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Question: Does it follow from the axioms for a direct limit that if $\mu_i(x_i)=0$ then there exists $j \geq i$ such that $\mu_{ij}(x_i)=0$?

Definitions and notation:

(Atiyah MacDonald, chapter 2, question 14 and 15 give the following construction for a direct limit of modules over a ring):

Begin with a directed set $I$. This is a partially ordered set $(I,\leq)$ such that for every $i,j \in I$, there exists $k \in I$ such that $i \leq k$ and $j \leq k$.

If $A$ is a ring, $I$ is a directed set, and $(M_i)_{i \in I}$ a family of modules with $A$-module homomorphisms $\mu_{ij}: M_i \rightarrow M_j$ for each pair $i \leq j$ such that the following axioms hold:

  1. $\mu_{ii}$ is the identity homomorphism on $M_i$ for each $i \in I$
  2. if $i \leq j \leq k$, then $\mu_{ik}=\mu_{jk}\circ\mu_{ij}$

then $(M_i,\mu_{ij})$ is called a direct system over $I$.

The direct limit of $(M_i,\mu_{ij})$ is constructed as follows:

Take $C=\bigoplus_{i \in I} M_i$, and let $D = \langle x_i - \mu_{ij}(x_i) | x_i \in M_i, i \leq j \rangle \leq C$. Identify each $M_i$ with its image in $C$. Form the quotient via $\mu: C \rightarrow M:=C/D$, and let $\mu_i$ be the restriction of $\mu$ to $M_i$.

The module $M$, together with the homomorphisms $\mu_i$, is the direct limit of the direct system $(M_i,\mu_{ij})$.

Note: This is part of problem 2.15 from Atiyah-MacDonald. There are several attempts at this problem available; for example,

The better solutions of these rewrite the axioms in terms of the construction of a stalk (for example, in Hartshorne II.1, or the second paragraph of the answer here).

I would like to know whether the property of a direct limit that if $\mu_i(x_i)=0$ then there exists $j \geq i$ such that $\mu_{ij}(x_i)=0$ follows directly from the axioms given, or whether the stalk construction can be shown to be equivalent to the axioms given above?

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In mathematics everything always follow from the axioms (note that a definition of a mathematical notion is actually a scheme of axioms, or equivalent axioms). –  Asaf Karagila May 31 '11 at 6:43
my previous response was incorrect: I'm sorry to say that I didn't carefully read the question. I'll try again later on... –  Pete L. Clark May 31 '11 at 18:15
It seems to me that the question has been misunderstood so far. I think that it is asked for a proof for this property "an element is zero in the direct limit iff it is eventually zero" which only uses the defining universal property of the direct limit. So no explicit construction (or cheating like in Dylan's answer) is allowed. For related questions, see mathoverflow.net/questions/10930 and mathoverflow.net/questions/86923 –  Martin Brandenburg Apr 14 '12 at 7:55

2 Answers 2

It might be clearer to say that colimits are characterized in a certain way, that (as usual with categorical things) proves uniqueness up to unique isomorphism. Existence is often proven by giving a construction, indeed. The kind of construction you gave succeeds in any category with coproducts, producing the colimit as a quotient. The property that vanishing in the colimit implies vanishing somewhere along the way ("in finite time"?) does hold in categories of modules: any relation in the colimit involves only finitely-many things, which appear in finite time.

But this property cannot be completely general, because it definitely fails in categories of topological vector spaces, where that quotient must be by the closure of all the relations, in order for the quotient to be Hausdorff.

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A direct limit of a system $(M_i, \mu_{ij})$ is an appropriate family of objects satisfying the universal property. Here Atiyah and Macdonald have constructed a $(M, \mu_i)$ which does the job. It seems like you're worried that a certain property of this entity might come from the particular construction given.

But if $N$ and $\nu_i\colon M_i \to N$ do the job just as well, then there is a (unique) isomorphism $\alpha\colon M \to N$ such that $\alpha \circ \mu_i = \nu_i$ for all $i$. If $\nu_i(x_i) = 0$ then $\alpha(\mu_i(x_i)) = 0$, and hence $\mu_i(x_i) = 0$ because $\alpha$ is an isomorphism. So you are back to exercise 15.

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