcal{C}(GB,G{B}^{\prime })$ is bijective; by naturality and the triangular identities this is the map induced by $G$ up to invertible natural conjugation, so $G$ is full and faithful. Thus all three conditions are equivalent.

Suppose $L⊣F⊣R$. If $L$ is full and faithful, the unit $1\to FL$ is an isomorphism. For $A,B$ in the codomain of $F$, the two adjunctions then give natural bijectionsBy the Yoneda lemma, the counit $FRB\to B$ is an isomorphism, so the fully faithful adjoint criterion makes $R$ full and faithful. The converse is dual.

Now assume $F$ is full and faithful. For $L⊣F$, its counit $ϵ:LF\to 1$ is invertible; for $F⊣R$, its unit $\alpha :1\to RF$ is invertible. DefineApplying $F$, then using naturality and the triangular identities, reduces this composite to the same map asSince $F$ is faithful, the two displayed composites are equal.

For a morphism $u:FB\to A$, let $\hat{u}=Ru{\alpha }_B:B\to RA$ be its transpose under $F⊣R$. The identities just proved giveIf every ${\theta }_A$ is monic and $Lu=Lv$, this equation gives $\hat{u}=\hat{v}$, hence $u=v$; thus $L$ is faithful on arrows whose domains are in the image of $F$. Conversely, if ${\theta }_{A}f={\theta }_{A}g$ for $f,g:B\to RA$, naturality of $ϵ$ and the second formula for $\theta$ giveBoth maps have domain $FB$, so the assumed faithfulness gives ${\beta }_{A}Ff={\beta }_{A}Fg$. These are the transposes of $f$ and $g$, hence $f=g$. Therefore $\theta$ is pointwise monic exactly under the stated faithfulness condition.

Solved by gpt-5.6-sol high.

The Yoneda lemma and the definition of the nerve of a category giveThus such a map is precisely a diagram of objects and composable morphismsin $\mathcal{C}$. The remaining edges and higher faces record the composites forced by this string.

Solved by gpt-5.6-sol high.