Vtyjkyuk

On page 4, of Atiyah's $K$ theroy
he stated

Suppose $V,W$ are real f.d. v.s, $E = X times V$ and $F= X times W$ are corresponding vector bundles. Then any homomoprhism $varphi:E rightarrow F$ determines a map
$$ Phi: X rightarrow Hom(V,W), text satisfying varphi(x,v) = (x, Phi(x)v). $$
where $Phi$ is continuous when $Hom(V,W)$ is considered the usual topology. Conversely, any such continuous map $Phi:X rightarrow Hom(V,W)$ determines a homomorphism $varphi:E rightarrow F$.

I would appreciate if someone can give a detailed explanation.

EDIT: Here is my attempt to spell out 3. of Bananeen's explanation.

$(Leftarrow)$ An open set in $Hom(V,W)$, containing $Phi(x)=(a_i,j(x))$, is set of matrices $A$, $|A-a_i,j(x)|<varepsilon$. Continuity at each $e_j$, shows exists open sets $U_1, ldots, U_n$ containing $x$ and that $$|(a_i,j(x))_i=1^n - (a_i,j(y))^i=1_n| < varepsilon/2n text for y in U_j.$$

Thus, taking $bigcap U_i=U$, we have found an nhood of $x$, such that for all $y in U$, $|Phi(y)-Phi(x)|<varepsilon$.

$(Rightarrow)$ The converse follows since on the basis $e_j$ vectors, $|Phi(x)e_j - Phi(y)e_j| le |Phi(x)-Phi(y)|$. For any arbitrary vector $v$, by triangle inequality, result follows.

1. Since $V,W$ are finite dimensional vector spaces, $Hom(V,W)$ is itself a finite dimensional vector space. Thus after choice of any bases for $V$ and $W$, $Hom(V,W)$ is identified with $mathbbR^nm$, and the standard Euclidean metric defines the "usual topology" (two matrices are "close" to each other, if their components are "close") . Then it is a standard exercise to show that topologies arising from norms on finite dimensional vector spaces are all equivalent.


2. $Phi(x)$ is defined as follows: vector $v in V$ is mapped to $W$ via the following composition: $$V to X times V xrightarrowphi X times W xrightarrowpr_2 W$$
   $$v mapsto (x,v) mapsto phi(x,v)=(x, Phi(x)v) mapsto Phi(x)v$$which is linear because of fiberwise linearity of $phi$.


3. To show that $Phi(x)$ is continuous, use the fact that map $Phi: X to Hom(V,W)$ is continuous iff $ev_v: X to W$ defined by $x mapsto Phi(x) v$ is continuous for any $vin V$. To see this, again choose bases for $V=span(e_1,...,e_n)$, $W=span(u_1,...,u_m)$, then $Phi(x)$ is a matrix valued function $x mapsto A(x)=( a_i,j(x) )$, and $x mapsto A(x)e_i$ just selects the $i$-th column of this matrix.

So to show continuity of $Phi(x)$ from 2 we just observe that $$x mapsto phi(x,v) mapsto pr_2 (phi(x,v) ) = Phi(x)v$$ is continuous for any $v in V$.