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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 $\pi_I$?

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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 $I$, $\iota_j\circ Df = \iota_i$.

Now let's address some of your questions.

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.

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    $\begingroup$ 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. $\endgroup$
    – user557
    Commented Feb 14, 2020 at 16:07
  • $\begingroup$ 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. $\endgroup$
    – user557
    Commented Feb 14, 2020 at 16:11

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