TITLE: Can cohomologies always be defined as the quotient of the kernel by the image of a certain operation?
QUESTION [8 upvotes]: For example, concerning de Rham cohomology as explained on the lower part of page 5 here, one considers the vector space of all forms on a manifold $M$, $A_{DR}(M)$ in the notation of the paper (?).
The $p$-th de Rham cohomology is then defined as the quotient of the kernel and the image of the exterior derivative $d$ acting on the space of $p$ and $p-1$ forms respectively as
$$
H_{DR}^p(M) = \frac{Ker(d: A_{DR}^{p}(M)\rightarrow A_{DR}^{p+1}(M))} {Im(d: A_{DR}^{p-1}(M)\rightarrow A_{DR}^{p}(M))}
$$
Can cohomology always be defined as the kernel divided by the image of a certain operation?
What does cohomology mean or "measure" in as simple as possible intuitive terms in the de Rham case but also more generally?
PS: I tried to read wikipedia of course, but it was (not yet?) very enlightening for me ...

REPLY [3 votes]: $\newcommand{\dd}{\partial}\newcommand{\Brak}[1]{\left\langle #1\right\rangle}$To give a more elementary answer intended to complement the existing answers, here are some multivariable calculus-type observations. Throughout, $M$ denotes a compact, smooth manifold of dimension $n$ (possibly disconnected), and differential forms are smooth. Briefly,

$p$-dimensional de Rham coholomogy measures "how many linearly-independent $p$-dimensional real homology classes $M$ contains". (William S. Massey's Algebraic Topology article in the Encyclopedia Britannica gives a clear, geometric introduction to homology.)

If $X$ is a smooth $k$-chain and $\omega$ is a smooth $p$-form on $M$, let
$$
\Brak{X, \omega} = \begin{cases}
  \int_{X} \omega & k = p, \\
  0 & k \neq p,
\end{cases}
$$
denote the "integration pairing". If $X$ is a $(p + 1)$-chain, Stokes's theorem gives
$$
\Brak{\dd X, \omega} = \int_{\dd X} \omega = \int_{X} d\omega = \Brak{X, d\omega}.
$$
With respect to the integration pairing, the boundary operator on chains and the exterior derivative operator on forms are formally adjoint.

A closed form $\omega$ (i.e., with $d\omega = 0$) satisfies
$$
\Brak{\dd X, \omega} = \Brak{X, d\omega} = 0.
$$
In words, the integral of $\omega$ over a $p$-chain bounding some $(p + 1)$-chain vanishes. Consequently, if $X_{1}$ and $X_{2}$ are homologous $p$-chains (i.e., whose difference is a boundary, say $X_{1} - X_{2} = \dd Y$), then
$$
\Brak{X_{1}, \omega} = \Brak{X_{2}, \omega}.
$$
An exact $p$-form $\omega$ (i.e., with $\omega = d\eta$ for some $(p - 1)$-form $\eta$) satisfies
$$
\Brak{X, \omega} = \Brak{X, d\eta} = \Brak{\dd X, \eta}.
$$
Particularly, the integral of $\omega$ over a boundaryless $p$-dimensional chain vanishes. Consequently, if $X_{1}$ and $X_{2}$ are $p$-chains having the same boundary, then
$$
\Brak{X_{1}, \omega} = \Brak{X_{2}, \omega}.
$$

Via the integration pairing, a $p$-form determines a real-valued function on $p$-chains (with real coefficients). Item 1 says that a closed $p$-form determines a real-valued function on real homology classes. Item 2 says this function is unchanged upon addition of an exact $p$-form, i.e., is constant on de Rham cohomology classes.