Let $\mathcal{D}$ be a finite category and let $F:\mathcal{D}\to\mathbf{Set}$ be a functor. Since $\mathbf{Set}$ has (co)equalizers and finite (co)products, by the Existence Theorem $F$ has a limit and (dually) a colimit, and the Theorem tells us how to construct them (see for instance this answer for an explicit construction of the colimit).

Here is another neat way to find the limit. Recall that, for $A$ a set and $\Delta_A:\mathcal{D}\to\mathbf{Set}$ the constant functor, we have a bijection $\hom(A,\lim F)=[\mathcal{D},\mathbf{Set}](\Delta_A,F)$. Setting $A=\{ 1 \}=\ast$ to be a one-element set, we therefore find $$ \begin{aligned} \lim F &= \hom(\ast,\lim F) = [\mathcal{D},\mathbf{Set}](\Delta_\ast,F)\\ &= \Big\{ \big(f_i:\ast\to F(i)\big)_i\in \prod_{i\in\mathcal{D}} \hom\big(\ast,F(i)\big) \;\Big\vert\; f_j=F(\alpha)\circ f_i \text{ for all }\alpha:i\to j \Big\}. \end{aligned} $$ Using that $\hom(\ast,F(i))=F(i)$ for each $i$, we obtain the usual description of the limit in $\mathbf{Set}$!

Question: Is there a similar trick, for example using $\hom(\text{colim} F,A)=[\mathcal{D},\mathbf{Set}](F,\Delta_A)$, to find a description of the colimit?

My approach so far has been to first reduce the question to the case $\mathcal{D}=\{1,2\}$ and to find the right set $A$ to deduce that $F(1)\sqcup F(2)=\big(F(1)\times\{1\}\big)\cup\big(F(2)\times \{2\}\big)$, but I haven't found such a set yet. I suspect that dividing out by the equivalence relation will somehow follow from the condition $f_i=f_j\circ F(\alpha)$ on a cone $(f_i:F(i)\to A)_i$.

  • 7
    $\begingroup$ Yes. The power set functor is representable and so you can compute the power set of the colimit as a set of cones. Then you can extract the actual colimit as the set of atoms. This is basically how the construction goes for a general elementary topos too. $\endgroup$
    – Zhen Lin
    Commented Jul 7, 2021 at 13:20
  • $\begingroup$ @ZhenLin Thank you, this sounds like it should work! I'm having trouble finding information online about this construction. Could you please explain how you get to the set of cones from $\hom(\{0,1\},-)$? $\endgroup$
    – Alexander
    Commented Jul 7, 2021 at 13:38
  • $\begingroup$ I'm talking about the contravariant power set functor, which is isomorphic to $\textrm{Hom}(-, \{ 0, 1 \})$, so you can just use the trick you wanted to use. $\endgroup$
    – Zhen Lin
    Commented Jul 7, 2021 at 16:38

2 Answers 2


The power set functor $P : \mathbf{Set}^{\mathrm{op}} \to \mathbf{Set}$ is monadic, for example by using the Crude Monadicity Criterion; see Theorem 5.1.1 in Toposes, Triples and Theories by Barr-Wells, where it is proven for every elementary topos. In particular, $P$ creates limits. This means that we can construct colimits in $\mathbf{Set}$ from limits in $\mathbf{Set}$. This is also the standard way to show that every elementary topos is finitely cocomplete (Corollary 5.1.2 in loc.cit.).


Here I work out @ZhenLin's comment.

The ncatlab atom page suggests that we have a functor $\DeclareMathOperator{\Atom}{Atom}\DeclareMathOperator{\BPos}{BPos} \DeclareMathOperator{\Set}{Set} \Atom : \BPos \to \Set$ from posets $(S,\leq)$ with bottom $\perp \in S$ to sets defined as $\Atom(S,\leq) := \cup\{ s \in S \mid s \neq \perp, \forall t \in S : t \leq s \Rightarrow t = s \lor t = \perp\}$. Then if $P : \Set \to \BPos$ is the power set functor we have $\Atom(P(S)) = S$.

Now suppose $F : \mathcal{D} \to \Set$ has a colimit $\varinjlim F$. Then we have $$\varinjlim F = \Atom(P(\varinjlim F)) = \Atom(\Set(\varinjlim F,\{0,1\})) = \Atom([\mathcal{D},\Set](F,\Delta_{\{0,1\}})).$$

Here we used the representability equalities $P = \Set(\bullet,\{0,1\})$ and $\Set(\varinjlim F, \bullet) = [\mathcal{D},\Set](F,\Delta_{\bullet})$.

So if we can define a bottom poset structure on $[\mathcal{D},\Set](F,\Delta_{\{0,1\}})$, then we can take as definition $\varinjlim F := \Atom([\mathcal{D},\Set](F,\Delta_{\{0,1\}})).$

We set $\sigma \leq \tau$ for $\sigma,\tau \in [\mathcal{D},\Set](F,\Delta_{\{0,1\}})$ if $\tau(d)(x) = 1 \Rightarrow \sigma(d)(x) = 1$ $\forall d \in \mathcal{D}, x \in F(d)$ and we set $\perp \in [D,\Set](F,\Delta_{\{0,1\}})$ by $\perp(d)(x) = 0$ $\forall d \in \mathcal{D},x \in F(d)$.

Then we get $\Atom([\mathcal{D},\Set](F,\Delta_{\{0,1\}}))$ to be all graphs $\sigma$ where only one path component is 1 and the rest is 0.

To get a bijection with the standard definition of colimit using equivalence relations $\Atom([D,\Set](F,\Delta_{\{0,1\}})) \leftrightarrows \coprod_{d \in \mathcal{D}} F(d) /\sim$ we can map a graph $\sigma$ to an element $x \in F(d)$ such that $\sigma(d)(x) = 1$ and conversely, we map an element $x \in F(d)$ to the graph $\sigma$ that is 1 on the path component of $x$ and 0 elsewhere.

So finally we have $$ \varinjlim F = \Atom([\mathcal{D},\Set](F,\Delta_{\{0,1\}})) \quad \text{and} \quad \varprojlim F = [\mathcal{D},\Set](\Delta_{\{*\}},F)$$ and what feels neat about this construction of $\varinjlim F$ is that we can completely describe it with subobjects so we don't need disjoint unions or equivalence relations.

  • $\begingroup$ Very clever! I was missing the idea to give the set of cones the appropriate poset structure. And this was exactly along the lines I was looking for too, so thank you! $\endgroup$
    – Alexander
    Commented Sep 13, 2021 at 21:40

You must log in to answer this question.

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