TITLE: Long exact sequence of cohomology group "without" Snake lemma
QUESTION [5 upvotes]: Let a short exact sequence
$$
0 \to L \to M \to N \to 0 
$$
is a short exact sequence of $G$-modules, then a long exact sequence is induced:
$$
0\longrightarrow L^G \longrightarrow M^G \longrightarrow N^G \overset{\delta^0}{\longrightarrow} H^1(G,L) \longrightarrow H^1(G,M) \longrightarrow H^1(G,N) \overset{\delta^1}{\longrightarrow} H^2(G,L)\longrightarrow \cdot
$$
The connecting homomorphism is,
$$
\delta^n : H^n (G,N) \to H^{n+1}(G, L)
$$

Question 1: How to prove the above long exact sequence is true? Is this simply based on the Snake Lemma [It's My Turn (1980)]? Or is there other simpler way to think about it, without using the Snake Lemma?

REPLY [6 votes]: Let's see that the long exact sequence theorem is equivalent to the snake lemma. 
From the short exact sequence we have an induced diagram
$$\require{AMScd}\begin{CD}&&L^n/B^n(G,L)@>>>L^n/B^n(G,M)@>>>L^n/B^n(G,N)@>>>0\\&@VVV@VVV@VVV\\0@>>>Z^{n+1}(G,L)@>{}>>Z^{n+1}(G,M)@>>>Z^{n+1}(G,N)\end{CD}$$
Applying snake lemma we have an exact sequence
$$H^n(G,L)\rightarrow H^n(G,M)\rightarrow H^n(G,N)\xrightarrow{\delta^n}H^{n+1}(G,L)\rightarrow H^{n+1}(G,M)\rightarrow H^{n+1}(G,N).$$
For other hand, a commutative diagram of $G$-modules
$$\require{AMScd}\begin{CD}0@>>>A^0@>>>B^0@>>>C^0@>>>0\\&@V{f}VV@V{g}VV@V{h}VV\\0@>>>A^1@>>>B^1@>>>C^1@>>>0\end{CD}$$
can be viewed as a short exact sequence between complexes $0\rightarrow A\rightarrow B\rightarrow C\rightarrow0$. Then there is an exact sequence between its cohomologies
$$0\rightarrow\ker(f)\rightarrow\ker(g)\rightarrow\ker(h)\xrightarrow{\delta^0}coker(f)\rightarrow coker(g)\rightarrow coker(h)\rightarrow0$$