Does the following property of the direct limit of a direct system follow from the axioms for a direct limit? 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:


*

*$\mu_{ii}$ is the identity homomorphism on $M_i$ for each $i \in I$

*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, 


*

*http://sierra.nmsu.edu/morandi/notes/DirectLimits.pdf, 

*http://dangtuanhiep.files.wordpress.com/2008/09/papaioannoua_solutions_to_atiyah.pdf, and 

*http://math.uchicago.edu/~allanaa/atiyah.pdf .
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?
 A: 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.
A: 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. 
A: One can look at Lemma 5.30 of An introduction to homological algebra by Joseph J. Rotman.
However, here is a proof: (notations are as in the question posed above)
First, we observe the following: Let $y_i\in M_i$ such that $y_i=x_{i1}+...+x_{ik}+z_{j_1}+...+z_{j_m}$ for some $i, j_1,...,j_m\in I$($k,m\in\mathbb{N}$), $x_{i1},...,x_{ik}\in M_i$ $z_{j_m}\in M_{j_m}$, $j_1,...,j_m\neq i$ and let $j\geq i$ in $I$. Then $\mu_{ij}(y_i)=\mu_{ij}(x_{i1}+...+x_{ik})+z_{j_1}+...+z_{j_m}$.
Now, let $x_i\in M_i$ for some $i\in I$ such that $\mu_i(x_i)=0$. Then $x_i\in D$. If $x_i=0$ then we are through. So let $x_i\neq 0$ and write $x_i=a_{1i_1}-\mu_{i_1j_1}(a_{1i_1})+...+a_{ni_n}-\mu_{i_nj_n}(a_{ni_n})...... (\ast)$ for some $i_1,...,i_n\in I$, $a_{ki_k}\in M_{i_k}$ for $1\leq k\leq n$ and  assume $j_k>i_k$ ($1\leq k\leq n$) (this assumption is without loss of generality, for if $i_k=j_k$ for some $k$, then $a_{ki_k}-\mu_{i_kj_k}(a_{ki_k})=a_{ki_k}-a_{ki_k}=0$.
case I: Let $x_i=a_j-\mu_{jk}(a_j)$ for some $k> j\in I$, $a_j\in M_j$. Since we are now working in the direct product $\bigoplus\limits_{l\in I}M_l$, either $j=i$ or $k=i$ for otherwise, $x_i=0$, contrary to our assumption. If $j=i$, then $x_i-a_i=\mu_{ik}(a_i)\in M_i\cap M_k$ (with $k>i$). Therefore, $\mu_{ik}(a_i)=0\Rightarrow x_i=a_i$ and hence $\mu_{ik}(x_i)=0$ as required. On the other hand if $j\neq i$, then $k=i$. Therefore, $x_i-\mu_{ji}(a_j)=a_j \in M_i\cap M_j$ which implies that $a_j=0$ and hence $x_i=\mu_{ji}(a_j)=0$, contrary to our assumption; so this possibility does not arise.
case II:(the general case) First, we rewrite $(\ast)$ as $x_i=y+z+w$, where
$$y=y_{i1}-\mu_{ij_1}(y_{i1})+...+y_{ir}-\mu_{ij_r}(y_{ir}),$$ $$z=z_{1i_1}-\mu_{i_1i}(z_{1i_1})+...+z_{si_s}-\mu_{i_si}(z_{si_s}),$$ $$w=w_{1k_1}-\mu_{k_1l_1}(w_{1k_1})+...+w_{tk_t}-\mu_{k_tl_t}(w_{tk_t}),$$ and $z_{si_s}\in M_{i_s}, w_{ki_k}\in M_{i_k}, y_{im}\in M_i$ for $1\leq m\leq r\in \mathbb{N}$, $i_1,....,i_s<i$ ($s\in \mathbb{N}$) and $r+s+t=n$. Then $\mu_{ij_1}(x_i)=y^\prime+z^\prime+w$, where $$y^\prime= \mu_{ij_1}(y_{i1})-\mu_{ij_1}(y_{i1})+\mu_{ij_1}(y_{i2})-\mu_{ij_2}(y_{i2})+...+\mu_{ij_1}(y_{ir})-\mu_{ij_r}(y_{ir})$$ and $$ z^\prime=z_{i_1}-\mu_{i_1j_1}(z_{i_1})+...+z_{i_s}-\mu_{i_sj_1}(z_{i_s}).$$
 Now let $k\in I$ such that $k\geq j_1,...,j_r,l_1,...,l_t$ and assume without loss of generality that $j_2,...,j_u\neq j_1$ and $j_{u+1}=...=j_r=j_1$ for some $2\leq u\leq r$. Then  $$\mu_{j_1k}(\mu_{ij_1}(x_i))=\mu_{ik}(x_i)=y^{\prime\prime}+z^{\prime\prime}+w^\prime,$$ where $$y^{\prime\prime}=-(\mu_{ij_2}(y_{i2})-\mu_{ik}(y_{i2})+...+\mu_{ij_u}(y_{iu})-\mu_{ik}(y_{iu})),$$ which implies that $$y^{\prime\prime}= -(\mu_{ij_2}(y_{i2})-\mu_{j_2k}(\mu_{ij_2}(y_{i2}))+...+\mu_{ij_u}(y_{iu})-\mu_{j_uk}(\mu_{ij_u}(y_{iu}))),$$ $$z^{\prime\prime}=z_{1i_1}-\mu_{i_1k}(z_{1i_1})+...+z_{si_s}-\mu_{i_sk}(z_{si_s})$$ and if $k_1=k_2=...=k_m=j_1=l_{m+1}=l_{m+2}=...=l_{m+p}$, then
$$w^\prime= \mu_{j_1k}(w_{1k_1})-\mu_{k_1l_1}(w_{1k_1})+...+\mu_{j_1k}(w_{mk_m})-\mu_{k_ml_m}(w_{mk_m})+w_{(m+1)k_{m+1}}-\mu_{k_{m+1}k}(w_{(m+1)k_{m+1}})+...+w_{(m+p)k_{m+p}}-\mu_{k_{m+p}k}(w_{m+pk_{m+p}}) +......$$
(the remaining terms from $w$).
Thus $\mu_{ik}(x_i)$ is a sum of at most $n-1$ many generators of $D$ (we started with $x_i$ as a sum of $n$ many generators of $D$). The result now follows by induction on $n$ (the case for $n=1$ is settled in \textbf{case I} above).
A much easier way to prove this result is via the alternate description of the direct limit as the disjoint union of $M_i$'s modulo an equivalence relation  as pointed out by Dylan above.
