Lemma 4.18.1. Let $\mathcal{I}$ be a category with

1. $\mathop{\mathrm{Ob}}\nolimits (\mathcal{I})$ is finite, and

2. there exist finitely many morphisms $f_1, \ldots , f_ m \in \text{Arrows}(\mathcal{I})$ such that every morphism of $\mathcal{I}$ is a composition $f_{j_1} \circ f_{j_2} \circ \ldots \circ f_{j_ k}$.

Then there exists a functor $F : \mathcal{J} \to \mathcal{I}$ such that

1. $\mathcal{J}$ is a finite category, and

2. for any diagram $M : \mathcal{I} \to \mathcal{C}$ the (co)limit of $M$ over $\mathcal{I}$ exists if and only if the (co)limit of $M \circ F$ over $\mathcal{J}$ exists and in this case the (co)limits are canonically isomorphic.

Moreover, $\mathcal{J}$ is connected (resp. nonempty) if and only if $\mathcal{I}$ is so.

Proof. Say $\mathop{\mathrm{Ob}}\nolimits (\mathcal{I}) = \{ x_1, \ldots , x_ n\}$. Denote $s, t : \{ 1, \ldots , m\} \to \{ 1, \ldots , n\}$ the functions such that $f_ j : x_{s(j)} \to x_{t(j)}$. We set $\mathop{\mathrm{Ob}}\nolimits (\mathcal{J}) = \{ y_1, \ldots , y_ n, z_1, \ldots , z_ n\}$ Besides the identity morphisms we introduce morphisms $g_ j : y_{s(j)} \to z_{t(j)}$, $j = 1, \ldots , m$ and morphisms $h_ i : y_ i \to z_ i$, $i = 1, \ldots , n$. Since all of the nonidentity morphisms in $\mathcal{J}$ go from a $y$ to a $z$ there are no compositions to define and no associativities to check. Set $F(y_ i) = F(z_ i) = x_ i$. Set $F(g_ j) = f_ j$ and $F(h_ i) = \text{id}_{x_ i}$. It is clear that $F$ is a functor. It is clear that $\mathcal{J}$ is finite. It is clear that $\mathcal{J}$ is connected, resp. nonempty if and only if $\mathcal{I}$ is so.

Let $M : \mathcal{I} \to \mathcal{C}$ be a diagram. Consider an object $W$ of $\mathcal{C}$ and morphisms $q_ i : W \to M(x_ i)$ as in Definition 4.14.1. Then by taking $q_ i : W \to M(F(y_ i)) = M(F(z_ i)) = M(x_ i)$ we obtain a family of maps as in Definition 4.14.1 for the diagram $M \circ F$. Conversely, suppose we are given maps $qy_ i : W \to M(F(y_ i))$ and $qz_ i : W \to M(F(z_ i))$ as in Definition 4.14.1 for the diagram $M \circ F$. Since

$M(F(h_ i)) = \text{id} : M(F(y_ i)) = M(x_ i) \longrightarrow M(x_ i) = M(F(z_ i))$

we conclude that $qy_ i = qz_ i$ for all $i$. Set $q_ i$ equal to this common value. The compatibility of $q_{s(j)} = qy_{s(j)}$ and $q_{t(j)} = qz_{t(j)}$ with the morphism $M(f_ j)$ guarantees that the family $q_ i$ is compatible with all morphisms in $\mathcal{I}$ as by assumption every such morphism is a composition of the morphisms $f_ j$. Thus we have found a canonical bijection

$\mathop{\mathrm{lim}}\nolimits _{B \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{J})} \mathop{\mathrm{Mor}}\nolimits _\mathcal {C}(W, M(F(B))) = \mathop{\mathrm{lim}}\nolimits _{A \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{I})} \mathop{\mathrm{Mor}}\nolimits _\mathcal {C}(W, M(A))$

which implies the statement on limits in the lemma. The statement on colimits is proved in the same way (proof omitted). $\square$

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