Proposition 1.2.10. Let $R$ be a ring and $(\seqi{A}, \bracs{T^i_j|i, j \in I, i \lesssim j)}$ be a downward-directed system of $R$-modules, then there exists $(A, \bracsn{T^A_i}_{i \in I})$ such that:

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

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

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

    \[\xymatrix{ & A_i \\ B \ar@{->}[r]_{S} \ar@{->}[ru]^{S^B_i} & A \ar@{->}[u]_{T^A_i} }\]

    for all $i \in I$.

Proof. Let

\[A = \bracs{x \in \prod_{i \in I}A_i \bigg | \pi_j(x) = T^i_j\pi_i(x) \forall i, j \in I, i \lesssim j}\]

For each $i \in I$, let $T^{A}_{i} = \pi_{i}$, then $(A, \bracsn{T^A_i}_{i \in I})$ satisfies (1) and (2) by definition of $A$.

(U): Let $(B, \bracsn{S^B_i}_{i \in I})$ satisfying (1) and (2). Let

\[S: B \to \prod_{i \in I}A_{i} \quad \pi_{i}(Sx) = S^{B}_{i}\]

then for any $x \in B$ and $i, j \in I$ with $i \lesssim j$,

\[\pi_{j} (Sx) = S^{B}_{j}x = T^{i}_{j} S^{B}_{i}x = T^{i}_{j} \pi_{i}(S x)\]

so $S \in \hom(B; A)$, and the diagram commutes. Since any map $f: B \to A$ is uniquely determined by its composition with the projections, $S$ is unique.$\square$