Direct Limits of Vector Spaces: Confusion about Definition of Mappings Given First, I give the definitions I am using for the question. They are essentially those found on the Wikipedia page concerning Direct Limits.

Let $\{V_i\}_{i\in I}$ be a family of vector spaces in $Vec_\mathbb{F}$, where $I$ is a directed set. For each $i\le j$, let $f_{ij}:V_i\to V_j$ be a morphism satisfying the following axioms:
$(1)$ $f_{ii}=Id_{V_i}$
$(2)$ $f_{ik}=f_{jk}\circ f_{ij}$ for all $i\le j\le k$. The pair $(\{V_i\},\{f_{ij}\})$ is called a direct system over $I$.
Let $C$ be the direct sum of the $V_i$'s:
$$C=\bigoplus_{i\in I}V_i $$
and let $\iota_i:V_i\to C$ be the natural inclusions of the $V_i$ into $C$. Let $D$ be the subspace of $C$ generated by elements of the form $x_i-f_{ij}(x_i)$ where $i\le j$ and $x_i\in V_i$. Let
$$\mathbf{V}=\lim_{\to}V_i:=C/D. $$
Let $\mu:C\to C/D$ be the quotient map and let $\mu_i=\mu\circ \iota_i:V_i\to\mathbf{V}$ be the restriction of $\mu$ to $V_i$. Since $D$ depends on the maps $\{f_{ij}\}$, so do $C/D$ and the maps $\{\mu_i\}$. The pair $(\mathbf{V},\{\mu_i\}_{i\in I})$ is called the Direct Limit of the system $(\{V_i\}_{i\in I},\{f_{ij}\}_{i\le j\in I}).$

Now that that is out of the way, the question I have been struggling with is the following:
Question: Show that each $[v]\in \mathbf{V}$ can be written as $[v]=\mu_i(v_i)$ for some $i\in I$ and $v_i\in V_i$.
Attempt: Let's assume $[v]$ is the equivalence class of some $v_i\in V_i$, one of the vector spaces in the direct summation. We can write this as $[v_i]$. Then, it is true by definition that $\mu_i(v_i)=[v_i]$.
Now, what if we have $v=v_i+v_j+v_k$, for $v_i\in V_i, v_j\in V_j, v_k\in V_k$, for $i\le j\le k\in I$. The enticing thing to do is to consider the morphisms $\mu_k:V_k\to C/D$. Because $V_k$ is the farthest along in the system, we can consider
$$\mu_k(v_{k'})=\mu_k(f_{ik}(v_i)+f_{jk}(v_j)+f_{kk}(v_k))=[v].$$
This construction should work for any such vector $v$, we just need to use the properties of a directed set to find $\ell\in I$ such that $\ell$ is $\ge$ than the rest of the indices of the vector spaces used in "creating" the element.
Thoughts: Truth be told I'm not certain about that attempt, but I suppose my confusion arises from how these $\{\mu_i\}$ can be made to act on elements of $V_i\oplus V_j\oplus V_k$, if not in this manner.
 A: An official answer after six years? Why not, it is nice to have a reference in this forum.
Let us first observe that

Let $D$ be the subspace of $C$ generated by elements of the form $x_i-f_{ij}(x_i)$ where $i\le j$ and $x_i\in V_i$

is imprecise. The formally correct definition is

Let $D$ be the subspace of $C$ generated by elements of the form $\iota_i(x_i)-\iota_j(f_{ij}(x_i))$ where $i\le j$ and $x_i\in V_i$.

For $i \le j$ we have
$$\mu_j \circ f_{ij} = \mu_i \tag{1} .$$
In fact, for $v_i \in V_i$ we have $\mu_i(v_i) = [\iota_i(v_i)]$ and $(\mu_j \circ f_{ij})(v_i) = [\iota_j(f_{ij}(v_i))]$ so that $\mu_i(v_i) - (\mu_j \circ f_{ij})(v_i) = [\iota_i(v_i) - \iota_j(f_{ij}(v_i))] = 0$, i.e. $(\mu_j \circ f_{ij})(v_i) = \mu_i(v_i)$.
Now let $[v] \in \mathbf V$. The element $v \in C$ has the form $v = (v_i)_{i \in I}$ with $v_i \in V_i$ and all but finitely many $v_i = 0$.
If $v = 0$, we have $[v] =  \mu_i(0)$ for each $i$.
If $v \ne 0$, let $i_1,\ldots,i_n$ be the indices such that $v_{i_k} \ne 0$. Then $v = \sum_{k=1}^n \iota_{i_k}(v_{i_k})$. Since $I$ is directed, we find $r \in I$ such that $i_k \le r$ for all $k$. Using $\text{(1)}$  we get
$$[v] = \sum_{k=1}^n \mu_{i_k}(v_{i_k}) =  \sum_{k=1}^n \mu_r(f_{i_k r}(v_{i_k})) = \mu_r\left(\sum_{k=1}^n f_{i_k r}(v_{i_k})\right) = \mu_r(v_r)$$
with $v_r =  \sum_{k=1}^n f_{i_k r}(v_{i_k}) \in V_r$.
Note that an alternative construction of the direct limit can be found in this wikipedia article.
