Let $V$ be a real, finite dimensional, vector space, let $\mathrm {Ind}_k(V)$ be the set of all linearly independent vectors $(v_1,...,v_k)\in V^k$ and let $\pi :\mathrm {Ind}_k(V) \rightarrow \mathrm {Grass}_k (V)$ be the function $$\pi (v_1,...,v_k)= <v_1,...,v_k>$$ Where $<v_1,...,v_k>$ is the subspace generated by $v_1,...,v_k$. I want to prove that for every smooth path $\sigma : [0,1] \rightarrow \mathrm {Grass}_k (V)$ there is a smooth path $\mu :[0,1] \rightarrow \mathrm {Ind}_k(V)$ such that $\pi \circ \mu = \sigma$.
I read somewhere that fibre bundles have the "path lifting" property. If I could prove that $\pi :\mathrm {Ind}_k(V) \rightarrow \mathrm {Grass}_k (V)$ is a fiber bundle, then Im done. Sadly, I know little about fibre bundles so Im not sure how to prove what I want. The only think I got is $$\pi ^{-1}(S) = \mathrm {Bases}(S) \cong \mathrm {GL}_k (\Bbb R) $$