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is it possible to characterize the elements of a (special) direct limit only using the universal property? in detail:

let's first concentrate on the category of sets. by an element, I mean a morphism defined on the terminal object $\{*\}$. let $A_1 \to A_2 \to ...$ be a sequence of sets, and $A$ their colimit which is defined by the universal property (morphisms on $A$ are compatible morphisms on the $A_n$). can we prove, without using an explicit construction of $A$, that every element of $A$ comes from an element in one of the $A_n$ and that two such elements are equal iff they are equal in some greater $A_m$?

for example, it can be shown without using an explicit construction that the coproduct $M+N$ of two sets $M,N$ (or an arbitrary family of sets) consists of the elements of $M$ and of $N$ and they exclude each other: denote $i_M,i_N$ the units ("inclusions"), then it is easy to see that $M,N \to im(i_M) \cup im(i_N)$ also satisfy the universal property so that $M + N = im(i_M) \cup im(i_N)$. now if $x \in im(i_M) \cap im(i_N)$, there are $m \in M, n \in N$ such that $x = i_M(m) = i_N(n)$. define $M \to \{0,1\}$ by sending $m$ to $0$, and the rest to $1$, and $N \to \{0,1\}$ by sending $n$ to $1$, and the rest to $0$. this yields $M + N \to \{0,1\}$ sending $x$ to $0$ and to $1$, contradiction.

perhaps this is not possible for directed limits because we have to use at least some specific properties of sets. but what about topoi or cosmoi (bicomplete closed symmetric monoidal categories)? I'm mainly interested in sets here, but perhaps more abstractions are needed to make clear what the goal is: a pure categorical approach to the usual constructions.

here is a functorial reformulation: let $... \to A_2 \to A_1$ be a sequence of (representable) endofunctors of sets and $A$ their explicitely constructed limit, i.e. $A(M)$ consists of compatible elements of the $A_n(M)$. how can we prove directly that every natural transformation $A \to id$ factors through one of the projections $A \to A_n$?

even if it is impossible, let's take this as a sort of axiom and consider algebraic structures (the usual ones, over sets). can we prove that the forgetful functor preserves direct limits of $A_1 \to A_2 \to ...$? again, without already knowing it by an explicit construction.

ps: there are lots of related questions. they are in the same spirit as my earlier one which has not been answered (although the rules for the bounty decided to accept an answer).

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Martin, you need to use two backslashes to write a set brace. –  Mariano Suárez-Alvarez Jan 6 '10 at 17:29
Martin, proving that these universal objects exist in a given (concrete) category pretty much requires an explicit construction. There are some time-saving lemmas that will help you save time. For example, limits exist iff equalizers and products exist, but at some point, you are forced to perform an explicit construction or some other kind of proof of existence (maybe by reductio). –  Harry Gindi Jan 6 '10 at 17:53
I think part of the problem is that it's not clear what the background theory of sets is supposed to be here, what the axioms are that enable you to say, "it is easy to see that..." To conclude for instance that every 1 -> X + Y factors through one of the coproduct inclusions, you need something more than the fact that Set is a topos (since this statement doesn't hold for all toposes). So, what is the background theory you are using to prove these assertedly clear statements? –  Todd Trimble Jan 6 '10 at 18:29
No, it's not. Take the topos of sheaves of sets on a space X with multiple path connected components. –  Reid Barton Jan 6 '10 at 18:47
Martin, it's false (externally) in Set/2, where 2 = 1+1. The terminal object T is id: 2 -> 2, and the unique map !: T --> T factors through neither or the subterminal objects given by the two coproduct inclusions 1 -> T. The discussion could be refined by taking into account the subtleties of internal logic in a topos, but this type of thing is not manifest in your account. –  Todd Trimble Jan 6 '10 at 18:48
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1 Answer

up vote 4 down vote accepted

As with that previous question, I don't understand precisely what the rules of the game are when you say "without using an explicit construction". But maybe I can say something useful.

First I'll answer the first part of the first question. Then I'll use it to explain a little of why I don't think there's going to be a precise way to formulate the "no explicit construction" rule.

We have a sequence $A_1 \to A_2 \to \cdots$ in the category of sets, and its colimit $A$, and we wish to show that every element of $A$ comes from an element of some $A_n$. And we wish to do it without using the explicit formula for colimits. Being a category theorist, I'll write 1 rather than $*$ for a one-element set.

I'll use the fact that the category of sets is a well-pointed topos, E say. (Those hypotheses can probably be weakened.) Take an element $a$ of $A$, that is, a map $a: 1 \to A$. Since colimits are stable under pullback in any topos, pulling the colimit cocone back along $a$ gives a sequential colimit $$ \mathrm{colim}(X_1 \to X_2 \to \cdots) = 1 $$ in the category E, together with a map $f_n: X_n \to A_n$ for each $n$, making the evident square commute. (Draw a diagram!) Since any colimit of initial objects is initial, and in a well-pointed topos $1$ is not initial, at least one $X_n$ is not initial. It's a fact that in a well-pointed topos, every object is either initial or admits an element (map from $1$). So, at least one $X_n$ admits an element, $x$ say. Then $f_n x$ is an element of $A_n$, and a quick diagram chase shows that it maps to $a \in A$.

So, I've answered the first question without apparently using any explicit constructions involving sets --- in the sense that I just assumed certain axioms on the category of sets and did the proof categorically. But the thing is, you can always do that. "Explicit constructions" can always be translated into categorical arguments (and though it's not apparent from the proof above, that's a totally mechanical process).

If your point of view on sets is that they are what ZFC says they are, then here's an equivalent categorical formulation: sets and functions form a well-pointed topos with natural numbers object and choice, satisfying a first-order axiom scheme of replacement. ZFC and this entirely categorical axiomatization are in a precise sense equivalent. Anything you can do in one context, you can do in the other.

On coproducts: it can be shown that in any topos, coproducts are disjoint and the coprojections are jointly epic. I think that's in Mac Lane and Moerdijk's book Sheaves in Geometry and Logic. In a well-pointed topos, epic = surjective and monic = injective (where by sur/injective I'm referring to the elementwise notion implicit in the question). Hence your statements on coproducts hold in any well-pointed topos.

On the question about algebraic structures:

Colimits (="direct limits") of sequences are an example of filtered colimits --- see Categories for the Working Mathematician Chapter IX, for instance. The forgetful functors Group$\to$Set, Ring$\to$Set, etc, all preserve filtered colimits. The jargon for this is: "the free group/ring/... monad is finitary". As the terminology hints, preservation of filtered colimits corresponds to the fact that the theory of groups, rings, etc. only involves finitary operations.

For example, the theory of groups has an operation with 2 arguments (multiplication), an operation with 1 argument (inverse), and an operation with 0 arguments (identity). The numbers 2, 1, and 0 are all finite, so the theory of groups is finitary, so the forgetful functor Group$\to$Set preserves filtered colimits.

(A non-example would be the theory of ordered sets in which every countable subset had a supremum, and maps that preserved those suprema. One of the operations in the theory is "take the supremum of a countably infinite subset", so this theory isn't finitary. Correspondingly, the forgetful functor from these ordered sets to Set won't preserve filtered colimits.)

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