Cubical cup poduct with a unit 1-cochain and the coboundary operator

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SUMMARY

The discussion focuses on the definition of the u-coboundary operator u-d in the context of cubic p-cochains, specifically defined as cup product multiplication U from the left with a unit 1-cochain, expressed as (u-d)f = 1Uf. It is established that due to the properties of the cup product, specifically that 1U1=0 and the associativity of U, the operator satisfies (u-d)^2=0. The participants explore the relationship between the u-coboundary operator u-d and the standard coboundary operator d, questioning whether they are equivalent.

PREREQUISITES
  • Cubic p-cochains
  • Cup product multiplication (U)
  • Unit 1-cochain
  • Standard coboundary operator (d)
NEXT STEPS
  • Research the properties of cup products in algebraic topology
  • Study the implications of (u-d)^2=0 in cohomology theories
  • Examine the differences between u-coboundary and standard coboundary operators
  • Explore applications of cochain complexes in homological algebra
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Mathematicians, algebraic topologists, and researchers interested in cohomology theories and the properties of coboundary operators.

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TL;DR
equivalence between cup product and coboundary operator
Define a u-coboundary operator u-d on a cubic p-cochain f as cup product multiplication U from the left with a unit 1-cochain: (u-d)f = 1Uf.
Because 1U1=0 and associativity of U, we have (u-d)^2=0.
What is the relation of u-d and the standard coboundary operator d?
Are they the same?
 

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