Cohomology of Cochain complex

Recall that the cochain $\mathcal{C}$ complex is not an exact sequence (condition $im\;d_{k-1}=\ker\:d_k$ won't hold). The following holds

$im\;d_{k-1} \subset ker\;d_k$.

This gives us an opportunity to define quotient space

$H^k(\mathcal{C})$ as $ker\;d_k / im\;d_{k-1}$ which measures cochain complex fails to be exact at $k$.

Terminology:

$ker\;d$ is k-cocyle or closed forms(DeRahm cohomology) and $im\;d$ is k-coboundary or exact forms (DeRahm cohomology). Elements of $H^k(\mathcal{C})$ are equivalent classes $[c]$ for $c \in ker\;d_k$ is called cohomology class.

cochain map:

Between any two cochain complexes $\mathcal{A}, \mathcal{B}$ one can define a cochain map $\phi:\mathcal{A} \rightarrow \mathcal{B}$ - a collection of linear maps $\phi_k:A_k \rightarrow B_k$. If $d_1,d_2$ are corresponding differential operators for $\mathcal{A},\mathcal{B}$, drawing a commuting diagram shows

$d_2\circ \phi_k = \phi_{k+1} \circ d_1$

Nice thing about this map is that induces map $\phi^{*}:H^k(\mathcal{A}) \rightarrow H^k(\mathcal{B})$ between cohomologies. This map is well defined because it takes exact forms to exact forms and closed forms to closed forms.

For $a \in Z^k(\mathcal{A})$ ie closed forms of $\mathcal{A}$ at $k$, $d(\phi(a))=\phi(d(a))=0$ and for $b \in \mathcal{A}^{k-1}$, easy to see that $\phi(d(b))=d(\phi(b))$.