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Mathematics Qualifying Exam Question Bank

Department of Mathematics

Question

Math 6510 • Differentiable Manifolds

Spring 2013 • Question 3

Question

Math 6510 • Differentiable Manifolds Spring 2013 • Question 3

The cup product for the de Rham cohomology of a smooth manifold \(M\) is defined as \[ \cup :H^k(M)\times H^\ell (M)\to H^{k+\ell }(M) \] where \([\omega ]\cup [\theta ]=[\omega \wedge \theta ]\). Show that the cup product is well-defined.

Raw LaTeX

The cup product for the de Rham cohomology of a smooth manifold \(M\) is
defined as
\[
  \cup:H^k(M)\times H^\ell(M)\to H^{k+\ell}(M)
\]
where \([\omega]\cup[\theta]=[\omega\wedge\theta]\). Show that the cup product is
well-defined.

Topic Tags

Cohomology and Cup Products Pullbacks and Wedge Products Smooth Manifolds and Embedded Submanifolds de Rham Cohomology