Continuity of a map: proving manifold structure of Grassmanian This post should be self contained - it is from a concern of page 13 of this notes.  

Let $q \ge n$, $V_n(\Bbb R^q)$ be the subspace of $n$ -linearly independent vectors in $(\Bbb R^q)^n$. 
We define $G_n(\Bbb R^q)$ to be the set of $n$ planes in $\Bbb R^q$. 
It is given the quotient topology of the map
$$ q:V_n(\Bbb R^q) \rightarrow G_n(\Bbb R^q)$$ 
sending $n$ vectors to the space spanend by it. 
Now fix $X \in G_n(\Bbb R^q)$, and let $U \subseteq G_n(\Bbb R^q)$, be the open set such that for $Y \in U$, the projection map $\pi:\Bbb R^q \rightarrow X$, induces an isomoprhism $\pi|_Y:Y \rightarrow X$. 

I want to prove the following map is continuous: 

Fix a basis $\{x_i \}$ of $X$
$$D:q^{-1}(U) \rightarrow q^{-1}(U)$$ 
  such that $(y_1, \ldots, y_n)$ is mapped to $(y'_1 ,\ldots, y'_n)$ where $\pi(y'_i) = x_i$ and $q(y'_1,\ldots, y'_n) = q(y_1,\ldots, y_n)$. 

It is claimed that (from the notes) this is continuous. But I do not see an easy expression of this map. 
 A: I think the author was not well-advised to omit the proof.
We begin by recalling some facts from linear algebra:
Let $V$ be a vector space over a field $K$. Each matrix $A = (a_{ij}) \in M_n(K)$ induces a linear map $A : V^n \to V^n, A(v_1,\dots,v_n) = (\sum_{j=1}^n a_{1j}v_j,\dots,\sum_{j=1}^n a_{nj}v_j)$. Note that in this definition $\dim(V)$ is arbitrary. If $\mathbf{v} = (v_1,\dots,v_n) \in V^n$ forms a basis of $V$ (which requires $\dim(V) = n$), then each $\mathbf{w}  = (w_1,\dots,w_n) \in V^n$ admits a unique $A(\mathbf{v},\mathbf{w}) \in M_n(K)$ such that $A(\mathbf{v},\mathbf{w})(\mathbf{v}) = \mathbf{w}$. We have $A(\mathbf{v},\mathbf{w}) \in GL_n(K)$ if and only if $(w_1,\dots,w_n)$ forms a basis of $V$. In that case $A(\mathbf{v},\mathbf{w})$ realizes the change of basis from $\mathbf{v}$ to $\mathbf{w}$.
Each linear map $f : V \to W$ induces a linear map
$$f^n : V^n \to W^n, f^n(v_1,\dots,v_n) = (f(v_1),\dots,f(v_n)) .$$
It is readily verified that
$$f^n(A(\mathbf{v})) = A(f^n(\mathbf{v})) .$$
We now come to the proof of continuity. Given a fixed basis $\mathbf{x} = (x_1,\dots,x_n)$ of $X$, the orthogonal projection $\pi : \mathbb{R}^q \to X$ can be written as $\pi(y) = \sum_{j=1}^n \xi_j(y) x_j$ with linear maps $\xi_j :\mathbb{R}^q \to \mathbb{R}$. This gives us a linear map
$$\phi : (\mathbb{R}^q)^n \to M_n(\mathbb{R}), \phi(y_1,\dots,y_n)_{ij} = \xi_j(y_i) .$$
For each $\mathbf{y} = (y_1,\dots,y_n) \in q^{-1}(U)$ the span $q(\mathbf{y})$ is mapped by $\pi$ isomorphically onto $X$. Hence $\pi^n(\mathbf{y}) = (\pi(y_1),\dots,\pi(y_n)) = (\sum_{j=1}^n \xi_j(y_1) x_j,\dots,\sum_{j=1}^n \xi_j(y_n) x_j)$ is a basis of $X$. Thus the matrix $\phi(\mathbf{y})$ realizes the change of basis from $\mathbf{x}$ to $\pi^n(\mathbf{y})$, i.e. we have $\phi(\mathbf{y})(\mathbf{x}) = \pi^n(\mathbf{y})$. We conclude
$$\phi(q^{-1}(U)) \subset GL_n(\mathbb{R}) .$$
Since linear maps between finite-dimensional vector spaces (endowed with any norm) are continuous and inverting matrices in $ GL_n(\mathbb{R})$ is continuous, we see that
$$\psi : q^{-1}(U) \to GL_n(\mathbb{R}), \psi(\mathbf{y}) = \phi(\mathbf{y})^{-1}$$
is continuous. The matrix $\psi(\mathbf{y})$ realizes the change of basis from $\pi^n(\mathbf{y})$ to $\mathbf{x}$, i.e. we have $\psi(\mathbf{y})(\pi^n(\mathbf{y})) = \mathbf{x}$.
For $\mathbf{y} \in q^{-1}(U)$ define
$$D(\mathbf{y}) = \psi(\mathbf{y})(\mathbf{y})$$
where we regard $\psi(\mathbf{y})$ as a linear map $q(\mathbf{y})^n \to q(\mathbf{y})^n$. Since $\mathbf{y}$ is a basis of $q(\mathbf{y})$, also $D(\mathbf{y})$ is a basis of $q(\mathbf{y}) \in U$. Hence $D(\mathbf{y}) \in q^{-1}(U)$, i.e. we have defined a function 
$$D : q^{-1}(U) \to q^{-1}(U) .$$
We have
$$\pi^n(D(\mathbf{y})) = \pi^n(\psi(\mathbf{y})(\mathbf{y})) = \psi(\mathbf{y})(\pi^n(\mathbf{y})) = \mathbf{x}$$
which shows that our $D$ is the same as the author's.
To see that $D$ is continuous note that the coordinate functions $D_i : q^{-1}(U) \to \mathbb{R}^q$ are given by $D_i(\mathbf{y}) = \sum_{j=1}^n \psi(\mathbf{y})_{ij}y_j = \sum_{j=1}^n \psi(\mathbf{y})_{ij}p_j(\mathbf{y})$ with (continuous!) coordinate projections $p_j : (\mathbb{R}^q)^n \to \mathbb{R}^q$.
