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First address the unnumbered preliminaries. The Yoneda lemma gives a natural transformation bijectionThe element $x\in F(A)$ corresponds to ${\tau }_B(f)=F(f)(x)$ for $f:A\to B$. These bijections are natural in $A$ and $F$; the contravariant form replaces $\mathcal{C}(A,-)$ by $\mathcal{C}(-,A)$.

Let $E=(1\downarrow F)$ be the category of elements. Its objects are $(A,x)$ with $x\in F(A)$; an arrow $u:(A,x)\to (B,y)$ is $u:A\to B$ with $F(u)(x)=y$. Since $\mathcal{C}$ is a small category, $E$ is small. Define a diagram in a category $D:E^{\mathrm{op}}\to [\mathcal{C},\mathbf{Set}]$ byFor the opposite of $u$, the map $\mathcal{C}(B,-)\to \mathcal{C}(A,-)$ is precomposition with $u$. The Yoneda lemma gives a cocone under a diagram $D\to F$ whose $(A,x)$ component sends $f:A\to B$ to $F(f)(x)$.

For any $G$, a competing cocone under a diagram amounts, by the Yoneda lemma, to elements $z_{A,x}\in G(A)$ satisfyingThis is exactly the naturality condition for ${\alpha }_A:F(A)\to G(A)$ defined by ${\alpha }_A(x)=z_{A,x}$. It gives a unique natural transformation $\alpha :F\to G$ factoring the cocone. Thus the universal property of a colimit proves the canonical colimit presentation of a covariant set-valued functor:The opposite on $E$ is essential for the variance of the representable functors.

We now prove the four conditions equivalent by the cycle $(i)\Rightarrow (ii)\Rightarrow (iii)\Rightarrow (iv)\Rightarrow (i)$. Suppose $F$ is a finite-limit-preserving set-valued functor. For a finite diagram in a category in the comma category $(S\downarrow F)$, write its objects as $(A_i,s_i:S\to F(A_i))$. Take $L={\lim }_i{A}_i$ in $\mathcal{C}$. Since $F$ preserves this finite limit, the compatible maps $s_i$ determine a unique $s:S\to F(L)$ whose composites with $F(L)\to F(A_i)$ are $s_i$. The pair $(L,s)$ has the required categorical limit property in $(S\downarrow F)$, by the same universal properties. For the empty diagram, $F(1)$ is a singleton, so there is exactly one map $S\to F(1)$; this provides the terminal object. Hence $(i)\Rightarrow (ii)$ for every set $S$.