# The “need” for cohomology theories

In many surveys or introductions, one can see sentences such as "there was a need for this type of cohomology" or "X succeeded in inventing the cohomology of...".

My question is: why is there a need to develop cohomology theories ? What does it bring to the studies involved ?

(I have a little background in homological algebra. Apart from simplicial homology and the fact that it allows to "detect holes", assume that I know nothing about more complicated homology or cohomology theories)

• See here, for example, for sheaf cohomology. – Zhen Lin Jul 18 '12 at 11:57
• math.stackexchange.com/questions/64064/… may be of interest – Matthew Towers Jul 18 '12 at 15:04
• The need for cohomology theories is (partially) a corollary of the fact chain complexes are interesting. Of course, that just means you should ask about the "need" for chain complexes.... :) – user14972 Jul 18 '12 at 15:21
• @Hurkyl: Ah but this raises another question I had, why cohomology and not homology ? For example, one never speaks about group homology... – MarcSimon Jul 18 '12 at 15:48
• I've actually used more group homology than I have group cohomology! (although not much of either) Anyways: en.wikipedia.org/wiki/Group_homology#Group_homology – user14972 Jul 18 '12 at 16:24

I am not an expert, but I think of a cohomology theory as a description of obstructions to solving some sort of equation. For example, if the first simplicial cohomology of a simplicial complex vanishes, there is no obstruction to assigning weights to edges and faces so that some equations relating these weights are satisfied. (see chapter 3 of Hatcher's Algebraic Topology for more).

So to answer your question, one might hope that a new cohomology theory would describe a new obstruction to solving some type of equation on a particular space. Solving equations on topological spaces is pretty useful, so cohomology theories are as well.

An example from complex analysis: let $f$ be a holomorphic on some open set $U \subset \mathbb{C}$. Perhaps you'd like to find a holomorphic antiderivative for $f$, i.e. a function $g$ which is holomorphic on $U$ and satisfies $$\frac{\partial}{\partial z} g = f.$$ It's a basic result in complex analysis that if $U$ is simply-connected, we can always find such a $g$. Let's look instead at $\mathbb{C} \setminus {0}$, the complex plane without the origin. Let $f = \frac{1}{z}$. This function is holomorphic on the punctured plane, but (again from basic complex analysis) it has no holomorphic antiderivative. So one cannot always solve the equation $\frac{d}{dz}g = f$ on this space. This can be seen by looking at the "sheaf cohomology of the sheaf of holomorphic differential forms." Moral: sheaf cohomology, a separate theory from simplicial homology, is useful.

• I once read that "cohomology is a generalisation of integration". I first understood this statement when taking a course in Complex Analysis, where (coutour) integration can be seen as a way of understanding the obstruction of a function to be holomorphic on a given domain. – M Turgeon Jul 18 '12 at 14:20
• Funny -- I was recently trying to understand a problem in algebraic geometry, and I said "this feels like an integral." In fact, the problem was solved by (sheaf) cohomology! – Adam Saltz Jul 18 '12 at 21:44

The point is that different cohomology theories are applicable in different situations and are computed from different data. For example, simplicial/singular cohomology is computed from a triangulation (or the map of a simplex) into your space, while, for example, Cech cohomology is computed from just the different open covers of your space.

Also, for certain algebraic-geometry applications, no matter how we define the topology on a variety, we never seem to get enough open sets to do any kind of homological computations with, and so the development of etale cohomology solved this obstacle and let us use easy to compute 'topological' tools to investigate geometric objects.

To complement Adam Saltz's nice answer (+1 BTW): cohomology theories give a way of proving that certain sequences are exact---for instance, of proving that certain maps are surjective (i.e., solving certain equations) or injective. More specifically, and towards your question "What does it bring to the studies involved?", given an exact sequence in an abelian category

$$0 \rightarrow A_1 \rightarrow A_2 \rightarrow A_3 \rightarrow 0$$ and and a left exact functor $F$ (simple example: take the abelian category to be $kG$-modules, where $G$ is a $p$-group and the field $k$ has characteristic $p$, and take $F$ to be the functor of $G$-fixed points), one wants to know under what circumstances the induced map $F(A_2) \rightarrow F(A_3)$ is surjective. Tautologically, cohomology gives a sufficient condition: it's enough to check that the first derived functor on $A_1$ vanishes: $R^1 F (A_1)=0$. This may seem like no real gain. However, one thing that the cohomological machinery provides is a new technique for proving this: try to show, by descending induction, that all the higher cohomology groups $R^i F(A_1)=0$ (here, you need some kind of vanishing theorem that shows they are zero for $i$ large enough, and then you can try to use all the long exact sequences and things like the 5 lemma and its relatives).

A good example of this type of argument to keep in mind is Grothendieck's proof of Zariski's main theorem, e.g. as reproduced in Hartshorne's algebraic geometry book: see the theorem on formal functions and its corollaries.