(I suspect this is a very easy question: I haven't spent much time thinking about category theory.) $\DeclareMathOperator{\colim}{colim}\DeclareMathOperator{\Dom}{Dom}\DeclareMathOperator{\im}{im}$

In The Rising Sea, Ravi Vakil notes that our intuition for limits and colimits of the diagram $$\lim{X_j}\to\dots\to X_{-2}\to X_{-1}\to X_0\to X_1\to X_2\to\dots\to\colim{X_j}$$ (the dots may be finite and diagonal morphisms were omitted to fit within MathJax) are very different. To wit, each element of $\lim{X_j}$ is a sequence of "compatible" elements from the $\{X_j\}_j$, whereas each element of $\colim{X_j}$ is a single distinguished element from $X_k$.

In my mind, this discrepancy occurs because of a fundamental asymmetry in the intuition regarding homomorphism functors. To wit, if we have $f\in X_0\to X_1$, then we automatically assume that there will be some remnant of $X_0$ in $\im{(f)}$. Conversely, given $f\in X_{-1}\to X_0$, we do not make the assumption that $\Dom{(f)}=X_{-1}$ contains all the properties of $X_0$; some may instead be "emergent."

To put it another way, it seems harder to construct domains than ranges for functions. Is this intuition rooted in truth; that is, does there exist a category with colimits but no limits?

  • 2
    $\begingroup$ There is no category with "no limits"; however there are categories that have "more" colimits than limits $\endgroup$ Commented Feb 18, 2019 at 21:39
  • $\begingroup$ @Max: That statement, along with an example of such a category, sounds like a complete answer I'd accept...*hint, hint* $\endgroup$ Commented Feb 18, 2019 at 21:41
  • 2
    $\begingroup$ I don't agree with your "that is" at all: your specific question at the end has almost nothing to do with whether your intuition is rooted in truth. $\endgroup$ Commented Feb 18, 2019 at 21:49
  • 2
    $\begingroup$ @Max: How about the empty category? :) Of course, that doesn't have any colimits either. $\endgroup$ Commented Feb 18, 2019 at 21:52
  • $\begingroup$ @EricWofsey : woops always forget that one; adding "nontrivial" should solve it (it always does) $\endgroup$ Commented Feb 18, 2019 at 21:53

2 Answers 2


Sure, take for example the category $$A \rightarrow B \leftarrow C$$

Other than the identities, there is exactly one arrow from $A$ to $B$ and one arrow from $C$ to $B$.

It has all colimits (almost all, that is; there's no initial object) -- the colimit of every diagram that contains two different objects is $B$, and of a diagram that contains only one object it is that object itself.

But it doesn't have all limits; for example there is no product of $A$ and $C$.

Of course, the opposite category of a category with all colimits and not all limits will be a category with all limits and not all colimits ...

  • 4
    $\begingroup$ Note that this example is a poset, where limits and colimits are just meets and joins. I add this since posets of form great examples and it's nice to think about what some fancy category theory terms mean for posets. $\endgroup$
    – Christoph
    Commented Feb 18, 2019 at 21:48
  • $\begingroup$ You and Eric Wofsey both provided great answers to this question. I'm giving you the checkmark because your answer matches the title best, and Google searchers will look at this question based on the title. $\endgroup$ Commented Feb 19, 2019 at 19:52

I object to the premise of your question, namely that the existence of a category with colimits but not limits has anything to do with the intuition you discuss. Note that in an arbitrary category, morphisms need not have any relation to functions. Moreover, given any category $C$, you can form the opposite category $C^{op}$ which reverses the directions of all the arrows. So, domains and codomains are really completely symmetric if you're just talking about arbitrary categories. For instance, if $C$ has all colimits but not all limits, then $C^{op}$ has all limits but not all colimits.

The intuition that Ravi Vakil is talking about is not applicable to arbitrary categories. Instead, it's applicable to the sort of "concrete" categories we tend to encounter in practice (but not their opposite categories), where morphisms are closely related to functions and limits and filtered colimits are constructed similarly to how they are in the category of sets.

As for the way you interpreted the intuition, I'm not entirely sure what you mean but it seems not entirely unreasonable. The way I might phrase it is that a map $X_n\to X$ gives you actual elements of $X$ for each element of $X_0$. So, if $X$ is the colimit of the sequence, it has specific elements which come from the elements of each $X_n$ (and it can be deduced from the universal property that every element of $X$ comes from some $X_n$, at least in the category of sets). On the other hand, a map $X\to X_n$ doesn't tell you any specific elements of $X$. So, you can't describe what an element of $X$ can look like using just a single $X_n$; you really need the whole sequence before you can even name a single element of $X$.

  • $\begingroup$ You and Henning Makholm both provided great answers to this question. I'm giving Makholm the checkmark because his answer matches the title best, and Google searchers will look at this question based on the title, but thank you nonetheless. $\endgroup$ Commented Feb 19, 2019 at 19:52

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .