Wednesday, September 03, 2014

Lex total categories and Grothendieck toposes III

Previous post in this series; next post in this series.

Today I want to give an application of the adjoint functor theorem for totally complete categories, namely to show that lex total categories are cartesian closed. The proof should be a generalization of the proof that locales are cartesian closed so we should look at that before attempting the case of lex total categories. We will be repeating some of the discussion of the first post in this series.

(i) By the adjoint functor theorem for complete posets it is sufficient to show that if $A$ is a locale and $x$ is an element then the functor $x\wedge -:A\to A$ preserves sups of downsets. That is, we need to show that if $U$ is a downset then $x\wedge sup(U)=sup(x\wedge u; u\in U)$.

(ii) We denote the downset generated by a subset $V$ as $\langle V\rangle $ and it is clear that $sup(V)=sup(\langle V\rangle )$.

(iii) Notice that  $\{x\wedge u;u\in U\}$ is not necessarily a downset. The crucial thing to prove is that $\langle  \{x\wedge u; u \in U\}\rangle =\langle x\rangle \cap U$,  since then, by the left exactness of $sup: PA\to A$ it follows that

 $sup(\{x\wedge u; u\in U\})=sup(\langle  \{x\wedge u; u \in U\}\rangle )$
$=sup(\langle x\rangle \cap U)=sup(\langle x\rangle )\wedge sup(U)$
$=x\wedge sup(U)$
 as required.

(iv) To prove that $\langle  \{x\wedge u; u \in U\}\rangle =\langle x\rangle \cap U$ notice that an element of the left-hand-side is an element $y\in A$ such that $y\leq x\wedge u; u \in U$; that is, such that $y\leq x$ and $y\leq u$ for some $u\in U$; that is, such that $y$ is in the right-hand-side. The converse is equally evident.

$\bf Generalizing\;\; to\;\; categories:$

 By the adjoint functor theorem for total categories it is sufficient to show that if $A$ is a lex total category and $x$ is an object then the functor $x\times -:A\to A$ preserves colimits of discrete fibrations $U:B\to A$ with small fibres. That is, we need to show that $x\times colim(U)=colim(x\times U(-))$ for such $U$.

Notice that  $x\times U(-):B\to A$ is not necessarily a discrete fibration, so we need the analogy of the downset generated by a subset, which for categories is achieved by the factorization into a cofinal functor followed by a discrete fibration.

$\bf Interlude\; about\; a\; factorization:$
The factorization of a functor $V:B\to A$ as $B\xrightarrow{V_1}\langle V\rangle \xrightarrow{V_2}A$ goes as follows. First I will describe the category $\langle V\rangle $.  $\langle V\rangle $ is the category with objects equivalence classes of triples $(a,\alpha :a\to V(b),b);a, \alpha\in A,b\in B$. Notice that this generalizes the requirement in a poset that $a\leq V(b)$ but in a category $a$ may (be less than or equal to =) have an arrow to many $V(b)$ and with different arrows. Hence the equivalence relation to be explained. We will denote the equivalence class of $(a,\alpha :a\to V(b),b)$ by $[(a,\alpha :a\to V(b),b)]$. Then the equivalence relation is specified by the requirement that $[(a,\alpha :a\to V(b),b)]=[(a,\alpha V(\beta) :a\to (b'),b')]$ for any $\beta : b\to b'\in B$. Finally an arrow in $\langle V\rangle $ is an arrow $\alpha':a'\to a: [(a',\alpha'\alpha ,b)]\to [(a,\alpha ,b)]]$. (I omit the verification that this definition is unambiguous.)

Now to describe the two functors of the factorization. $V_2:\langle V\rangle \to A$ is defined by $V_2([(a,\alpha ,b)])=a$ and $V_2(\alpha':a'\to a: [(a',\alpha'\alpha,b)]\to [(a,\alpha,b)])=\alpha'$. Again I omit the check that it is unambiguous. Given that, it is clear that $V_2$ is a discrete fibration. Finally the definition of $V_1$: $V_1(b)=[V(b), 1_{V(b)}, b]$ and
$V_1(\beta':b\to b')=V(\beta'):[V(b),1_{V(b)},V(b)]$
$=[V(b), V(\beta'), V(b')]\to [V(b'),1_{V(b')},V(b')]$.

O.K. I have left out a lot of checking. I should also explain why $F_1$ is cofinal. What we need here is a little less, namely that $colimit(V)=colimit(V_2:\langle V\rangle \to A)$. The check is straightforward.

$\bf Return\; to \; the\; proof:$ We have now reached point (iii) of the proof for posets which we would like to generalize to  categories.
The crucial point to prove is that the discrete fibration $\langle (x\times U(-)\rangle \to A$ is $Elements(Hom(-,x))\times_A U(-)\to A$, because then the left exactness of $colim : PA\to A$ allows the following calculation:
$colim(x\times U(-))=colim(\langle x\times U(-)\rangle )$
$= colim(Elements(Hom(-,x))\times U(-) )$
$=colim(Elements(Hom(-,x))\times colim(U(-))$
$=x\times colim(U)$ as required.

So now to the crucial point, which should be a jazzed up categorical version of (iv) for locales above. We have two discrete fibrations which we want to show isomorphic (and at the same time verify that they have small fibres - I almost forgot to say that! In fact, $Elements(Hom(-,x))\times_A U(-)\to A$ clearly has small fibres).
I will only describe the two functors (and on objects only) which I claim form an isomorphism and leave the rest to you.

The functor $\langle (x\times U(-)\rangle \to Elements(Hom(-,x))\times_A U(-)$.

An object over $a$ of the left hand side is an equivalence class of the form $[(a, \alpha:a\to x\times U(b),b)]$. From a representative of the equivalence class we can extract two arrows $\alpha_1\to x$ and $\alpha_2:a\to U(b)$.
But since $U$ is a discrete fibration  there is a unique object $b'\in B$ over $a$ and an arrow $\beta:b'\to b$ such that $U(\beta)=\alpha_2$.  Clearly
$[(a, \alpha:a\to x\times U(b),b)]=$
$[(a,a\xrightarrow{\Delta}a\times a\xrightarrow{\alpha_1\times 1_a}x\times U(b')\xrightarrow{1_x\times U(\beta)}x\times U(b),b]=$
$[(a,a\xrightarrow{\Delta}a\times a\xrightarrow{\alpha_1\times 1_a}x\times U(b'),b']$.

So from each equivalence class over $a$ we can extract an arrows from $a$ to $x$, and an object $b'\in B$ over $a$. But the elements over $a$ on the right hand side over $a$ are exactly such pairs.

 The functor $ Elements(Hom(-,x))\times_A U(-)\to \langle (x\times U(-)\rangle$.

 An object of the left hand side over $a$ consists of an arrow $alpha:a\to x$ and an object $b$ of $B$ such that $U(b)=a$. The functor takes this object to the equivalence class $[(a,a\xrightarrow{\Delta}a\times a\xrightarrow{\alpha\times 1_a}x\times U(b),b)]$.

 I have run out of steam! I hope you see the analogy with the locale result.

$\bf  References$

Ross Street and I announced the stronger result that well-powered lex total categories are elementary toposes in the paper
R.H. Street and R.F.C. Walters. Yoneda structures on 2-categories. Journal of Algebra, 50:350--379, 1978.
Looking around in subsequent literature however I cannot see where a proof has been published.

A less elementary description of the comprehensive factorization was given in the original paper:
R. H. Street and R.F.C. Walters. Comprehensive factorization of a functor. Bull. Amer. Math. Soc., 79:936-941, 1973.

Other papers on total and lex total categories:
Walter Tholen, Note on total categories, Bulletin of the Australian Mathematical Society, 21 (1980 ), 169 - 137

Ross Street, Notions of topos, Bull. Austral. Math. Soc. VOL. 23 (1981) , 199-208.

R.J Wood, Some remarks on total categories, J. Algebra 75:2, 1982, 538–545

Ross Street, The family approach to total cocompleteness and toposes, Trans. Amer.Soc., 284 (1984) 355-369

Max Kelly, A survey of totality for enriched and ordinary categories, Cahiers de Topologie et Géométrie Différentielle Catégoriques, 27 no. 2 (1986), p. 109-132

Brian Day, Further criteria for totality, Cahiers de Topologie et Géométrie Différentielle Catégoriques, 28 no. 1 (1987), p. 77-78



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