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A cocone under a diagram $D:\mathcal{J}\to \mathcal{C}$ with vertex $X$ is a natural family ${\lambda }_j:Dj\to X$, so ${\lambda }_{k}D(u)={\lambda }_j$ for every $u:j\to k$. A colimit of $D$ is an initial such cocone: for every cocone $\lambda :D\Rightarrow \Delta X$, there is a unique $h:\mathrm{colim}D\to X$ with $h{\iota }_j={\lambda }_j$ for all $j$.

Let $F:\mathcal{I}\to \mathcal{J}$ be a final functor and let ${\lambda }_i:D(Fi)\to X$ be a cocone under $DF$. For each $j$, choose an object $(i,u:j\to Fi)$ of the nonempty comma category $(j\downarrow F)$ and defineA morphism $(i,u)\to (i^{\prime },u^{\prime })$ in the comma category gives $F(v)u=u^{\prime }$, and the cocone identity shows that the two resulting maps are equal. Because $(j\downarrow F)$ is a connected category, a zigzag proves independence of the chosen object. The same construction applied after a morphism $j\to k$ proves naturality, so $\bar{\lambda }$ is a cocone. Any extension must have this value because it must satisfy the cocone identity along $u$, proving uniqueness. This is cocone extension along a final functor.

If $L$ is a colimit of $DF$, its universal cocone extends uniquely to $D$. Restriction and extension give mutually inverse correspondences between cocones from $D$ and from $DF$, so the extended cocone is a colimit of $D$. Therefore the existence of all colimits of shape $\mathcal{I}$ implies the required colimits of shape $\mathcal{J}$.

Now suppose $\mathcal{J}$ is a sifted category. For $D,E:\mathcal{J}\to \mathbf{Set}$, the product functor $-\times A$ is a left adjoint, so it preserves colimits. Applying this once in each variable givesThe diagonal $\Delta :\mathcal{J}\to \mathcal{J}\times \mathcal{J}$ is final, so the right side isThus ${\mathrm{colim}}_{\mathcal{J}}$ preserves binary products in a category. Since $\mathcal{J}$ is connected, the colimit of the constant singleton diagram is a singleton, so it also preserves the terminal object. It therefore preserves finite products.

A diagram in a category is a functor $D:\mathcal{J}\to \mathcal{C}$. A cone over a diagram with vertex $A$ is a familysuch that $D(u){\gamma }_i={\gamma }_j$ for every $u:i\to j$. A categorical limit is a terminal cone: for every cone $(A,\gamma )$ there is a unique map $A\to \lim D$ commuting with all legs.

Suppose $\mathcal{C}$ has small products and equalizers. For a small diagram $D$, formThere are two maps $r,s:P\rightrightarrows Q$. In the coordinate indexed by $u:i\to j$, letThe equalizer $E\to P$ imposes exactly the cone equations. Maps $A\to E$ are therefore naturally the same as cones from $A$ to $D$, so $E\cong \lim D$. This is the construction of small limits from products and equalizers.

Let $F:\mathcal{I}\to \mathcal{J}$ be initial, so every $(F\downarrow j)$ is nonempty and connected. Restriction sends a cone $(A,{\gamma }_j)$ over $D$ to $(A,{\gamma }_{Fi})$ over $DF$. Conversely, given a cone $(A,{\delta }_i)$ over $DF$, choose an objectand defineA morphism in the comma category shows that this expression is unchanged along one edge, and connectedness makes it independent of the chosen object. The cone equations follow by choosing $(i,vu)$ for an arrow $v:j\to j^{\prime }$. This construction is inverse to restriction and acts identically on vertex maps, proving the cone restriction along an initial functor isomorphism.

Terminal objects in the two cone categories therefore correspond. Whenever the $\mathcal{I}$-shaped limit exists,naturally in $D$. Equivalently, the triangle formed by precompositionand the two limit functors commutes up to natural isomorphism.

For the converse, suppose this commutation holds for $\mathcal{C}={\mathbf{Set}}^{\mathrm{op}}$. Passing to opposite categories says that restriction along $F^{\mathrm{op}}$ preserves all set-valued colimits. Fix $j\in \mathcal{J}$ and take the representable functorIts colimit is a singleton: the category of its elements has the initial object $(j,1_j)$. The restricted colimit iswhose elements are precisely the connected components of $(F\downarrow j)$. By the assumed comparison this set is also a singleton. Thus $(F\downarrow j)$ is nonempty and connected for every $j$, so $F$ is initial.