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  • So what is Cohomology? - Mathematics Stack Exchange
    On the most basic level, you can think of cohomology as a fancy way of counting classifying holes in an underlying space (although modern offshoots are a bit more general) There are many versions of cohomology which all use the same basic approach, but the most intuitive version for someone who has gone through the usual calculus sequence
  • algebraic topology - Intuitive Approach to de Rham Cohomology . . .
    De Rham cohomology studies these differential forms and a so called exterior derivative $\mathrm{d}$ In the first and third cases, $\mathrm{d}\alpha_1=\mathrm{d}\alpha_3=0$ and that’s why small closed curves have zero value; we say that $\alpha_1$ and $\alpha_3$ are closed forms
  • algebraic topology - Difference between Homology and Cohomology . . .
    If $\phi$ induces homology isomorphisms in all dimensions, then $\phi$ induces cohomology isomorphism in all dimensions That is to say, if we can not use homology to distinguish two space, so do cohomology However my teacher told us that the essential difference between them is the ring structure on the cohomology So my questions are
  • soft question - Surprising applications of cohomology - Mathematics . . .
    The concept of cohomology is one of the most subtle and powerful in modern mathematics While its application to topology and integrability is immediate (it was probably how cohomology was born in the first place), there are many more fields in which cohomology is at least a very interesting point of view
  • general topology - Why are we interested in cohomology? - Mathematics . . .
    There are duality theorems (all kind of variants of Poincaré duality) that make use of cohomology, so even if ultimately you are interested in homology, studying cohomology can be useful Cohomology natually carries a sort of algebra structure given by the cup product, which is really helpful in a lot of situations
  • How is de Rham cohomology useful? - Mathematics Stack Exchange
    De Rham cohomology is invariant under homotopy equivalence (in the smooth category), which allows us to easily prove that two manifolds are not homeomorphic or homotopy equivalent Furthermore, cohomology comes with a ring structure that gives an extra condition to require for equivalence
  • algebraic topology - References for calculating cohomology rings . . .
    Maybe I will at least show how the cohomology ring for the sphere works, based on the above I will leave as community wiki, so it can be tidied up if something is not quite right If I work out any other spaces, I will try explain them here as well
  • How to understand the definition of cohomological functor?
    Stack Exchange Network Stack Exchange network consists of 183 Q A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers
  • algebraic topology - Is homology determined by cohomology . . .
    Can we compute the homology groups from the cohomology groups? I am aware of Poincare duality - this only applies to "nice" spaces - seeing as homology determines cohomology in the above sense for general spaces, I was wondering if such a fact still holds switching the roles of homology and cohomology





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