Lemma 4.17.5. Let $F : \mathcal{I} \to \mathcal{I}'$ be a functor. Assume

the fibre categories (see Definition 4.32.2) of $\mathcal{I}$ over $\mathcal{I}'$ are all connected, and

for every morphism $\alpha ' : x' \to y'$ in $\mathcal{I}'$ there exists a morphism $\alpha : x \to y$ in $\mathcal{I}$ such that $F(\alpha ) = \alpha '$.

Then for every diagram $M : \mathcal{I}' \to \mathcal{C}$ the colimit $\mathop{\mathrm{colim}}\nolimits _\mathcal {I} M \circ F$ exists if and only if $\mathop{\mathrm{colim}}\nolimits _{\mathcal{I}'} M$ exists and if so these colimits agree.

**Proof.**
One can prove this by showing that $\mathcal{I}$ is cofinal in $\mathcal{I}'$ and applying Lemma 4.17.2. But we can also prove it directly as follows. It suffices to show that for any object $T$ of $\mathcal{C}$ we have

\[ \mathop{\mathrm{lim}}\nolimits _{\mathcal{I}^{opp}} \mathop{\mathrm{Mor}}\nolimits _\mathcal {C}(M_{F(i)}, T) = \mathop{\mathrm{lim}}\nolimits _{(\mathcal{I}')^{opp}} \mathop{\mathrm{Mor}}\nolimits _\mathcal {C}(M_{i'}, T) \]

If $(g_{i'})_{i' \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{I}')}$ is an element of the right hand side, then setting $f_ i = g_{F(i)}$ we obtain an element $(f_ i)_{i \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{I})}$ of the left hand side. Conversely, let $(f_ i)_{i \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{I})}$ be an element of the left hand side. Note that on each (connected) fibre category $\mathcal{I}_{i'}$ the functor $M \circ F$ is constant with value $M_{i'}$. Hence the morphisms $f_ i$ for $i \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{I})$ with $F(i) = i'$ are all the same and determine a well defined morphism $g_{i'} : M_{i'} \to T$. By assumption (2) the collection $(g_{i'})_{i' \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{I}')}$ defines an element of the right hand side.
$\square$

## Comments (2)

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