Isomorphism between homotopy groups of Lie group, Grassmann manifold It is asserted without proof in a book edited by Novikov and Rokhlin that
$$\pi_{k - 1}(\text{GL}_n^+(\mathbb{R})) \cong \pi_k(\tilde{G}_n).$$
I know how to show that these two spaces are bijective. But I can't see for the life of me why they're isomorphic, i.e. why 
$$f \mapsto \text{clutching function of } f^* \gamma_n(\tilde{G}_n)$$
is a homomorphism. Here, $\gamma_n$ is the tautological bundle. Could anybody help?
 A: The oriented grassmanian $\tilde{G}_n$ is the classifying space of $\operatorname{GL}^+_n(\mathbb{R})$, i.e. $\tilde{G}_n = \operatorname{BGL}^+_n(\mathbb{R})$. So there is a principal $\operatorname{GL}^+_n(\mathbb{R})$-bundle $\operatorname{EGL}^+_n(\mathbb{R}) \to \operatorname{BGL}^+_n(\mathbb{R})$; the associated vector bundle is $\gamma_n$. Applying the long exact sequence in homotopy to the $\operatorname{GL}^+_n(\mathbb{R})$-bundle, we get the exact sequence
$$\dots \to \pi_k(\operatorname{EGL}^+_n(\mathbb{R})) \to \pi_k(\operatorname{BGL}^+_n(\mathbb{R})) \to \pi_{k-1}(\operatorname{GL}^+_n(\mathbb{R})) \to \pi_{k-1}(\operatorname{EGL}^+_n(\mathbb{R})) \to \dots$$
As $\operatorname{EGL}^+_n(\mathbb{R})$ is contractible, $\pi_k(\operatorname{BGL}^+_n(\mathbb{R})) \cong \pi_{k-1}(\operatorname{GL}^+_n(\mathbb{R}))$. Replacing $\operatorname{BGL}^+_n(\mathbb{R})$ by $\tilde{G}_n$, you obtain the desired result.
Note that one could instead pass to the compact subgroup $\operatorname{SO}_n(\mathbb{R}) \subset \operatorname{GL}^+_n(\mathbb{R})$ as the two spaces are homotopy equivalent. In particular, $\tilde{G}_n = \operatorname{BSO}_n(\mathbb{R})$.

The argument given above can be used to show that $\pi_k(BG) \cong \pi_{k-1}(G)$ for any topological group $G$.
