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In the paper Mapping class group and function spaces: a survey, F. Cohen, M.A. Maldonado, 2014, page 3, Section 3:

for a $m$-manifold $M$, consider the disc bundle $D(M)$ in the tangent bundle $T(M)$, and define $\dot T(M)$ by the identification of two vectors that lie in the same fibre and outside $D(M)$. This gives a $S^m=D^m/\partial D^m$-bundle $\dot T(M)\to M$.

For a pointed topological space $X$, denote by $\dot T(M;X)$ the fibrewise smash product of the bundle $\dot T(M)\to M$ with $X$ to obtain a bundle $$ \dot T(M;X)\to M$$ with fibre $S^m\wedge X=\Sigma^mX$.

Let $\Gamma(M;X)$ be the space of sections of $\dot T(M;X)$.

Question: When $M$ is $\mathbb{R}^m$, I want to prove (as given in The homology of $\mathcal{C}_{n+1 }$–spaces, $n ≥ 0$. F. Cohen, page 225) $$ \Gamma(\mathbb{R}^m;X)\simeq \Omega^m\Sigma^mX. $$ How to prove this?

I have tried but failed: $$ \Omega^m\Sigma^m X=\Omega^m(S^m\wedge X)=\text{Map}_*(S^m,S^m\wedge X), $$ $$ \Gamma(\mathbb{R}^m;X)=\text{Map}(\mathbb{R}^m,S^mX). $$ They are different.

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    $\begingroup$ I agree with your calculation. However I'm pretty sure that $\Gamma$ ought to denote the sections with compact support (i.e. the sections that coincide with the base point outside a compact space). In that case the formula $\Gamma(\mathbb{R}^m;X)\simeq \Omega^m\Sigma^mX$ is correct. $\endgroup$
    – Geoffroy Horel
    Commented Jul 13, 2015 at 6:37
  • $\begingroup$ Under the assumption that $\Gamma$ is the collection of cross-sections with compact support, I still have difficulties. $$\Gamma(\mathbb{R}^n;X)=\{f\in \text{Map}(\mathbb{R}^n, S^n\wedge X)\mid f \text{ has compact support}\}\\=\{f\in \text{Map}(\mathbb{R}^n\cup\infty, S^n\wedge X)\mid \text{ outside a compact subset of $\mathbb{R}^n$, } f \text{ takes values at base point } *\} $$ On the other hand, $$ \Omega^n\Sigma^nX=\text{Map}(S^n,S^n\wedge X)\\ =\{f\in \text{Map}(\mathbb{R}^n\cup \infty, S^n\wedge X)\mid f(\infty)=*\} $$ (continued) $\endgroup$
    – Shiquan Ren
    Commented Jul 13, 2015 at 8:22
  • $\begingroup$ If $ \text{ outside a compact subset of $\mathbb{R}^n$, } f \text{ takes values at base point } *$, then $f(\infty)=*$. However the converse is not true. Hence $$ \Gamma(\mathbb{R}^n;X)\subsetneq \Omega^n\Sigma^nX. $$ Since they are not equal, how to show that they are homotopy equivalent? $\endgroup$
    – Shiquan Ren
    Commented Jul 13, 2015 at 8:22
  • $\begingroup$ @GeoffroyHorel. Dear Prof. Horel, thanks so much. How to solve the difficulty as in the following answer? $\endgroup$
    – QSR
    Commented Jul 13, 2015 at 11:05

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