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Suppose that we have the following commutative diagram of graded Lie algebras

$$\begin{array} A 0& {\longrightarrow} & C_n & {\longrightarrow} & A_{n+1} &{\longrightarrow} & A_n & {\longrightarrow} &0 \\ & & \downarrow{} & &\downarrow{}& &\downarrow{}\\ 0& {\longrightarrow}&D_n & \stackrel{}{\longrightarrow} &B_{n+1} & \stackrel{}{\longrightarrow} & B_n & {\longrightarrow}&0 & \end{array}$$ for all $n\in\mathbb{Z}^+$, where the both rows are split exact sequences and every vertical map is onto.

My question is, suppose that we know $A_n$, $C_n$, $D_n$ for all $n\in\mathbb{Z}^+$ and $B_1$, do we have enough information to uniquely determine up to isomorphism every $B_n$ (likely by induction)?

Here by "know", I mean that we can write $A_n$ (similarly $C_n, D_n, B_1$) as the free Lie algebra modulo some known relations $$B_n=L[S_n]/I_n,$$ where $L[S_n]$ is the free Lie algebra generated by a set $S_n$ and $I_n$ is 2-sided Lie ideal generated by some known relations.

If the answer to my question is positive, then is there a systematic way to determine the structure of $B_n$?

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It looks to me like it is true that $B_n$ is uniquely determined. Let $K_n$ be the kernel of the map $A_n\to B_n$ and let $E_n$ be the kernel from $C_n$ to $D_n$. Then we have an exact sequence $0\to E_n\to K_{n+1}\to K_n$. $E_n$ is known and $K_n$ is known inductively. That means that $K_{n+1}$ can be determined as the subspace of $A_{n+1}$ generated by all preimages of $K_n$ and images f $D_n$. Therefore $B_{n+1}=A_{n+1}/K_{n+1}$ is uniquely determined.

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  • $\begingroup$ Thank you for your answer! May I clarify two things? 1. Why do we have the exact sequence $0\to D_n\to K_{n+1}\to K_n\to 0$? 2. You mentioned that $K_n$ is known inductively (from the exact sequence?). If we can find $K_n$ inductively from the exact sequence $0\to D_n\to K_{n+1}\to K_n\to 0$, then why not finding $B_n$ directly from the exact sequence $0\to D_n\to B_{n+1}\to B_n\to 0$? $\endgroup$ – Zuriel Nov 22 '14 at 7:28
  • $\begingroup$ @Zuriel: I've sketched the inductive step. Assume $K_n$ is known, and then this argument tells you how to find $K_{n+1}$. (Try it out with $n=1$.) $\endgroup$ – Cheerful Parsnip Nov 22 '14 at 11:50
  • $\begingroup$ @Zuriel: the reason you can't do this directly with $B_{n+1}$ is that an exact sequence with known first and last term does not determine the middle term! However, it does when we are looking at subspaces of a known exact sequence. $\endgroup$ – Cheerful Parsnip Nov 22 '14 at 11:51
  • $\begingroup$ @Zuriel: also, there was a typo! The kernel should have been $E_n$. $\endgroup$ – Cheerful Parsnip Nov 22 '14 at 11:56
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    $\begingroup$ If you have an exact sequence $0\to A\to B\to C\to 0$ which includes in an exact sequence $0\to A'\to B'\to C'\to 0$, then $B<B'$ is the set of all preimages of elements of $C$ and images of $A$. $\endgroup$ – Cheerful Parsnip Nov 24 '14 at 11:23

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