Mittag-Leffler Condition and fiber products

Question

For a general inverse system of short exact sequences $ 0\rightarrow A_i\rightarrow B_i \rightarrow C_i \rightarrow 0$ for some index $I$, we only get a left exact sequence $$0\rightarrow \lim A_i \rightarrow \lim B_i \rightarrow \lim C_i.$$ If $I$ is ordered, and not just partially ordered, there exists a sufficient condidtion on $(A_i)_i$ which ensures that the above left exact sequence is in fact exact. My question is whether there are some special cases where we can make a similar statements of only partial ordered $I$. The case I'm most interested about is the following: Consider $$A_1\rightarrow A_0 \leftarrow A_2$$ $$B_1\rightarrow B_0 \leftarrow B_2$$ $$C_1 \rightarrow C_0 \leftarrow C_2$$ and short exact sequences $0\rightarrow A_i\rightarrow B_i \rightarrow C_i \rightarrow 0$ of abelian groups. We get a left exact sequence $$0\rightarrow A_1\times_{A_0} A_2\rightarrow B_1\times_{B_0}B_2 \rightarrow C_1\times_{C_0} C_2.$$ When is this sequence exact?

Answer 1

First of all, a convenient way to look at fibered products in this context is the following:

Denoting $\alpha_i: A_i \rightarrow A_0 , \;i=1,2$, the fibered product $A':=A_1\times_{A_0}A_2$ can be described as the kernel of the map $f_A:A_1 \oplus A_2 \stackrel{\begin{pmatrix}\alpha_1 \\ -\alpha_2\end{pmatrix}}{\longrightarrow} A_0$, so that one has an exact sequence $$0 \rightarrow A' \rightarrow A_1 \oplus A_2 \rightarrow A_0,$$ and similarly for $B$'s and $C's$.

Now, your assumption implies that we have a commutative diagram with exact rows

\begin{array} \\ 0 & \rightarrow & A_1\oplus A_2 & \rightarrow & B_1\oplus B_2 & \rightarrow &C_1\oplus C_2 & \rightarrow & 0 \\ & & f_A \downarrow & & f_B \downarrow & & f_C \downarrow & & \\ 0 & \rightarrow & \;\;\;\;A_0 & \rightarrow &\;\;\;\; B_0 & \rightarrow & \;\;\;\;C_0 & \rightarrow & 0, \\ \end{array}

So by the snake lemma, we obtain an exact sequence

$$0 \rightarrow A' \rightarrow B' \rightarrow C' \stackrel{\delta}{\rightarrow} \mathrm{Coker}\,f_A \cdots$$

So, the sufficient and necessary condition is that $\delta$ is the zero morphism. This is not easy to control for, so a perhaps-better condition, sufficient for exactness, is that $f_A$ is surjective. This is equivalent to $A_1 \rightarrow A_0$, $A_2 \rightarrow A_0$ being "jointly surjective" - meaning that the sum of images of the two maps is the whole group $A$.

(Note: This argument is actually not that far from justification for why the Mittag-Leffler conditions works for inverse ($\mathbb{N}$-indexed) limits.)