# Long exact sequence of cohomology group "without" Snake lemma

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?

• This theorem is equivalent to Snake Lemma. Sep 17, 2018 at 1:06
– Pedro
Sep 17, 2018 at 10:19

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

$$0\rightarrow\ker(f)\rightarrow\ker(g)\rightarrow\ker(h)\xrightarrow{\delta^0}coker(f)\rightarrow coker(g)\rightarrow coker(h)\rightarrow0$$

• What the connecting morphism $\delta^n : H^n (G,N) \to H^{n+1}(G, L)$ does in terms of cocycles, here? Jan 18, 2019 at 20:31