Proposition 1.2.9. Let $R$ be a ring and $(\seqi{A}, \bracsn{T^i_j| i, j \in I, i \lesssim j})$ be an upward-directed system of $R$-modules, then there exists $(A, \bracsn{T^i_A}_{i \in I})$ such that:

  1. For each $i \in I$, $T^{i}_{A} \in \hom({A_i, A})$.

  2. For any $i, j \in I$ with $i \lesssim j$, the following diagram commutes:

    \[\xymatrix{ A_i \ar@{->}[rd]_{T^i_A} \ar@{->}[r]^{T^i_j} & A_j \ar@{->}[d]^{T^j_A} \\ & A }\]
  3. For any pair $(B, \bracsn{S^i_B}_{i \in I})$ satisfying (1) and (2), there exists a unique $S \in \hom({A, B})$ such that the following diagram commutes

    \[\xymatrix{ A_i \ar@{->}[d]_{T^i_A} \ar@{->}[rd]^{S^i_B} & \\ A \ar@{->}[r]_{g} & B }\]

    for all $i \in I$.

Proof. Let $M = \bigoplus_{i \in I}A_{i}$. For any $i, j \in I$ with $i \lesssim j$ and $x \in A_{i}$, let $x_{i, j}\in M$ such that for any $k \in I$,

\[\pi_{k}(x_{i, j}) = \begin{cases}x&k = i \\ T^{i}_{j} x&k = j \\ 0&k \ne i, j\end{cases}\]

Let $N \subset M$ be the submodule generated by $\bracs{x_{i, j}|i, j \in I, i \lesssim j, x \in A_i}$, $A = M/N$, and $\pi: M \to M/N$ be the canonical map.

(1): For each $i \in I$, let

\[T^{i}_{M}: A_{i} \to M \quad \pi_{k} T^{i}_{M} x = \begin{cases}x &k = i \\ 0 &k \ne i\end{cases}\]

and $T^{i}_{A} = \pi \circ T^{i}_{M}$.

(2): Let $i, j \in I$ with $i \lesssim j$, then for any $x \in A_{i}$, $T^{i}_{M}x - T^{j}_{M} T^{i}_{j} x \in N$. Hence $T^{i}_{A}x = T^{j}_{A} T^{i}_{j}x$.

(U): Let

\[S_{0}: M \to B \quad x \mapsto \sum_{i \in I}S^{i}_{B} \pi_{i} x\]

then $S_{0}$ is the unique linear map such that $S_{0} \circ T^{i}_{M} = S^{i}_{B}$ for all $i \in I$. For any $i, j \in I$ with $i \lesssim J$, $S^{i}_{B} x = S^{j}_{B} T^{i}_{j} x$, so $\ker S_{0} \supset N$. By the first isomorphism theorem, there exists a unique $S \in \hom(A; B)$ such that $S_{0} = S \circ \pi$.$\square$