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I´m reading the Zig Zag lemma in Cohomology and i want to prove the exactness of cohomology sequence at $ H^k(A)$ and $H^k(B)$ :

A short exact sequence of cochain complexes $ 0 \to A \ \xrightarrow{i} \ B \ \xrightarrow{j} \ C \to 0$ gives rise to a long exact sequence in cohomology:

$ ... \ \xrightarrow{j^*} \ H^{k-1}(C) \ \xrightarrow{d^*} \ H^k(A) \ \xrightarrow{i^*} H^k(B) \ \xrightarrow{j^*} H^k(C) \ \xrightarrow{d^*} H^{k+1}(A) \ \xrightarrow{i^*} ...$

where $i^∗$ and $j^∗$ are the maps in cohomology induced from the cochain maps i and j,and $d^∗$ is the connecting homomorphism.

I think first i need to prove that $im(d^∗) = ker(i^∗)$ for exactness in $H^k(A)$ . Help please…..

I prove the exactness of $H^k(C)$:

First I prove that $im( j^*)\subseteq ker (d^*)$. Let $[b]\in H^k(B) $ then $d^* j^* [b] = d^*[j(b)]$. In the recipe above for $d^*$ , we can choose the element in $B^k$ that maps to $j(b)$ to be b. Then $db \in B^{k+1}$. Because b is a cocycle, $db=0$. Following the Zig-Zag diagram we see that since $i(0) = 0 = db$, we must have $d^*[j(b)] = [0]$, so $j^*[b]\in ker(d^*)$.

The other way, i.e., $ker(d^*) \subseteq im(j^*)$: suppose $d^*[c] = [a]=0$, where $[c] \in H^k(C) $, this means that $a=da´$ for some $ a´ \in A^k$.i calculate the $d^*$ again by the diagram and take an element $ b \in B^k$with $j(b) = c$ and $i(a) = db$. Then $b - i(a´)$ is a cocycle in $B^k$ that maps to c under j:

$d(b - i(a´)) = db-di(a´) = db - id(a´) = db - ia = 0$, $j(b - i(a´)) = db-ji(a´) = j(b) = c$ Therefore, $ j^*[b - i(a´)]= [c]$.

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The interested reader can find this proof of exactness at $H^k(C)$ in An introduction to manifolds by Loring W. Tu. – Vladhagen Nov 11 '14 at 23:55
up vote 1 down vote accepted

First of all the proofs aren't very difficult and go through without any complications as well as your proof the exactness at $H^k(C)$ given. For a proof of this I would recommend any standard-literature such as Neukirch, p. 24 (in German) or NSW p.26 using the Snake lemma.

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Yes thanks I could prove what I needed just taking elements in the sequence , thanks :) – Knight Oct 17 '13 at 21:09

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