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This is probably a question I should have asked myself a bit earlier. For some reason I always thought I knew the answer so I did not bother, but now that I actually need to use it (I am studying the $+$ construction on presheaves) I realize I am not really familiar with this.

So, what is a colimit of sets, formally?! Let $F:\mathcal{D}\to \mathbf{Set}$ be a diagram in $\mathbf{Set}$. What is $\mathrm{Colim}\;F$ ?

I believe (am I right?) that if $\mathcal{D}$ is filtered, then the colimit coincides with the direct limit, but what is it in the general case? Thanks!

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Colimits are more general than direct limits on filtered preorders or filtered categories. Look at the definition and you'll see that they're the same thing. –  egreg Nov 15 '13 at 10:39
Of course colimits are more general than direct limits in general. But in the case of a functor from a filtered category to $\mathbf{Set}$, do they not coincide? –  user121314 Nov 15 '13 at 10:46
They do coincide. –  egreg Nov 15 '13 at 10:55

2 Answers 2

Limits and colimits don't coincide in $Set$. Just like any colimit, it suffices to understand coproducts and coequalizers. A coproduct of a family $\{A_i\}_{i\in I}$ of sets is their disjoint union. More formally, take $$X=\bigcup_{\{i\in I\}}\{i\}\times A_i$$with the obvious injections $A_i\to X$. Then $X$ is a coproduct. Let me know if you need help figuring out coequalizers.

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Did I say colimits coincide with limits? I asked if colimits coincide with direct limits when $\mathcal{D}$ is filtered, which are actually colimits... en.wikipedia.org/wiki/Direct_limit I know, I hate the terminology... –  user121314 Nov 15 '13 at 10:33
I see the confusion. –  Ittay Weiss Nov 15 '13 at 19:45

To make my answer clearer, firstly I describe some basic concepts in category theory.

  1. Suppose you have two categories $A$ and $B$ and functor $T\colon A\to B$. Then we can take its limit $\varprojlim T$ and its colimit $\varinjlim T$(see definitions here: limit and colimit). They may not exist(or one of them may not exists), but if they exist, then they are objects in $B$, defined up to isomorphism. Note that a colimit of $T$ is nothing but a limit of the dual functor $T^{op}\colon A^{op}\to B^{op}$.
  2. Suppose you have a graph $D$, a category $B$ and a diagram $F\colon D\to B$. Then you can take a limit(respectively, colimit) of the diagram $F$, which is nothing but a limit(respectively, colimit) of the corresponding functor $C[F]\colon C[D]\to B$, where $C[D]$ is a free category on the grath $D$.
  3. Well, now we can take limits and colimits of $\mathbf{Set}$-valued functors $T\colon A\to\mathbf{Set}$ and $\mathbf{Set}$-valued diagrams $F\colon D\to\mathbf{Set}$. From 2 we understand that we can reduce our problem to (co)limits of $\mathbf{Set}$-valued functors. There is an important result about such (co)limits: Theorem. Let $A$ be a small category and $T\colon A\to\mathbf{Set}$ be a functor. Then the colimit(respectively, limit) of $T$ exists and given by a formula(in the case of colimit): $$ \varinjlim T=\coprod_{a\in A}T(a)/\sim $$ where $\sim$ is the minimal equivalence relation on $\coprod_{a\in A}T(a)$, contains all pairs $((a,x),(a',x'))\in\coprod_{a\in A}T(a)\times\coprod_{a\in A}T(a)$, such that there exists a morphism $f\in Arr(A)$, $f\colon a\to a'$ satisfying $(T(f))(x)=x'$. The proof is straightforward.
  4. Direct limit in modern mathematics is nothing but a colimit(but in some literature you can find this term in the meaning of directed colimit, see 5).
  5. There is a notion of directed colimit which is a colimit of a functor from a preorder, corresponding to some directed set. This is, of course, the special case of the aforementioned construction.
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I wouldn't say "some". I cannot recall a single example of "direct limit" which does not mean "directed colimit". –  Zhen Lin Nov 15 '13 at 13:55
@ZhenLin: e.g. MacLane, "Categories for the working mathematician" - as a synonym for "colimit" - but only in the definition. –  Oskar Nov 15 '13 at 14:07
Thanks! In Kashiwara-Schapira, Categories and Sheaves, the word "colimit" does not appear anywhere. They only use "inductive" and "projective" limits. This is where my confusion came from. –  user121314 Nov 15 '13 at 15:51

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