A continuous map arising from vector bundles I can't see why something Atiyah says in page 27 of his book on K-theory is true. The context is the following. Let V be a complex vector space, and denote by $G_n(V)$ the set of all subspaces of $V$ of codimension $n$. If V is given a Hermitian metric, each subspace of V determines a projection operator. This defines a map $G_n(V)\to End(V)$. Giving $End(V)$ its usual topology, this map induces a topology on $G_n(V)$ (i.e. it's open sets are exactly the preimages of open sets in End(V)).
Now suppose $E$ is a vector bundle over a space $X$, and $\varphi:X\times V\to E$ is an epimorphism of bundles. Why is the map $X\to G_n(V)$ given by $x\mapsto ker(\varphi_x)$ continuous regardless of the choice of metric on $V$? Atiyah just claims it. The map $\varphi_x$ is of course defined through $\varphi_x(v)=\varphi(x, v)$. 
Given the definition of the topology on $G_n(V)$, my question is equivalent to why is the map $P:X\to End(V)$ continuous, where $P(x)$ is the orthogonal projection onto $ker(\varphi_x)$. 
 A: Since the question is local on $X$, we can assume $E$ is trivial.  So, choosing bases for $V$ and $E$, we can represent each $\varphi_x$ as a matrix, and the map taking $x$ to the matrix of $\varphi_x$ is continuous (since the columns of the matrix are just $\varphi_x$ evaluated at the basis elements of $V$).  So, it suffices to show the map $f:U\to G_n(V)$ taking a full rank $m\times k$ matrix to the projection onto its kernel is continuous (here $k=\dim V$ and $m=k-n$ is the rank of $E$, and $U\subset\mathbb{C}^{m\times k}$ is the space of all full rank $m\times k$ matrices).
So, suppose $A$ is a full rank $m\times k$ matrix.  We can find $m$ of its columns which are linearly independent, without loss of generality the first $m$ columns.  Note moreover that the first $m$ columns will then be linearly independent for any matrix sufficiently close to $A$, since they determine a nonvanishing minor and taking determinants is continuous.  But now, we have a canonical way to construct an orthonormal basis for the kernel of $A$.  Let us write $e_1,\dots,e_k$ for our basis for $V$.  Note that for each $j>m$, $A(e_j)=\sum_{i=1}^m c_{ij} A(e_i)$ for some unique scalars $c_{ij}$ which are continuous in $A$ (since they can be obtained by inverting the matrix formed by the first $m$ columns of $A$).  So, the vectors $e_j-\sum c_{ij}e_i$ for $j=m+1,\dots,k$ form a basis of $\ker(A)$.  Orthonormalizing this basis by Gram-Schmidt, we get an orthonormal basis for $\ker(A)$ which is continuous in $A$.  Finally, the projection onto $\ker(A)$ can be computed continuously from this orthonormal basis.
