# A Very basic question about cohomology.

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1. Jan 25, 2017

### lichen1983312

I am self leaning some basic cohomology theory and I managed to go through from the definition to the universal coefficient theorem. But I don't think I get the main point of this theory, I like to ask this questions:
Is such an abstract theory practical?

I would say that homology is practical, because the chain groups could be built on the basis of maps from simplexes to the space, which is intuitive and easy to operate. Since the boundary operator also has clear geometric meaning, at least in theory one can just write down the chain group and follow a well defined procedure to compute the homology group.

For the cochain groups $C_n^ * = Hom({C_n},G)$ , the elements are just homomorphisms. This construction is too abstract that I cannot see what is this group look like and how is cohomology gorup $\ker /im$ computed. If one cannot easily do these things, why would we even need this theory ?

Last edited by a moderator: Jan 27, 2017
2. Jan 26, 2017

### Staff: Mentor

3. Jan 26, 2017

### lavinia

Cohomology groups have a natural multiplication called the cup product that turns them into a graded algebra. These products give more information about a space than the homology groups. There are even examples of spaces with identical homology (with any coefficients) but whose cup product structure is not the same. This is one reason why cohomology is important.

Last edited: Jan 27, 2017
4. Jan 27, 2017

### lichen1983312

5. Jan 27, 2017

### lavinia

For oriented compact manifolds, integer cohomology classes can be expressed as intersections of cycles of complementary dimension. If you think of trying to completely separate two submanifolds of complementary dimension then the intersection number (the oriented count of the number of intersection points) is an obstruction to being able to do this. If you ignore the orientation then you get a cohomology class with $Z_2$ coefficients. This tells you something about the topology of the manifold.

Last edited: Jan 27, 2017
6. Jan 27, 2017

### lichen1983312

Thanks very much, I need time to understand all this. But before that, given a space, is there a general procedure to find the co-chain group or cohomology group?

7. Jan 27, 2017

### lavinia

Not that I have seen. One technique that often works is to divide the space up into subspaces whose cohomology is easy and then fit the pieces together. This can often be done with a Mayer-Vietoris sequence. This sequence can be used for both homology and cohomology.

8. Jan 28, 2017

### lavinia

A simple case is the torus. Its first homology group is generated by two orthogonal circles that intersect in a single point. These circles can be taken to be the equator of the torus and the orthogonal circle that cycles through the hole. Intersecting homology classes with either one of the circles and counting the number of intersection points (with orientation) produces a homomorphsim of the first homology group into the integers.

The homomorphism corresponding to one of the circles maps the homology class of that circle to zero and the homology class of the orthogonal circle to 1. The interesting thing to see is that the intersection count is always the same for any representative of either homology class as long as the intersection is transversal - that is: as long as the two curves cross over each other at each intersection point.

For a homomorphisminto into $Z_{2}$ assign 1 if the number of intersections is odd and 0 if the number of intersections is even. Note for instance that if one distorts one of the two orthogonal circles so that it intersects the other in more than one point, an even number of new intersection points is created so the mod 2 count is unchanged. The oriented count is also unchanged.

Last edited: Feb 5, 2017