# Induced short exact sequence $L_i / L_{i-1} \to M_i / M_{i-1} \to N_i / N_{-1}$ from $L \to M \to N$ and $M_{i-1} < M_i < M$

Let $$L \xrightarrow{f} M \xrightarrow{g} N$$ be a short exact sequence of $$R$$-modules, and assume that there is a chain of submodules $$0 = M_0 < M_1 < \dotsb < M_n = M$$ in which the quotient $$M_i / M_{i-1}$$ is simple for each $$i$$. Setting $$L_i = f^{-1}(M_i)$$ and $$N_i = g(M_i)$$. Show that there is short exact sequence $$L_i / L_{i-1} \to M_i / M_{i-1} \to N_i / N_{i-1}$$ for each $$i$$.

Can I approach the prove using the canonical homomorphisms, and can I have a chain, say $$0 = L_0 < L_1 < \dotsb < L_n = L$$ from $$f^{-1}$$ and $$0 = N_0 < N_1 < \dotsb < N_n = N$$ from $$g$$.

I am stuck because everywhere I found short exactness, it is like this $$0 \rightarrow L \xrightarrow{f} M \xrightarrow{g} N \rightarrow 0$$. I get confused. Please help.

• math.stackexchange.com/questions/629735 Commented Jan 9, 2014 at 15:04
• If you're told that such a sequence as yours is short exact, you can simply add the initial and final 0 and the appropriate arrows; these are implied.
– Nick
Commented Jan 9, 2014 at 15:07
• @MartinBrandenburg I have seen it but there were no comment on it. I need some ideas or hints. Commented Jan 9, 2014 at 15:10

By constructions of $$L_i$$ and $$N_i$$, you have exact sequences $$L_i \rightarrow M_i \rightarrow N_i \rightarrow 0 \,,$$ obtained by restricting $$f$$ and $$g$$. This fits in a commutative diagram: $$\require{AMScd} \begin{CD} 0 @>>> L_{i-1} @>>> L_i @>>> L_i / L_{i-1} @>>> 0 \\ @. @VVV @VVV @VVV @. \\ 0 @>>> M_{i-1} @>>> M_i @>>> M_i / M_{i-1} @>>> 0 \\ @. @VVV @VVV @VVV @. \\ 0 @>>> N_{i-1} @>>> N_i @>>> N_i / N_{i-1} @>>> 0 \\ @. @VVV @VVV @VVV @. \\ {} @. 0 @. 0 @. 0 @. {} \end{CD}$$