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Clarification on smooth de Rham theorem
I am misunderstanding something in Theorem 2.1.9 in Dimca’s Sheaves in Topology:
Let $X$ be a real smooth manifold. Then the natural morphism from the constant sheaf to the de Rham complex
$$\mathbb{R}...
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In terms of sheaf cohomology, what does Bott & Tu's relative de Rham cohomology $H^\bullet(f)$ compute for $f: S \to M$ a smooth map?
Given a map $f: S \to M$ of smooth manifolds, Bott & Tu define on page 78 a complex by $\Omega^q(f)=\Omega^q(M) \oplus \Omega^{q-1}(S)$ and $d(\omega, \theta)=(d\omega, f^*\omega - d\theta)$ where ...
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