Let an $R$-module $C$ be a direct limit of finitely presented $R$-modules $C_i$, and we have a short exact sequence as follows:
$$0→A↪B\stackrel{\pi}→C→0.\qquad (S)$$
From each $C_i$ to the direct limit $C$ we have a homomorphism $γ_i$ which we suppose that could be lifted to a homomorphism $λ_i:C_i→B$. Let $B_i=\{(x,y)∈B⊕C_i : π(x)=γ_i(y)\}$ be the pullback of $π$ and $γ_i$. The rule $ρ_i(y)=(λ_i(y),y)$ for $y$ in $C_i$ defines a splitting from $C_i$ to $B_i$ for $π_i: B_i→C_i$. Now we get split s.e.s.:
$$0→A→B_i\stackrel{\pi_i}→C_i→0.\qquad (S_i)$$
My question is how could one prove that $B$ is the direct limit of $B_i$? If so, $(S)$ would be the direct limit of the sequences $(S_i)$.
THANKS IN ADVANCE!
For every $i$ we have $C_i=B_i/A$. Taking direct limit, when $B'$ denotes the limit of $(B_i)$, we get $$C=B'/A,$$ and $B'$ is a submodule of $B$. But we know that $C=B/A$, hence $B'=B$.