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. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)