Lemma 4.22.9. Let $\mathcal{C}$ be a category. Let $M : \mathcal{I} \to \mathcal{C}$ be a diagram with filtered index category $\mathcal{I}$. The following are equivalent

1. $M$ is an essentially constant ind-object, and

2. $X = \mathop{\mathrm{colim}}\nolimits _ i M_ i$ exists and for any $W$ in $\mathcal{C}$ the map

$\mathop{\mathrm{colim}}\nolimits _ i \mathop{\mathrm{Mor}}\nolimits _\mathcal {C}(W, M_ i) \longrightarrow \mathop{\mathrm{Mor}}\nolimits _\mathcal {C}(W, X)$

is bijective.

Proof. Assume (2) holds. Then $\text{id}_ X \in \mathop{\mathrm{Mor}}\nolimits _\mathcal {C}(X, X)$ comes from a morphism $X \to M_ i$ for some $i$, i.e., $X \to M_ i \to X$ is the identity. Then both maps

$\mathop{\mathrm{Mor}}\nolimits _\mathcal {C}(W, X) \longrightarrow \mathop{\mathrm{colim}}\nolimits _ i \mathop{\mathrm{Mor}}\nolimits _\mathcal {C}(W, M_ i) \longrightarrow \mathop{\mathrm{Mor}}\nolimits _\mathcal {C}(W, X)$

are bijective for all $W$ where the first one is induced by the morphism $X \to M_ i$ we found above, and the composition is the identity. This means that the composition

$\mathop{\mathrm{colim}}\nolimits _ i \mathop{\mathrm{Mor}}\nolimits _\mathcal {C}(W, M_ i) \longrightarrow \mathop{\mathrm{Mor}}\nolimits _\mathcal {C}(W, X) \longrightarrow \mathop{\mathrm{colim}}\nolimits _ i \mathop{\mathrm{Mor}}\nolimits _\mathcal {C}(W, M_ i)$

is the identity too. Setting $W = M_ j$ and starting with $\text{id}_{M_ j}$ in the colimit, we see that $M_ j \to X \to M_ i \to M_ k$ is equal to $M_ j \to M_ k$ for some $k$ large enough. This proves (1) holds. The proof of (1) $\Rightarrow$ (2) is omitted. $\square$

Comment #8282 by Et on

It seems aa though in (2) => (1) you don't actually use the colimit aasumption, right? That is to say it is enough to assume we have a cocone X for which the map in (2) is bijective.

The reason I'm asking is that in lemma 13.14.6, you spend the last part of the proof proving (in a purely formal manner) the colimit assumption on the object denoted C in that lemma, and it seems as though that may be unecessary.

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