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.
He gives a proof for the exactness at $H^k(\mathcal{C})$ as follows.
First claim: im$(j^*) \subset $ ker$(d^*)$.
Proof: Let $[b] \in H^k(\mathcal{B})$. 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
(here he includes a diagram which I was unable to create. The diagram is the same as in the post I have attached below, with the vertical arrow on the left removed, $a$ is replaced with $0$, $db$ is replaced with $db = 0$, and $c$ is replaced with $j(b)$)
we see that since $i(0) = 0 = db$, we must have $d^*[j(b)] = [0]$. So $j^*[b] \in$ ker$d^*$.
The post I have linked below contains a screenshot of the reverse inclusion.
I am new to cohomology and these types of diagrams. How did he conclude that $0$ goes in the upper left hand corner of the diagram? That is, how did we know that $0 = i(0)$?
Best Answer
Here's the diagram you wanted to include:
This represents how the connecting homomorphism $d^{*} \colon H^{k}(\mathcal{C}) \to H^{k+1}(\mathcal{A})$ is defined: you take a representative cocycle $c$ of an element $[c] \in H^{k}(\mathcal{C}),$ you find an element $b \in B^{k}$ such that $j(b) = c$ (this is possible because $j$ is surjective, by exactness). Then, you show that $j(d(b)) = 0,$ hence there exists $a \in A^{k+1}$ such that $i(a) = d(b)$ (by exactness, $\ker j = \text{im} \; i$). Finally, you check that $d(a) = 0,$ hence $a$ is a cocycle, and $[a]$ is an element of $H^{k+1}(\mathcal{A}).$ You define $$d^{*}([c]) = [a].$$
The details of the above definition can be found in Section 25.3 of Tu. A very important exercise is 25.5, which asks you to show that the above definition is well-defined. Indeed, we've made a number of choices: there are possibly multiple $b, b'$ such that $j(b) = j(b') = c,$ and there are possibly multiple $c, c'$ such that $[c] = [c'].$ It is very important that you check for yourself that these choices end up giving the same $[a]$ at the end. If you need any hints, I can include some in this answer.
Going back to your question, the well-definedness of the above process is key. Since the result of the function $d^{*}$ is going to be the same no matter which representative $c$ we choose and which element $b$ we choose, we might as well pick convenient ones.
Consider $[b] \in H^{k}(\mathcal{B}).$ Then, $$d^{*}(j^{*}([b])) = d^{*}([j(b)]).$$ Remember that to define $d^{*}([c]),$ we first pick a representative for $[c].$ Here, our $[c]$ is $[j(b)],$ so we might as well pick the representative $j(b)$. Since $d^{*}$ is well-defined, we can make this choice to be the most convenient for us.
Now, following the definition of $d^{*}$, we want to find some element of $B^{k}$ such that $j$ sends that element to our representative $j(b).$ We might as well pick our element to be $b,$ because $j$ sends $b$ to $j(b).$ Once again, since $d^{*}$ is well-defined, we can make this choice to be the most convenient for us.
Then, we send $b$ to $d(b),$ and we find an $a \in A^{k+1}$ such that $i(a) = d(b).$ Note that $b$ is a cocycle, so $d(b) = 0.$ So, we want to find $a \in A^{k+1}$ such that $i(a) = 0.$ Since $i$ is a homomorphism, $a = 0$ does the trick (in fact, since $i$ is injective, this is the only possible $a$).
Therefore, $$d^{*}([j(b)]) = [a] = [0] = 0.$$ The conclusion follows.