Assume that the complex $$\cdots \xrightarrow{c_{i+2}} L \xrightarrow{c_{i+1}} M \xrightarrow{c_i} N \xrightarrow{c_{i-1}} \cdots$$
is exact and $L$ and $M$ are Noetherian. Prove that $M$ is Noetherian.
Attempt: I know there are two ways I can try to prove $M$ is Noetherian. Every submodule is finitely generated or every increasing chain of submodules of $M$ terminates. I will try to prove that any increasing sequence of submodules of $M$ terminates. This is a problem from the book I am using, Algebra Chapter 0. It gives a hint to solve this problem. It says to decompose the exact sequence into short exact sequences.
Doing this,I get $$0 \rightarrow \text{ker}c_{i+1}=\text{im}c_{i+2} \rightarrow L \rightarrow \text{im}c_{i+1}=\text{ker}c_i \rightarrow 0$$
$$0 \rightarrow \text{ker}c_i=\text{im}c_{i+1} \rightarrow M \rightarrow \text{im}c_i=\text{ker}c_{i-1} \rightarrow 0$$
Can I combine the two sequences to get a short exact sequence $$0 \rightarrow L \xrightarrow{\alpha} M \xrightarrow{\beta} N \rightarrow 0$$
and does this simplify the problem at all? Or if this isn't so, does this yield a short exact sequence with a submodule of $L$ embedding into $M$ and a submodule of $N$ being isomorphic to $M$ mod the submodule of $L$? If I can combine the two sequences in this way I can take a sequence $\{M_k\}$ of submodules of $M$ such that $M_1 \subset M_2 \subset \cdots$. Then $$\alpha^{-1}(M_1) \subset \alpha^{-1}(M_2) \subset \cdots $$ terminates for some $N$. Similarly $$\beta(M_1) \subset \beta(M_2) \subset \cdots$$ terminates for some $N'$. Choose $K=\max\{N,N'\}$
Then applying the five lemma to
$$\begin{array}{c} 0 & \longrightarrow & \alpha^{-1}(M_K) & \longrightarrow & M_K & \longrightarrow & \beta(M_K) & \longrightarrow & 0 \\ & & \downarrow && \downarrow && \downarrow & \\ 0 & \longrightarrow & \alpha^{-1}(M_{K+1}) & \longrightarrow & M_{K+1} & \longrightarrow & \beta(M_{K+1}) & \longrightarrow & 0.\end{array}$$
it follows that $M_K=M_{K+1}=\cdots$ for all $n\geq K$, since the map $M_K \rightarrow M_{K+1}$ is an isomorphism.
Is this the correct way to solve this problem? I am curious as to whether I am allowed to conclude the existence of the short exact sequence $$0 \rightarrow L \xrightarrow{\alpha} M \xrightarrow{\beta} N \rightarrow 0$$
and if this was what the author of Chapter 0 intended when giving the hint to consider the decomposition of the exact sequence into short exact ones. If this is wrong, what is the correct way to use the hint to solve this problem?
The point is that you already have the exact sequence you need. As you say, you have an exact sequence
$$0\to\operatorname{im}(c_{i+1})\to M\to\ker(c_{i-1})\to0.$$
Now $\operatorname{im}(c_{i+1})\simeq L/\ker(c_{i+1})$, which is a Noetherian module because $L$ is, and $\ker(c_{i-1})\subseteq N$ is Noetherian because $N$ is. Now you should recall (or prove for yourself) the standard fact that if one has an exact sequence $$0\to M'\to M\to M''\to0$$
with $M'$ and $M''$ Noetherian then $M$ is Noetherian as well.