Let $D \colon \mathbf{J} \to \mathbf{Cat}$ be a filtered diagram of categories and functors. It has a colimit $\mathbf{C} = \mathrm{colim}\;D$. If you replace the diagram by a naturally isomorphic one $D' \colon \mathbf{J} \to \mathbf{Cat}$, then the colimit
$\mathbf{C'} = \mathrm{colim}\;D'$ is isomorphic to $\mathbf{C}$. What happens if you replace the diagram by a naturally *equivalent* one $D'' \colon \mathbf{J} \to \mathbf{Cat}$, is the colimit $\mathbf{C''} = \mathrm{colim}\;D''$ still equivalent to $\mathbf{C}$?


(I'm interested in $\mathbf{Cat}$, or in fact the category $\mathbf{MonCat}$ of monoidal categories, but suspect that there is a general statement in weak 2-categories. Any references are appreciated.)



Are equivalences of categories stable under filtered colimits?
For filtered diagram (as asked in the question) the answer is **yes**. Of course this fails for general diagram as mentioned in Harry's answer.


Of course the "equivalence" has to be implemented by a pseudo-natural equivalence $f\_i:F\_i \to F'\_i$ otherwise it is not really an equivalence of diagram.


**First a very concrete proof:** one constructs "by hand" a fully faithful and essentially surjective functor:


For each object $x \in \text{colim }F\_i$, one chose a representaive $(i,x\_0 \in F\_i)$ and one define $f(x)$ to be $(i,f\_i(x\_0)) \in \text{colim }F'\_i $.


One then define $f$ on arrows:


given $t:x \to y$ in $\text{colim }F\_i$, $t$ one chose $i$ such that $t$ is defined at level $F\_i$, and which is larger than the $i\_x$ and $i\_y$ used to define $f(x)$ and $f(y)$, ideally, we would like to define $f(t)$ as $f\_i(t)$. But the source and target are not quite right $f\_i(t) : f\_i(x) \to f\_i(y)$, while we want a map $f\_{i\_x}(x) \to f\_{i\_y}(y)$. The trick is that the pseudo-natrulity of $(f\_i)$ gives us canonical isomorphism in $F'\_i$, $f\_{i\_x}(x) \simeq f\_i(x)$ here I'm using an abuse of notation, the $x$ on the left is in $F\_{i\_x}$ and on the right it is in $F\_i$, this iso is really just the pseudo-naturality). We define $f(t)$ as the transport of $f\_i(t)$ along these isomorphisms.


We then show by usual methods that this defines a functors, that it is fully faithful and essentially surjective.


The argument can be generalized to the monoidal case by a painful treatment "by hand" of the monoidal coherence.


**A more general argument:**


One can show very generally that in a combinatorial model category that has a set of generating cofibrations that are between $\kappa$-presentable objects the class of weak equivalences is closed under $\kappa$-filtered colimits. (see proposition 4.1 in this paper of [Raptis and Rosicky](https://arxiv.org/pdf/1403.3042.pdf))


This applies to the folk model structure on Cat is (for $\kappa=\omega$) i.e. it shows that equivalences of categories are stable under $\omega$-filtered (i.e. filtered) colimits. The same applies for the analogue of the folk model structure on monoidal categories (the one obtained for e.g. by transfer from the folk model structure).


Well, to be precise, that does not quite give us the result we want yet: it says that the colimit of a strictly natural equivalence is an equivalence, but we want it from pseudo-natrual transformation (in the monoidal case it gives us something about strictly monoidal functors).


The trick to get the result we want it to use the notion of "flexible replacement" (as for example [here](https://www.sciencedirect.com/science/article/pii/0022404989901606) ) of a diagram of categories that allows to turn a pseudo-natural transformation into a span of strictly natural transformation.


In very short, given any diagram $ F:I \to $ Cat, there is a diagram $\overline{F}:I \to $ Cat, such that pseudo-natural transformation $F \to G$ are the same as strictly natural transformation $\overline{F} \to G$ (in particular one has a strictly natural equivalence $\overline{F} \to F$ and a lax transformation $F \to \overline{F}$) and every lax transformation $F \to G$ is equivalent to the span of strict transformation $\overline{F} \to F$ and $\overline{F} \to G$ where the first one is an equivalence.


A final remarks: the first argument use the axiom of choice. The second argument avoids it by only constructing a span of equivalence. But one can also write a more natural version of the first argument by only constructing an "anafunctor" in the sense of makkai, and considering all $i$ at the same time everytime.


