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Let $\pi: E \to M$ a fiber bundle and $\Gamma(M,E)$ the space of smooth sections of the bundle with topology induced by the Whitney topology on $C^{\infty}(M,E)$. Assume that each fiber is contractible. Is $\Gamma(M,E)$ necessarily contractible?

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I first give a solution in the context of CW-complexes, topological bundles and continuous sections. In this setting, using contractibility of the fibers and induction on skeleta, one constructs a (continuous) section $$ s: M\to E $$ of the bundle. Next, again using contractibility of the fibers and induction on skeleta, one constructs a (continuous) map $$ f: E\times [0,1]\to E $$ such that $f(x,0)=x, f(x, 1)=s(\pi(x))$ and $\pi\circ f(x,t)=\pi(x)$ for every $x$ and $t$. Using the map $f$ one defines a contraction $F$ on the space $\Gamma=\Gamma(M,E)$ of (continuous) sections by the formula $$ F(\sigma(m),t)=f(\sigma(m), t), m\in M, t\in [0,1] $$ for sections $\sigma: M\to E$. Clearly, $F$ is continuous on the product $\Gamma\times [0,1]\to \Gamma$ and $F(\sigma, 0)=\sigma, F(\sigma, 1)=s$. Thus, $\Gamma$ is indeed contractible.

Now, to the smooth case. There is an approximation procedure for continuous sections via smooth sections described in Section 6.7 of Steenrod's "The Topology of Fibre Bundles". Using this approximation scheme, one can replace "continuous" with "smooth" in the a above proof provided that $\pi$ is a smooth fiber bundle.

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  • $\begingroup$ I have never studied topological bundles, but I'll try to catch up. But its the approximation theory necessary to conclude the proof? Isn't the smooth case, in particular, continuous? $\endgroup$ – 115465 Mar 24 '14 at 21:43
  • $\begingroup$ @Léo: The topological construction I described will not preserve smoothness of sections since $f$ is merely continuous. $\endgroup$ – Moishe Kohan Mar 24 '14 at 22:45

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