How $\operatorname{Iso}(V,W)$ becomes open subset of $\operatorname{Hom}(V,W)$? In Atiyah's K-Theory it is given that   Iso(V,W)  is an open subset of Hom(V,W). Here V and W are finite dimensional vector spaces. Let dimension of V be m and dimension of W be n. Hom(V,W)  is set of all homomorphism between vector spaces. Iso(V,W) is set of all isomorphism between V and W.  
Before that he says that we are giving usual topology on Hom(V,W) .  I think the natural way of defining a topology is by considering set  the corresponding matrices. Then the  topology will be same  mn dimensional Euclidean space (or $F^{mn}$)
How can we prove  Iso(V,W) is an open subset of Hom(V,W). What will be the properties of the matrix corresponding to an element in Iso(V,W) ?
How it will be an open subset of Hom(V,W) ?
 A: There is a map $\operatorname{det}\colon \operatorname{Hom}(V,W) \to k$ sending a linear map to its determinant. One checks that this map is continuous. When $k$ is a field, $\operatorname{Iso}(V,W) = \operatorname{det}^{-1}(k \setminus\{0\})$, so is the preimage of an open set under a continuous function, and is thus an open set.
A: First, an extended comment: It's true that $\operatorname{Hom}(V, W)$ inherits a natural topology from $F^{n \times m} \cong F^{mn},$ but we should say a little more.
In the particular case $V = F^m, W = F^n$ we can identify a linear map $\phi \in \operatorname{Hom}(V, W)$ with the matrix whose $j$th column is the image $\phi(E_j)$ of the $j$th standard basis vector $E_j = (0, \ldots, 0, 1, 0, \ldots, 0)$, where the $1$ is the $j$th entry.
On the other hand, for general vector spaces $V, W$ there are no preferred bases, and so there isn't a (basis-independent) notion of "matrix corresponding" to a linear map. But if we pick bases $(X_a)$ of $V$ and $(Y_b)$ of $W$ we can exploit the above special case to produce such a matrix. The bases $(X_a)$ and $(Y_b)$ separately define linear isomorphisms $\Bbb F^m \cong V$ and $W \cong \Bbb F^n$ by mapping $E_a \mapsto X_a$ and likewise $Y_b \mapsto E_b$. But composing these maps with our linear map $\phi : V \to W$ gives a map $\Bbb F^m \to \Bbb F^n$, and we can associate to $\phi$ the matrix $[\phi]$ associated to this latter map.
So, given a choice of bases the isomorphism $\operatorname{Hom}(V, W) \to F^{n \times m} \cong F^{mn}$ defined by $\phi \mapsto [\phi]$ determines a topology by declaring the isomophism to be a homeomorphism. A priori, however, this topolgy could depend on the choice of bases. It turns out not to be---this is a consequence of the fact that linear maps $\Bbb F^k \to \Bbb F^k$ are continuous functions with respect to the topology on $F^{k \times k} \cong F^{k^2}$. Put another way, $\operatorname{Hom}(V, W)$ inherits a topology from $F^{n \times m}$ that does not require a choice of basis and hence entails no concept of "corresponding matrix".

For the question itself: First, it follows from the definition of dimension that if $\dim V \neq \dim W$ there are no isomorphisms $V \to W$ and so $\operatorname{Iso}(V, W) = \emptyset$, which is open.
This leaves the case $\dim V = \dim W$. If $V = W = \Bbb F^m$, we know that a map is an isomorphism iff its determinant is nonzero. On the other hand, the determinant map $\det : \operatorname{Hom}(\Bbb F^m, \Bbb F^m) \cong \Bbb F^{m \times m} \to \Bbb F$ is continuous---it's a polynomial in matrix entries, so the set $\operatorname{Iso}(\Bbb F^m, \Bbb F^m)$ of invertible linear maps is the preimage $\det^{-1}(F \setminus \{0\})$ of the open set $F \setminus \{0\}$ under the continuous map $\det$ and so it is an open subset of $\operatorname{Hom}(\Bbb F^m, \Bbb F^m)$.
For general $V, W$, a choice of bases identifies linear maps $\phi : V \to W$ with their respective matrix representations $[\phi]$ with respect to those bases, and by construction, the map $\Phi : \operatorname{Hom}(V, W) \to \operatorname{Hom}(\Bbb F^m, \Bbb F^m)$ is continuous. Unwinding definitions we have $$\operatorname{Iso}(V, W) = \Phi^{-1}(\operatorname{Hom}(\Bbb F^m, \Bbb F^m)) ,$$ so again: $\operatorname{Iso}(V, W)$ is the preimage of an open set under a continuous map and hence $$\boxed{\operatorname{Iso}(V, W) \textbf{ is open in } \operatorname{Hom}(V, W) \textbf{.}}$$
Remark We can show the claim without resorting to choosing a basis, too. Briefly: The map $\phi \in \operatorname{Hom}(V, W)$ induces a map $\det \phi \in \operatorname{Hom}\left(\bigwedge^m V, \bigwedge^m W\right)$, and the map $\det : \operatorname{Hom}(V, W) \to \operatorname{Hom}\left(\bigwedge^m V, \bigwedge^m W\right)$ is continuous. A map $\phi$ is invertible iff $\det \phi \neq 0$, so by definition $$\operatorname{Iso}(V, W) = \det{}^{-1}\left(\operatorname{Hom}\left(\bigwedge^m V, \bigwedge^m W\right) \setminus \{0\}\right) .$$ In particular, $\operatorname{Iso}(V, W)$ is the preimage of the open set $\operatorname{Hom}\left(\bigwedge^m V, \bigwedge^m W\right) \setminus \{0\}$ under the continuous map $\det$ and so is open in $\operatorname{Iso}(V, W)$.
