# General description of colimits in $\mathbf{Set}$ - 2

I've previously posted a question about the example below, but this question is different.

Example 5.2.16. The colimit of a diagram $$D \colon \mathbf{I} \to \mathbf{Set}$$ is given by $$\lim_{\to \mathbf{I}} D = \left. \left( \sum_{I \in \mathbf{I}} D(I) \right) \middle/ {\sim} \right.$$ where $$\sim$$ is the equivalence relation on $$\sum D(I)$$ generated by $$x \sim (Du)(x)$$ for all $$u \colon I \to J$$ in $$\mathbf{I}$$ and $$x \in D(I)$$. To see this, note that for any set $$A$$, the maps $$\left. \left( \sum D(I) \right) \middle/ {\sim} \right. \to A$$ correspond bijectively with the maps $$f \colon \sum D(I) \to A$$ such that $$f(x) = f( (Du)(x) )$$ for all $$u$$ and $$x$$ (by Remark 5.2.8). These in turn correspond to families of maps $$( f_I \colon D(I) \to A )_{I \in \mathbf{I}}$$ such that $$f_I(x) = f_J( (Du)(x) )$$ for all $$u$$ and $$x$$; but these are exactly the cocones on $$D$$ with vertex $$A$$.

First, let's assume, as suggested in the answer to the question referred to above, that all sets $$D(I)$$ are disjoint and the sum is the union.

To prove that the quotient is a colimit, we need to prove two things:

(1) $$Du\circ \pi_J=\pi_I$$ for all $$u:I\to J$$ where $$\pi_I$$ are the projections of the cone which is claimed to be the colimit (which aren't defined, as far as I can see);

(2) If there is a cone $$(A, g_I:D(I)\to A)_{I\in\mathbf I}$$ then there is a unique $$\alpha:co\lim D\to A$$ such that $$\alpha\circ \pi_I= g_I$$.

As far as I can see, the example is only discussing (2). So let's start with (2). Here I have two questions. First, is the last sentence (namely, the fact that the maps $$\cup D(I)\to A$$ with $$f(x)=f((Du(x))$$ correspond bijectively to families of maps $$(f_I:D(I)\to A)$$ such that $$f_I(x)=f_J((Du)(x))$$) a general result that has nothing to do with $$D, u$$? What would be a precise general statement? Secondly, how does this prove (2) (if it does)?

And lastly, why does (1) hold? To begin with, what are $$\pi_I$$?

Remember that the cones are for limits. The dual notion, cocones, are for colimits, the accompanying maps are called the canonical injections, and I often label them $$\iota_i$$ for $$i\in I$$. With this notation, I would write a cocone with apex $$X\in C$$ to a functor $$D:I\to C$$ as the pair $$(X,\{\iota_i\})$$, where the $$\iota_i$$ are morphisms $$\iota_i:Di\to X$$ such that for all $$f:i\to j$$ in $$I$$, $$\iota_j\circ Df = \iota_i$$.

Sums

First, suppose $$I$$ is a discrete category (no nonidenty morphisms), $$D:I\to\newcommand\Set{\mathbf{Set}}\Set$$ is it clear that we can construct the sum $$\newcommand\colim{\operatorname{colim}}\sum_I D$$? I believe a traditional construction is to take $$\bigcup_{i\in I} \newcommand\set[1]{\left\{{#1}\right\}}\set{(i,a) : a\in Di}$$. The canonical injections $$\iota_i : Di\to \sum_I D$$ being defined by $$\iota_i a = (i,a)$$.

Given a cocone to $$D$$, $$(X,\set{f_i})$$, we define $$f : \sum_I D\to X$$ by $$f(i,a) = f_i(a)$$. You can check that this works and is unique.

General colimits

Now to construct a general colimit, we no longer suppose $$I$$ is discrete. It still makes sense to construct $$\sum_I D$$ by the same formula as before. However, now $$I$$ has nonidentity morphisms, so the cocone $$(\sum_I D,\set{\iota_i})$$ might not be a colimit cocone. The problem is that we might not have $$\iota_j\circ Df = \iota_i$$ for all morphisms $$f:i\to j$$.

How do we fix this? Well we just impose these relations. Define $$\sim$$ to be the smallest equivalence relation on $$\sum_I D$$ such that for all $$a\in Di$$, $$\iota_j (Df(a))\sim \iota_i a.$$

We have the quotient map $$q:\sum_I D\to \sum_I D/\sim$$. Let $$\iota_i' = q\circ \iota_i$$. The claim is that $$(\sum_I D/\sim, \set{\iota_i'})$$ is a colimit cocone.

Given any other cocone $$(X,\set{g_i})$$, as noted above, we have an induced map $$g:\sum_I D\to X$$ such that $$g\circ \iota_i =g_i$$. Moreover, since the $$g_i$$ form a cocone, we have for all $$f:i\to j$$ in $$I$$, $$g_j\circ Df = g_i$$. Thus combining the last two relations, for all $$i\in I$$, and all $$a\in Di$$ $$g_j(Df(a)) = g(\iota_j(Df(a)))=g_i(a) = g(\iota_i(a)).$$ Thus $$g$$ respects the equivalence relation $$\sim$$, and hence defines a well defined map $$g' : \sum_I D/\sim \to X$$. Again, you can check that this satisfies the correct relations and is unique.

• Your answers are very easy to follow, and because you give some background and motivation, I can predict what your next step would be, work it out myself and then check against your answer. Commented Feb 14, 2020 at 16:07
• Along the way, you (partially) answered the first part of my question (which I cited in my post): the equivalence relation is not generated by $(x, (Du)(x))$, as Leinster says, but by $(\iota_i(x), \iota_j((Du)(x))$. I guess Leinster assumes that $\iota_i$ are "true inclusions", which they are not, and this makes his exposition quite confusing for me. Commented Feb 14, 2020 at 16:11