Interpreting Diagrams About Power Object Functor

Question

I'm reading Sheaves In Geometry And Logic and this diagram confuses me:

First, do those diagrams really live in $\mathcal{E}$ resp. $\mathcal{E}^\text{op}$ or rather in $[J^\text{op},\mathcal{E}]$ resp. $[J,\mathcal{E}^\text{op}]$? Otherwise, wouldn't we have $H_j$ instead of $H$?

And trying to interpret them in Set, where I assume for simplicity that $J_0=\mathbf{N}$ and that we interpret the limit $t$ as a set of sequences $(X_i)_i$ with $X_i\subseteq H_i$ and the usual projections $\pi_i$, would it be correct to say that $h$ maps a set $Z\subseteq P^\text{op} t$ to the sequence $(A_i)_i$ where $A_i$ is the set of $a\in X_i$ such that there exists $Y_{i,a}\in Z$ consisting of exactly those sequences $(X_i)_i\in t$ with $a\in X_i$. Would that be about right?

Another question about the second diagram, although maybe a duplicate, what would $\ell$ look like in Set? To me it looks like it contains all the $Y\subseteq t$ such that $Z\subseteq P^\text{op}t$ contains $Y$ if and only if $h(Z)$ is contained in $Y$. So especially $h(\{Y\})\in Y$ which means $Y=Y_{i,a}$ for some $a\in X_i$, but also vice versa, for all $Y_{i, a}$ we have $Y_{i,a}\in Z$ if and only if $h(Z)\in Y_{i,a}$.

This would match with the fact that elements of colimits in Set can be described as equivalence classes in $\cup_i H_i$ (assuming all $H_i$ being disjoint) and the description above doesn't rule out $Y_{i,a}=Y_{j,b}$ since we only look at sequences in $t$ which is already a limit hence not necessarily containing all sequences (which would be the case if $J$ is discrete).

I don't expect a full proof in the answer, just tell me whether I'm on the right track...

Answer 1

$\newcommand{\E}{\mathcal{E}}\newcommand{\set}{\mathsf{Set}}\newcommand{\op}{{^{\mathsf{op}}}}\newcommand{\P}{\mathscr{P}}$The diagrams can be interpreted in $\E,\E\op$ just fine, one only needs to read “$H$” as a placeholder for $H(j)$ for some object $j$ of $J$. One could try to view these as diagrams in the functor categories but that would require effort to justify everything being natural and it would also be an arguably worse abuse of notation, e.g. $\ell$ would have to be identified with the constant functor to $\ell$. Please note that the arrows $\tau,\sigma$ will vary (naturally) as $H(j)$ varies, they just have not been given a subscript for brevity. Alternatively you can view e.g. $\tau:t\implies\P H$ as a natural transformation if $t$ is abused with the constant functor to $t$.

I trust you understand the diagrams in the sense that you agree they exist and that what is said to commute really does commute. In comments there was some confusion about $\ell,\sigma$. $\ell$ is computed as the coequaliser of the parallel pair: $$\P\op\P\P\op t\overset{\P\op h}{\underset{\epsilon_{\P t}}{\rightrightarrows}}\P\op t$$In the dual category $\E\op$, that is, $\ell$ is the limit in $\E$ of the parallel pair. This is important because we are trying to form colimits in $\E$ given limits in $\E$, so our constructions must use limits in $\E$ only.

Now to your simplification where $J=\Bbb N$ (as a discrete category?) and where $\E=\set$ (the only topos I know the slightest thing about). The composite functors of $\P$ can be quite confusing to handle. I can’t parse what your suggested $h$ is doing; your definitions don’t quite feel like definitions. I’ve done the definition chasing myself and I believe that we have:

For a subset $Z$ of $\P\op(t)$ we have that: $$h(Z)=(A_n)_{n\in\Bbb N}$$Where for each $n\in\Bbb N$ we have: $$A_n:=\{x\in H_n: \{(X_j)_{j\in\Bbb N}\in t: x\in X_n\}\in Z\}$$

Using this explicit description of $h$ we may identify a quite confusing description of $\ell$:

$\ell$ contains precisely those $Y\subseteq t$ which satisfy, quantifying over $Z\subseteq\P(t)$: $$Y\in Z\\\iff\\\left(\{x\in X_n:\{(X_j)_{j\in\Bbb N}\in t:x\in X_n\}\in Z\}\right)_{n\in\Bbb N}\in Y$$

Less confusingly, it is as you say; $Y\in Z$ iff. $h(Z)\in Y$.

The quotient maps $\sigma$ act as follows:

For any $n\in\Bbb N$; $$\sigma_n:H_n\to\ell\\x\mapsto\{(X_j)_{j\in\Bbb N}\in t:x\in X_n\}$$

Define $Z$ to be the set of all $Y\in\ell$ which are not of the form $\sigma_n(x)$ for some $n$ and $x$. By the characterisation of elements of $\ell$ we infer that the empty sequence is in each element of $Z$. From the same characterisation it would further follow that each element of $Z$ is in the empty set; thus, $Z$ is empty. So all elements of $\ell$ are in the image of the $\sigma$ as one would expect for a colimit.

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We can summarise all of this into a new, explicit construction of colimit. I change notation slightly, and use $\P$ for the power set (class) function $\set\to\set$.

Let $J$ be any small category and $H:J\to\set$ any diagram. Define:

$$t:=\{X\in\prod_{j\in J}\P(H_j):\forall(f:i\to j\text{ in }J), H(f)^{-1}(X_j)=X_i\}$$

$$\ell:=\{\{X\in t:x\in X_j\}:j\in J,\,x\in H(j)\}$$

$$\sigma_j:H_j\to\ell,\,\,x\mapsto\{X\in t:x\in X_j\}$$

The sets $\{X\in t:x\in X_j\}$ serve as a nice "explicit" description of the colimit equivalence classes. Let $R$ be the usual colimit relation on $\bigsqcup_{j\in J}H_j$. By that I mean, the relation such that $\varinjlim_JH\cong\bigsqcup_{j\in J}H_j/R$, the transitive closure of $(x,i)\sim(y,j)$ iff. there is some arrow $f:i\to j$, $H(f)(x)=y$.

We can define a new relation $R'$ on $\bigsqcup_{j\in J}H_j$ given by $(x,i)\sim(y,j)$ iff. $\sigma_i(x)=\sigma_j(y)$ in $\ell$. To claim $(\ell;\sigma_\bullet)$ really is the colimit (without resorting to abstract nonsense) we need only show $R=R'$.

First note that $R'$ is a genuine equivalence relation. Suppose $(x,i)R(y,j)$ in a single step, i.e. $H(f)(x)=y$ for some arrow $f:i\to j$ in $J$. Let $X\in t$ have $y\in X_j$. Then: $$X_i=H(f)^{-1}(X_j)\supseteq H(f)^{-1}\{y\}\ni x$$So $x\in X_i$ as well. Conversely, suppose $x\in X_i$. Then from $X_i=H(f)^{-1}(X_j)$, we must have $H(f)(X_i)\subseteq X_j$; in particular, $y=H(f)(x)\in H(f)(X_i)\subseteq X_j$. Therefore $X\in t$ has $y\in X_j$ iff. it has $x\in X_i$; therefore $\sigma_i(x)=\sigma_j(y)$.

It follows $R'$ contains $R$. Suppose some $(x,i)$ is not $R$-related to $(y,j)$. For every $k\in J$, define $X_k:=\{z\in H(k):xRz\}$. It is not too hard to check, by similar logic to the above paragraph, that $X:=(X_k)_{k\in J}$ is an element of $t$. Clearly $x\in X_i$ but $y\notin X_j$. Therefore, $\sigma_i(x)\neq\sigma_j(y)$; therefore $(x_i)$ is not $R'$-related to $(y,j)$.

$R=R'$ follows as desired.