I am in the following situation: I have a category $A$ and an increasing chain of full subcategories $A\_0\subseteq A\_1\subseteq\, ...\subseteq A\_n\subseteq \, ...$ inside $A$, such that any object of $A$ belongs to some $A\_n$. Roughly speaking, $A$ is the union of the $A\_n$. 


I would like to construct a functor from $A$ to a category $B$, and what I can do is the following: for any $n\in \mathbb N$ I can construct a functor $F\_n\colon A\_n\to B$ and prove that the restriction of $F\_{n+1}$ to $A\_n$ is naturally isomorphic to $F\_n$.


The question is, can we use these functors $F\_n$ to construct a functor $F\colon A\to B$ in some canonical way? Of course one basic requirement would be that $F(a)\cong F\_n(a)$ if $a\in A\_n$.


I think I can give a construction, but it is quite tedious and definitely it is not nice to give this construction in the context I am writing. So, is there any reference I can quote? Or, do you have any one-line argument for the existence of $F$?



Piecewise definition of a functor
Basically you want to prove that $A = \mathrm{colim}\_n A\_n$ in the $2$-categorical sense. There is a general construction of $2$-colimits; we may use this and check that it is equivalent to $A$. But I think it is a good idea to just wrote down the functor.


Let us denote by $\theta\_n : F\_n \to F\_{n+1}|{A\_n}$ the given isomorphisms. These may be composed to compatible isomorphisms $\theta\_{n,m} : F\_n \to F\_m |{A\_n}$ for $n \leq m$.


Define $t : \mathrm{Ob}(A) \to \mathbb{N}$ by $t(a) = \min\{n \in \mathbb{N} : a \in A\_n\}$. Define $t(a,b) := \max(t(a),t(b))$. The functor $F : A \to B$ is defined as follows: An object $a \in \mathrm{Ob}(A)$ is mapped to $F(a):=F\_{t(a)}(a)$. A morphism $f:a \to b$ is mapped to the composition
$$F(f) : \quad F\_{t(a)}(a) \xrightarrow{~\large \theta\_{t(a),t(a,b)}(a)~} F\_{t(a,b)}(a) \xrightarrow{~\large F\_{ t(a,b)}(f)~} F\_{t(a,b)}(b) \xrightarrow{~\large \theta\_{ t(a),t(a,b)}^{-1}(b)~} F\_{t(b)}(b).$$
It is easy to check that $F$ is, in fact, a functor. If $a \in A\_n$, then $t(a) \leq n$. Hence $\theta\_{t(a),n} : F(a)=F\_{t(a)}(a) \to F\_n(a)$ is an isomorphism. It is natural in $a$; this follows from the construction of $F(f)$. Hence, there is an isomorphism $\vartheta\_n : F|{A\_n} \cong F\_n$. Moreover, these isomorphisms are compatible with the $\theta\_n$: The diagram


$$\begin{array}{cc} F|A\_n & \xrightarrow{\large \vartheta\_n} & F\_n\\ \mathrm{id}\downarrow ~~ && ~~\downarrow \theta\_n \\ F|A\_{n+1}|A\_n & \xrightarrow{\large \vartheta\_{n+1}|A\_n} & F\_{n+1}|A\_n\end{array}$$


commutes.


