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

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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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