Continious subbundle 

Let $W$ be a vector bundle with base $\Omega$ and projection $p$. A continious subbundle of $W$ is a subset $W_0$ of $W$ such that $p|W_0$ defines a vector bundle over $\Omega$.


Now here is something I do not understand:


If $W=\Omega\times\mathbb{R}^n$, a continious subbundle (of dimension $r$) is a continious map from $\Omega$ to the Grassmannian $G_{n,r}$.


Could you please explain this sentence to me?
 A: The Grassmannian $G_{n,r}$ is the space of $r$-dimensional subspaces of $\mathbb{R}^n$. Given a continuous rank $r$ subbundle $V$ of $\Omega\times\mathbb{R}^n$, for each point $x\in\Omega$, the definition implies that the fiber of $V$ over $x$, i.e. $(\{x\}\times\mathbb{R}^n) \cap V$, is a subspace of $\mathbb{R}^n$ of dimension $r$. Define a map $f:\Omega\to G_{n,r}$ by assigning to $x$ the point in $G_{n,r}$ defined by the subspace $(\{x\}\times\mathbb{R}^n) \cap V$. Then the fact that $V$ is a continuous subbundle can be used to show that $f$ is continuous. Intuitively, the fact that $V$ is itself a vector bundle over $\Omega$ (and in particular is locally trivial) means that if $x$ is close to $y$ in $\Omega$, then the fiber $f(x)$ must be close to the fiber $f(y)$ in $G_{n,r}$. The details are a little cumbersome and come down to describing the topology of $G_{n,r}$, but it would be a good exercise if you haven't worked with Grassmannians before.
For the other direction, if you are given a continuous map $f:\Omega\to G_{n,r}$, then for each $x\in\Omega$, the point $f(x)$ is an $r$-dimensional subspace of $\mathbb{R}^n$. Define a subset $V$ of $\Omega\times\mathbb{R}^n$ by setting
$$
  V = \{ (x, v) \in \Omega \times\mathbb{R}^n : v \in f(x) \}.
$$
That is, the fiber of $V$ over the point $x$ is the set of all vectors $v$ in the $r$-dimensional subspace given by $f(v)$. In the same way as the other direction, you can use the fact that $f$ is continuous to show that $V$ is a continuous subbundle of $\Omega\times\mathbb{R}^n$. In particular, the fact that $f(x)$ varies continuously with $x$ can be used to show local triviality.
Very intuitively, this statement says: "a continuous rank-$r$ subbundle of a trivial bundle is given by assigning an $r$-dimensional subspace of the fiber to each point $x$ of the base in a way that depends continuously on $x$." The hidden details are in what is meant by an assignment of subspaces that depends continuously on $x$, and are given by the topology of the Grassmannian.
