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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.

A categorical limit of $D:\mathcal{J}\to \mathcal{C}$ is a terminal cone: it consists of an object $L$ and compatible maps $L\to D(j)$ through which every other cone factors uniquely. For a finite diagram, take the product $P={\prod }_{j}D(j)$ and the product $Q={\prod }_{u:i\to j}D(j)$. There are two maps $P\rightrightarrows Q$: on the $u:i\to j$ coordinate one uses respectively $D(u)$ after projection to $D(i)$ and direct projection to $D(j)$. Their equalizer is exactly the compatible-cone object. Hence finite products and equalizers construct every finite limit.

In the category of metric spaces and non-expansive maps, give $X\times Y$ the maximum metricThe projections are non-expansive, and a pair of non-expansive maps into $X$ and $Y$ induces a non-expansive map into this product, proving the universal property.

Let two maps $1\rightrightarrows X$ select $a,b\in X$. On the set quotient identifying $a$ and $b$, define the quotient metric by shortest paths that may jump from $a$ to $b$ at zero cost. Explicitly,This is the largest metric making the quotient map non-expansive. A map $X\to Z$ equalizing $a$ and $b$ factors through the set quotient and remains non-expansive by the path formula, so this is the coequalizer. Its underlying set is the set-theoretic coequalizer.

For $X=\{0,1,5,6\}$ with $0\sim 6$, write $\ast =\{0,6\}$. The quotient hasThus in $Q\times Y$, with $Y=\{0,3\}$,After first taking $X\times Y$, the product of the parallel pair identifies $(0,y)$ with $(6,y)$ separately for $y=0,3$. In its quotient metric, every path from $(1,0)$ to $(5,3)$ has length at least $4$, and length $4$ is attained either directly or via one of those identifications. The canonical bijection from this coequalizer to $Q\times Y$ is therefore not an isometry. Hence $-\times Y$ does not preserve this coequalizer. In a cartesian closed category, $-\times Y$ is a left adjoint and preserves all colimits, so $\mathbf{Met}$ is not cartesian closed.

Solved by gpt-5.6-sol high.

An object $A$ of a cartesian closed category is tiny when exponentiation $(-{)}^A$ itself has a right adjoint.