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]$.
Best Answer
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.