Definition 4.26.1. Let \mathcal{C} be a big1 category. An object X of \mathcal{C} is called a categorically compact if we have
for every filtered diagram M : \mathcal{I} \to \mathcal{C} such that \mathop{\mathrm{colim}}\nolimits _ i M_ i exists.
A little bit about “small” objects of a category.
Definition 4.26.1. Let \mathcal{C} be a big1 category. An object X of \mathcal{C} is called a categorically compact if we have
for every filtered diagram M : \mathcal{I} \to \mathcal{C} such that \mathop{\mathrm{colim}}\nolimits _ i M_ i exists.
Often this definition is made only under the assumption that \mathcal{C} has all filtered colimits.
Lemma 4.26.2. Let \mathcal{C} and \mathcal{D} be big categories having filtered colimits. Let \mathcal{C}' \subset \mathcal{C} be a small full subcategory consisting of categorically compact objects of \mathcal{C} such that every object of \mathcal{C} is a filtered colimit of objects of \mathcal{C}'. Then every functor F' : \mathcal{C}' \to \mathcal{D} has a unique extension F : \mathcal{C} \to \mathcal{D} commuting with filtered colimits.
Proof. For every object X of \mathcal{C} we may write X as a filtered colimit X = \mathop{\mathrm{colim}}\nolimits X_ i with X_ i \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C}'). Then we set
in \mathcal{D}. We will show below that this construction does not depend on the choice of the colimit presentation of X.
Suppose given a morphism \alpha : X \to Y of \mathcal{C} and X = \mathop{\mathrm{colim}}\nolimits _{i \in I} X_ i and Y = \mathop{\mathrm{colim}}\nolimits _{j \in J} Y_ i are written as filtered colimit of objects in \mathcal{C}'. For each i \in I since X_ i is a categorically compact object of \mathcal{C} we can find a j \in J and a commutative diagram
Then we obtain a morphism F'(X_ i) \to F'(Y_ j) \to F(Y) where the second morphism is the coprojection into F(Y) = \mathop{\mathrm{colim}}\nolimits F'(Y_ j). The arrow \beta _ i : F'(X_ i) \to F(Y) does not depend on the choice of j. For i \leq i' the composition
is equal to \beta _ i. Thus we obtain a well defined arrow
by the universal property of the colimit. If \alpha ' : Y \to Z is a second morphism of \mathcal{C} and Z = \mathop{\mathrm{colim}}\nolimits Z_ k is also written as filtered colimit of objects in \mathcal{C}', then it is a pleasant exercise to show that the induced morphisms F(\alpha ) : F(X) \to F(Y) and F(\alpha ') : F(Y) \to F(Z) compose to the morphism F(\alpha ' \circ \alpha ). Details omitted.
In particular, if we are given two presentations X = \mathop{\mathrm{colim}}\nolimits X_ i and X = \mathop{\mathrm{colim}}\nolimits X'_{i'} as filtered colimits of systems in \mathcal{C}', then we get mutually inverse arrows \mathop{\mathrm{colim}}\nolimits F'(X_ i) \to \mathop{\mathrm{colim}}\nolimits F'(X'_{i'}) and \mathop{\mathrm{colim}}\nolimits F'(X'_{i'}) \to \mathop{\mathrm{colim}}\nolimits F'(X_ i). In other words, the value F(X) is well defined independent of the choice of the presentation of X as a filtered colimit of objects of \mathcal{C}'. Together with the functoriality of F discussed in the previous paragraph, we find that F is a functor. Also, it is clear that F(X) = F'(X) if X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C}').
The uniqueness statement in the lemma is clear, provided we show that F commutes with filtered colimits (because this statement doesn't make sense otherwise). To show this, suppose that X = \mathop{\mathrm{colim}}\nolimits _{\lambda \in \Lambda } X_\lambda is a filtered colimit of \mathcal{C}. Since F is a functor we certainly get a map
On the other hand, write X = \mathop{\mathrm{colim}}\nolimits X_ i as a filtered colimit of objects of \mathcal{C}'. As above, for each i \in I we can choose a \lambda \in \Lambda and a commutative diagram
As above this determines a well defined morphism F'(X_ i) \to \mathop{\mathrm{colim}}\nolimits _\lambda F(X_\lambda ) compatible with transition morphisms and hence a morphism
This morphism is inverse to the morphism above (details omitted) and proves that F(X) = \mathop{\mathrm{colim}}\nolimits _\lambda F(X_\lambda ) as desired. \square
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