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]$.
Tu=[Tu, L. W.: Introduction to Manifolds, 2nd ed., New York: Springer, 2011.]
Spivak=[Spivak, M.: A Comprehensive Introduction to Differential Geometry, Vol. 1, Berkeley: Publish or Perish, 1979.]
I use the notations in Tu, p.285, Theorem 25.6.
The exactness at $H^k(\mathcal{C})$ [$\text{ im }j^*=\ker d^*$] is proved in Tu, p.286, l.5--l.$-$4. $\ker i^*\subset \text{ im }d^*$ [one part of exactness at $H^k(\mathcal{A})$] is proved in Spivak, p.576, l.$-$7--l.$-$1.
Proof of $\text{ im }d^*\subset \ker i^*$ [the other part of exactness at $H^k(\mathcal{A})$]:
Let $c\in \mathcal{C}^{k-1}$ such that $dc=0$. Then
$\exists b\in \mathcal{B}^{k-1}: j(b)=c$ [since $j$ is surjective].
$db=i(a)$ [Tu, p.284, l.$-$2]
$d^*[c]=[a]$ [Tu, p.285, l.3].
$i^*[a]=[i(a)]$ [Tu \cite [p.283, (25.2)]
$=[db]=0$. This completes the proof.
For the proof of exactness at $H^k(\mathcal{B})$ and other related material about the zigzag lemma, see Example 6.157 in https://sites.google.com/view/lcwangpress/%E9%A6%96%E9%A0%81/papers/mathematical-methods.