Atiyah's K Theory, pg 4 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.

 A: *

*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 $\mathbb{R}^{nm}$, 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.

*$\Phi(x)$ is defined as follows: vector $v \in V$ is mapped to $W$ via the following composition: $$V \to X \times V \xrightarrow{\phi} X \times W \xrightarrow{pr_2} W$$
$$v \mapsto (x,v) \mapsto \phi(x,v)=(x, \Phi(x)v) \mapsto \Phi(x)v$$which is linear because of fiberwise linearity of $\phi$.

*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 $v\in V$. 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 $v \in V$.
