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

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The discussion centers on the definition of a u-coboundary operator u-d for a cubic p-cochain f, expressed as cup product multiplication with a unit 1-cochain. It is established that (u-d)f equals 1Uf, leading to the conclusion that (u-d)^2 equals zero due to the properties of the cup product. A key question arises regarding the relationship between the u-coboundary operator u-d and the standard coboundary operator d, specifically whether they are equivalent. The conversation seeks to clarify this relationship, emphasizing the mathematical implications of their definitions. Understanding the distinctions and connections between these operators is crucial for further exploration in the field.
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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?
 

Thread 'About the definition of topological manifold using closed sets'

We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.
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