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
\[ F(X) = \mathop{\mathrm{colim}}\nolimits F'(X_ i) \]
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
\[ \xymatrix{ X_ i \ar[r] \ar[d] & X \ar[d]^\alpha \\ Y_ j \ar[r] & Y } \]
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
\[ F'(X_ i) \to F'(X_{i'}) \xrightarrow {\beta _{i'}} F(Y) \]
is equal to $\beta _ i$. Thus we obtain a well defined arrow
\[ F(\alpha ) : F(X) = \mathop{\mathrm{colim}}\nolimits F(X_ i) \to F(Y) \]
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
\[ \mathop{\mathrm{colim}}\nolimits _\lambda F(X_\lambda ) \longrightarrow F(X) \]
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
\[ \xymatrix{ X_ i \ar[rr] \ar[rd] & & X_\lambda \ar[ld] \\ & X } \]
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
\[ F(X) = \mathop{\mathrm{colim}}\nolimits _ i F'(X_ i) \longrightarrow \mathop{\mathrm{colim}}\nolimits _\lambda F(X_\lambda ) \]
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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