A Continuous Choice of $k$-Subspaces of a Vector Space Gives a Continuous Choice of Bases

Question

$\newcommand{\R}{\mathbf R}$ The Grassmannian $G_k(\R^n)$ as a topoplogical space is defined in the following way:

Let $F_k(\R^n)$ be the collection of all the linearly independent lists of size $k$ of $\R^n$. For two members $\mathbf v=(\mathbf v_1, \ldots, \mathbf v_k)$ and $\mathbf u=(\mathbf u_1,\ldots, \mathbf u_k)$ of $F_k(\R^n)$ write $\mathbf u\sim \mathbf v$ iff $\text{span}(\mathbf u)=\text{span}(\mathbf v)$. It is clear that $\sim$ is an equivalence relation.

Note that $F_k(\R^n)$ can be thought of all the set of all the $k\times n$ matrices having rank $k$. So it is an open subspace of $\R^{nk}$.

We define $G_k(\R^n)$ as the quotient $F_k(\R^n)/\sim$ in the quotient topology.

Suppose we have a continuous function $f:\R\to G_k(\R^n)$. Can we choose a continuous function $F:\R\to F_k(\R^n)$ such that $f(t)=\text{span}(F(t))$ for all $t$?

Loosely, suppose we have a continuous choice of $k$-dimensional subspaces of $\R^n$, then can we have a continuous choice of bases of these $k$-subspaces?

Thanks.

Answer 1

Yes. Basically this works because $\mathbb R$ is simply connected. See e.g. here.