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For an initial functor $F:\mathcal{I}\to \mathcal{J}$, restriction gives an isomorphism between the category of cones over $D:\mathcal{J}\to \mathcal{C}$ and that over $DF$. Given a cone over $DF$, choose $(i,u:Fi\to j)$ and define its $j$-leg as $D(u){\gamma }_i$; connectedness of $(F\downarrow j)$ makes the result independent of the choice.

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.