I would like to compute the de Rham cohomology of the tangent bundle $TM$ of a manifold $M$. It seems to me that we can just homotopy each fibre to a point, and that this would give a homotopy equivalence between $TM$ and $M$. Since de Rham cohomology is a homotopy invariant, this would imply that $H^*_{\text{dR}}(TM) \cong H^*_{\text{dR}}(M)$.

Is it always true that the tangent bundle of a manifold is homotopic to the manifold itself by the homotopy of each fibre with a point?

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    $\begingroup$ Yes. Actually, every vector bundle (not only the tangent one) over a manifold is homotopy equivalent to the manifold. $\endgroup$ – Amitai Yuval Feb 5 '15 at 11:44
  • $\begingroup$ Is the vector bundle homotopic 1-1 with each fibre? $\endgroup$ – JMP Feb 5 '15 at 12:04

Yes, that's true. In fact every vector bundle is homotopy equivalent to the base space, and in fact deformation retracts onto the base space. To prove this:

Let $p : E \to M$ be a smooth vector bundle, and $s : M \to E$ the zero section. It's a general fact that $s$ is an embedding, so we can identify $M$ with the image of $s$. Define a deformation retraction $H_t : E \to E$ by: $$H_t(\xi) = t \cdot \xi.$$

It follows directly from the definition of a vector bundle that each $H_t$ is smooth. Indeed look at a trivializing chart $(U, \varphi)$ where $U \subset M$ is an open set and $\varphi : p^{-1}(U) \to V \times \mathbb{R}^k$ is a homeomorphism (with $V \subset \mathbb{R}^n$ an open set) Then if $\varphi(\xi) = (x, \vec{v})$, then $\varphi(H_t(\xi)) = (x, t \vec{v})$. This is obviously smooth on the chart, so by definition it's smooth on the whole bundle.

Besides, $H_0$ is the projection onto $M$, $H_1$ is the identity, and each $H_t$ is the identity on $M$ (remember that it's the zero section of $E$, and $t \cdot \vec{0} = \vec{0}$). So $E$ deformation retracts onto $M$, thus $E$ and $M$ are homotopy equivalent.


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