# How necessary is the filteredness condition in the result “a filtered diagram in which all morphisms are isomorphisms has a colimit”?

It can be proven that if $$F:\mathcal{I}\to\mathcal{C}$$ is a filtered diagram (i.e., $$\mathcal{I}$$ is a filtered category) such that $$Ff$$ is an isomorphism for every morphism $$f$$ of $$\mathcal{I}$$, then $$\operatorname{colim}_{i\in\mathcal{I}}F(i)$$ exists and in fact it equals $$F(i)$$, for any $$i\in\mathcal{I}$$. Dually, any cofiltered diagram has always a limit.

So I was wondering: is the filteredness condition necessary? Can we find a non-filtered diagram $$F:\mathcal{I}\to\mathcal{C}$$ such that $$Ff$$ is always an iso but such that $$\not\exists\operatorname{colim}_{i\in\mathcal{I}}F(i)$$? The answer is yes, because it suffices to consider the discrete category $$\mathcal{D}_2$$ of two points and the identity functor of this category as diagram. But the thing is that this isn't the flavour of counterexample I was looking for.

We define a componentwise filtered category to be a category whose connected components are filtered. Then, from the result above, it follows that if $$F:\mathcal{I}\to\mathcal{C}$$ is a componentwise filtered diagram (meaning $$\mathcal{I}$$ is componentwise filtered) that sends all morphisms of $$\mathcal{I}$$ to isos and if $$\mathcal{C}$$ has all coproducts, then $$\operatorname{colim}_{i\in\mathcal{I}}F(i)$$ exists and it is equal to the coproduct of the colimits of the diagrams $$F|_\mathcal{J}$$, where $$\mathcal{J}$$ moves over the family of connected components of $$\mathcal{I}$$. So now I can better restate my question: how necessary is the componentwise filtered condition? Can we find a diagram $$F:\mathcal{I}\to\mathcal{C}$$ for which the following holds?

1. $$\mathcal{I}$$ is not componentwise filtered,
2. $$F$$ sends all morphisms of $$\mathcal{I}$$ to isos,
3. $$\mathcal{C}$$ has all coproducts, and
4. $$F$$ has no colimit.

If we want to forget about the coproducts, I would be also satisfied if instead we could find a diagram $$F:\mathcal{I}\to\mathcal{C}$$ such that the following holds:

1. $$\mathcal{I}$$ is connected,
2. $$\mathcal{I}$$ is not filtered,
3. $$F$$ sends all morphisms of $$\mathcal{I}$$ to isos, and
4. $$F$$ has no colimit.

For a simple example, you can take a coequalizer of two different isomorphisms. For instance, let $$\mathcal{C}$$ be the full subcategory of $$\mathtt{Set}$$ consisting of sets that have an even or infinite number of elements, and consider the coequalizer of the identity $$\{0,1\}\to\{0,1\}$$ and the swap map $$\{0,1\}\to\{0,1\}$$. This coequalizer does not exist in $$\mathcal{C}$$, since the coequalizer would have to have exactly $$n$$ maps to a set with $$n$$ elements (coming from the constant maps out of $$\{0,1\}$$ that coequalize the two isomorphisms) and no object of $$\mathcal{C}$$ has this property.