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.

1 Answer 1


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.


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