In Tu's book on manifolds he gives the Zig-Zag Lemma as:
A short exact sequence of cochain complexes $$0 \to \mathcal{A} \xrightarrow{i} \mathcal{B} \xrightarrow{j} \mathcal{C} \to 0$$ gives rise to a long exact sequence in cohomology $$\cdots \xrightarrow{j^*} H^{k-1}(\mathcal{C}) \xrightarrow{d^*} H^k(\mathcal{A}) \xrightarrow{i^*} H^k(\mathcal{B}) \xrightarrow{j^*} H^k(\mathcal{C}) \xrightarrow{d^*} \cdots$$ where $i^*$ and $j^*$ are the maps in cohomology induced from the cochain maps $i$ and $j$, and $d^*$ is the connecting homomorphism.
I posted this question yesterday which went over part of the proof Tu gives for the exactness at $H^k(\mathcal{C})$. In the exercises he asks the reader to show exactness at $H^k(\mathcal{A})$ and $H^k(\mathcal{B})$. This is what I am working on now.
Starting with $H^k(\mathcal{A}$):
First we show Ker$(i^*) \subset $ Im$(d^*)$. Suppose $[a] \in$ Ker$(i^*)$. Then $i^*([a]) = [b] = 0$. Since $[b]$ is a cocycle, $b = db'$ for some $b' \in B^{k-1}$. Then $j(b') \in C^{k-1}$, and creating a diagram similar to the one in the post above, we have that the connecting homomorphism maps $c = j(b) \mapsto a$. In other words,
$$d^*([c]) = [a]$$
and hence $[a] \in $ Im$(d^*)$.
Next we show Im$(d^*) \subset$ Ker$(i^*)$. Suppose $[a] \in $ Im$(d^*)$.Then $[a] = d^*([c])$ for some $c \in C^{k-1}$. By the surjectivity of $j$, $c = j(b)$ for some $b \in B^{k-1}$. But then
$$j(d(b)) = d(j(b)) = d(c) = 0$$
since $c$ is a cocycle. Moreover, since Ker$(j) = $ Im$(i)$ and by the injectivity of $i$, there exists a unique $a' \in A^k$ such that $i(a') = db$.
To complete the proof I need to show that
- $a' \in [a]$
- $[db] = 0$
This is where I am stuck. I assume (2) holds by definition, since $db \in $ Im$(d_{k-1}$), and in each cocycle class of degree $k$ we are modding out everything in Im$(d_{k-1}$).
Can anyone suggest how to proceed? I am uneasy about the rest of the proof too, as it feels like a lot of symbol pushing. I assume the proof for $H^K(\mathcal{B})$ is similar.
Best Answer
I also struggled to prove this whilst reading this section of Tu, and after banging my head against it, I've managed to fill in all the details. Throughout the proof I will drop all the $k's$, and all occurences of $a,b,c$ will refer to elements of a suitable grade in the cochain complexes.
This is usually the easier direction. Suppose $[a]$ (with $da=0$) lies in the image of $d^*$. Then there exists $c$ with $dc = 0$ and $d^*[c]=[a]$. In other words, there exists $c$ with $dc=0$, $b$ such that $jb=c$, and $ia = db$.
Now by definition, $i^*d^*[c]=i^*[a]=[ia]=[db]=[0]$, where the final equality follows from the definition of cohomology.
Let $[a]$ be the in the kernel of $d^*$. Then $da = 0$ and $[ia] =[0]$. In particular, $ia=db$ for some $b$. Let $c := jb$. Note that $dc = jdb=jia=0$, since $\mathrm{im}(i) \subseteq \ker(j)$. Then $d^*[c]=[a]$ by construction.
Let $[b]$ (with $db=0$) be in the image of $i^*$. Then for some $a$ with $da=0$ we have $[b]=[ia]$. In particular $ia = b + db'$ for some $b'$. Note $d(ia)=ida=0$ too. Then $j^*[b]=[jb]=[j(ia-db')]$. Now $jia = 0$ for any $a$. Thus $j^*[b] = [-j(db')] = [d(-jb')] = [0]$. Thus $[b]$ lies in the kernel of $j^*$.
Let $[b]$ (with $db=0$) lie in the the kernel of $j^*$. Then $j^*[b] := [jb] = [0]$. Thus $jb = dc$ for some $c$. Since $j$ is surjective, there exists some $b'$ with $jb'=c$. Also, $j(db') = djb' = dc = jb$. Thus $j(b-db')=0$ and $b-db'$ lies in the kernel of $j$ which coincides with the image of $i$. Thus there is an $a$ such that $ia = b-db'$.
We now show $da=0$. Note $i(da) = dia = d(b-db') = 0$. Since $i$ is injective, it follows that $da=0$. Finally, $i^*[a] := [ia] = [b-db'] = [b]$, establishing that $[b]$ lies in the image of $i^*$.