## Equalizer completion

Can anybody give a definition of the equalizer completion of a cartesian category?

Is the method to get more or less as the regular and exact completions in the way that are given in: http://ncatlab.org/nlab/show/regular+and+exact+completions?

How is in that case the behaviour of the forgetful functor FL-->FP (where FL are categories with finite limits and FP categories with finite products)?

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There are general results about how to freely add limits or colimits to a category. They are formally dual, but people normally state the colimit variety because they involve the category of presheaves.

To freely add some class of colimits to a given category $C$, you form the closure of the representables in the presheaf category $[C^{op},\mathsf{Set}]$ under the given type of colimit.

If you want to do this in such a way that certain existing colimits are preserved, then you form the closure not in the presheaf category, but in the full subcategory of those presheaves which send the existing colimits in C into limits in Set.

So to freely complete a category $C$ with finite products to a category with finite limits, you should look at the Yoneda embedding of $C$ into the opposite $FP(C,Set)^{op}$ of the category $FP(C,\mathsf{Set})$ of finite-product-preserving functors from $C$ to $\mathsf{Set}$. Now take the closure of the representables under finite limits. That is, finite limits in $FP(C,\mathsf{Set})^{op}$, or finite colimits in $FP(C,\mathsf{Set})$. This then gives the value at $C$ of a left biadjoint to the forgetful 2-functor from the 2-category LEX of categories with finite limits to the 2-category FP of categories with finite products.

There is an explicit, syntactic description, due to Andy Pitts. It can be found, for example, in the paper

M. Bunge and A. Carboni, The symmetric topos, Journal of Pure and Applied Algebra, 1995.

An object is a pair of maps $f,f':X \to X'$ equipped with a common retraction $r$. A morphism from such a pair to $g,g':Y \to Y',s$ consists of an equivalence class, under a suitably defined equivalence relation, of pairs $(a,a')$ where $a:X \to Y$, $a':X'\to Y'$, and the evident diagrams commute.

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 Welcome Steve. You can use LaTeX in your answers (and it compiles on the fly as you type). I did it for you this time. – Andrej Bauer Nov 16 2010 at 10:22 Thanks, Andrej - Steve. – Steve Lack Nov 17 2010 at 0:28

The exact and regular completions don't help because they take a category that already has all finite limits (including equalisers) and because their purpose is to add colimits rather than limits. There is also the Yoneda embedding into the presheaf topos, which does not require limits in the given category but add all set-indexed colimits.

The standard way in which to approach such a question would be to try to define a category whose objects are formal equaliser diagrams $X\rightrightarrows Y$. However, it is not clear how to define morphisms between diagrams $f:(X_1\rightrightarrows Y_1)\to(X_2\rightrightarrows Y_2)$. The obvious way is as an equivalence class of maps $X_1\to X_2$, but that results in making the images of the objects in the given category injective in the new one.

Consider, for example, the categories of sober topological spaces and of locales. In each of these, the full subcategory of locally compact objects (or indeed of algebraic lattices with the Scot topology) generates the whole category via equalisers. Which of these, however, would you regard as the universal completion?

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Is there an a general reason why the completion should exist? – Andrej Bauer Nov 15 2010 at 13:13

I try:

Let $\mathcal{C}$ a category , by induction we build $Y^n\mathcal{C}\subset \mathcal{C}^>$ :

$Y^0\mathcal{C}$ is the image of the Yoneda immersion $h_-: \mathcal{C}\to \mathcal{C}^>$, let $Y^{n+1}\mathcal{C}$ the subcategory of $\mathcal{C}^>$ generated by $Y^{n}\mathcal{C}$ and by the ker diagrams $K \to A \rightrightarrows B$ (by $A, B\in Y^n\mathcal{C}$) and with the arrow induced by morphisms of pairs diagram in $Y^{n}\mathcal{C}$. Let $\mathcal{C}_{K}:= \bigcup_n Y^n\mathcal{C}$. Then for $F: \mathcal{C}\to \mathcal{B}$, $\mathcal{B}$ by Kernels, has for induction a extentions to a Ker-preserving funtor to any $Y^n\mathcal{C}$ (with reference to the diagram above let $F(K):= Ker(F(A) \rightrightarrows F(B))$ and this extention in unique but isomorphisms.

Then $h_-: \mathcal{C} \to \mathcal{C}_K$ is the Ker-completions.

For duality $(h^-)^{ op }: \mathcal{C} \to ((C^{op})_K)^{ op }$ is the Coker-completion.

If above you consider $A$ and $B$ as (finite) product of objects of $Y^n\mathcal{C}$ you get the (finite)limits completion. Dually you can build the (finite)colimit completion.

Alternatively in the proof above you can consider $K$ as coker (instead Ker) of arrow pairs: $A \rightrightarrows B \to K$, then you get $\mathcal{C}_{CK}:= \bigcup_n Y^n\mathcal{C}$ and this has the cokernel's, (or (finite)colimits taking for $A$ and $B$ the (finite) coproduct of objects of $Y^n\mathcal{C}$). But the yoneda functor $h_-: \mathcal{C}\to \mathcal{C}_{CK}$

don't preserve cokernel's , but there is advantage: the Yoneda immersion is full and dense then $\mathcal{C}_{CK}$

PS. I did this proof for the first time here (no read anywhere) I hope this work, and dont get mistake.

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 What is Yoneda immersion what is $\mathcal{C}^{>}$? Also, for this to be called a proof you'd have to show that $\mathcal{C}_K$ has the appropriate universal property. – Andrej Bauer Nov 15 2010 at 13:13 Of course $\mathcal{C}^>$ is the category of presheaves on $\mathcal{C}$: $[\mathcal{C}^{op}, Set]$ this convention is well known and used by Grothendieck and other texts on category theory. That $C_K$ has the universal propriety follow because $F$ has a Ker-preserving extention to any $Y^n\mathcal{C}$ unique but isomorphisms. – Buschi Sergio Nov 16 2010 at 14:20

I suspect that this paper does what you need:

\bib{MR569571}{article}{
author={Kaphengst, H.},
author={Reichel, H.},
title={Finite limit-colimit completions of small categories},
conference={
title={Algebraische Modelle, Kategorien und Gruppoide},
},
book={
series={Stud. Algebra Anwendungen},
volume={7},
place={Berlin},
},
date={1979},
pages={21--33},
review={\MR{569571 (81d:18004)}},
}


I also think that the key construction is explained somewhere in this book:

\bib{MR2178101}{article}{
author={Barr, Michael},
author={Wells, Charles},
title={Toposes, triples and theories},
note={Corrected reprint of the 1985 original [MR0771116]},
journal={Repr. Theory Appl. Categ.},
number={12},
date={2005},
pages={x+288},
review={\MR{2178101}},
}


However, I do not have my copy to hand so I cannot easily check this. The main idea is that the problem can be formulated equationally. For each parallel pair $f,g\:X\to Y$ we need an object $E(f,g)$ and a morphism $k(f,g)\:E(f,g)\to X$. There should be a map $X\to E(1_X,1_X)$ that is inverse to $k(1_X,1_X)$ and also maps $m(f,g,h)\:E(f,g)\to E(hf,hg)$ for all $h\:Y\to Z$. By some cunning means that I don't remember one can write down a few equations for the operators $k(-,-)$ and $m(-,-,-)$ that encapsulate the universal property of equalisers.

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