Suppose $\mathcal{C}$ is a category with small colimits, and $G \in \mathrm{Ob}(\mathcal{C})$ is a strong generator. (This means that $f: X \to Y$ is an isomorphism iff $f_{*}: \mathrm{Hom}(G,X) \to \mathrm{Hom}(G,Y)$ is a bijection.)

How does one prove that every object $X$ is a colimit of copies of $G$ (i.e., a quotient of coproducts of $G$)?

I think $X \simeq \mathop{\mathrm{colim}}_{G \to X} G$. I can see there is a natural map $e: \mathop{\mathrm{colim}}_{G \to X} \to X$, and that the induced map $e_{*}: Hom(G,\mathop{\mathrm{colim}}_{G \to X}) \to Hom(G, X)$ is surjective, but I cannot see that it map is injective.

  • $\begingroup$ I know the following definition of a generator (from Mac Lane's book): For every distinct $f,g : X \to Y$, there is some $h : G \to X$ such that $fh, gh$ are also distinct. I doubt that this is equivalent to your one. $\endgroup$ Jun 6, 2010 at 8:34
  • 2
    $\begingroup$ What you called a "generator" is usually called a strong generator. Although a few people do confusingly call it merely a "generator," and instead say "separator" for the notion mentioned by Martin (which is what is more commonly called a "generator"). $\endgroup$ Jun 6, 2010 at 14:18
  • 1
    $\begingroup$ The colimit is taken over the slice category as $$\varinjlim_{f\in(G\downarrow X)}s(f)$$ where $s$ is the canonical functor $(G\downarrow X)\to C$ sending every object to its source (in this case G) and every commutative triangle to the corresponding morphism in $C$. Joey's notation is a bit too abusive for my tastes, but I understood what he meant. $\endgroup$ Jun 6, 2010 at 16:09
  • 1
    $\begingroup$ If that's what is meant, then the statement is false. In the category of compact Hausdorff spaces, the terminal object is a strong generator. But since it has no endomorphisms, that slice category is discrete, so the colimit is just the coproduct of 1 over the set Hom(1,X), and this is the Stone-Cech compactification of the underlying set of X -- generally not the same as X. $\endgroup$ Jun 6, 2010 at 18:34
  • 4
    $\begingroup$ Sorry, Joey -- compact Hausdorff spaces are a cocomplete category. This is well-known. (You are right however that colimits there are not computed as they would be in $Top$.) I don't think however the question has been totally settled by Mike's example, because one can still present a compact Hausdorff space as a colimit of a diagram consisting totally of 1's (essentially because compact Hausdorff spaces are monadic over sets). $\endgroup$
    – Todd Trimble
    Jun 6, 2010 at 19:44

2 Answers 2


I think there are actually three possible things that you might be asking, but the answer to all of them is no. Suppose that G is a strong generator in a cocomplete category C. Then you can ask:

  1. Is every object X of C the colimit of G over the canonical diagram of shape $(G\downarrow X)$? (If so, then G is called dense in C.)

  2. Is every object some colimit of a diagram all of whose vertices are G? (If so, then G is called colimit-dense in C.)

  3. Is C the smallest subcategory of itself containing G and closed under colimits? (If so, then G is a colimit-generator of C.)

The category of compact Hausdorff spaces is a counterexample to the first two. It is monadic over Set (the monad is the ultrafilter monad, aka Stone-Cech compactification of the discrete topology), and hence cocomplete, and the one-point space is a strong generator. But any colimit of a diagram consisting entirely of 1-point spaces must be in the image of the free functor from Set (the one-point space being its own Stone-Cech compactification), and hence (since that functor is a left adjoint and preserves colimits) must be the free object on some set. However, not every compact Hausdorff space is the Stone-Cech compactification of a discrete set.

As Todd pointed out in the comments, though, CptHaus is the colimit-closure of the one-point space, since every object is a coequalizer of maps between free ones (because the category is monadic over Set); thus it isn't a counterexample to the third question. Counterexamples to the third question are actually much harder to come by, and in fact if you assume additionally that C has finite limits and is "extremally well-copowered", then it is true that any strong generator is a colimit-generator. I think there's a proof of this somewhere in Kelly's book "Basic concepts of enriched category theory," and a brief version can be found here. However, without these assumptions, one can cook up ugly and contrived counterexamples, such as example 4.3 in the paper "Total categories and solid functors" by Borger and Tholen.

  • $\begingroup$ +1 Nice answer, Mike. Now I don't have to struggle with this any more! :-) $\endgroup$
    – Todd Trimble
    Jun 7, 2010 at 0:12

See cor. 4.4 in "Strong regular and dense generators" Tholen , Borger:



Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.