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?

  • 3
    $\begingroup$ This theorem is equivalent to Snake Lemma. $\endgroup$
    – Rafael
    Sep 17, 2018 at 1:06
  • 1
    $\begingroup$ @RafaelHolanda Please make that into an answer! We should avoid answering questions in the comments. $\endgroup$
    – Pedro
    Sep 17, 2018 at 10:19

1 Answer 1


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


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


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

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

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

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