Existence of Boundary Homomorphisms for Cohomology

Question

I am just starting to learn the basics of cohomology and am confused about the construction of the cohomology groups. So given a group $G$, the idea is you take a projective resolution of $P_0 = \mathbb{Z}$ consisting of free $G$-modules $P_1, P_2, ...$ $$ 0 \leftarrow \mathbb{Z} \leftarrow P_1 \leftarrow P_2 \leftarrow \cdots$$ and let $H^q(G,-) = Hom_G(P_q,-)$ for $q = 0, 1, 2, ...$. Then the $q$th cohomology group of a $G$-module $A$ is $Hom_G(P_q,A)$. The thing I need to verify is the existence of so called "boundary homomorphisms." Given an exact sequence of $G$-modules $$ 0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0 $$ there should be some way to construct abelian group homomorphisms $\delta_q: H^q(G,C) \rightarrow H^{q+1}(G,A)$ such that the infinite sequence $$ 0 \rightarrow H^0(G,A) \rightarrow H^0(G,B) \rightarrow H^0(G,C) \xrightarrow{\delta_0} H^1(G,A) \rightarrow \cdots $$ is exact. How in the world do I go about doing this? Since $P_q, q \geq 1$ is projective I know the covariant functor $H^q(G,-)$ is exact. However $H^0(G,-)$ is only left exact. Is there some trick with the snake lemma that gives an abstract nonsense argument?

Answer 1

You are missing an important intermediate step. You first create a dual complex by applying Hom functor to each member. Then you give a boundary map and compute the homology. This homology is what we call cohomology.

The critical notion for boundary maps is that you want the composition of consecutive boundary maps to be the zero map. The way this is done in cohomology theory is to define a dual boundary map, $\delta$, from the boundary map $\partial$ of the original chain complex. For any $\phi$ sitting in a cochain group, you take $\delta (\phi)$ to be $\phi \circ \partial$.

It is not necessary to venture into the territory of short exact sequences and how they imply the existence of a long exact sequence of homology groups. The boundary maps are more fundamental than that.