topology on $\hom_{\mathbb C}(V,W)$ Let's $V,W$ be finite-dimensional complex vector spaces.
How to define topology on $\hom_{\mathbb C}(V,W)$? As I see we can define metric in this $(\dim_{\mathbb C}V\times\dim_{\mathbb C}W)-$dimensional vector space and induced topology from it.
Is there any way to define topology in terms of topologies on $V$ and $W$? Or more "canonical" way, then by induction from metric?
How much more complicated situation if I don't require finiteness of $\dim$'s?
UPD: why the space  of all isomorphisms is open in $\hom_{\mathbb C}(V,W)$?
Thanks a lot!
 A: There is a canonical way of defining a topology on $C(X, Y)$ where $X$ and $Y$ are topological spaces. Just pick as a basis $[K, A] = \{f \in C(X, Y) ;f(K) \subset A\}$ where $K$ is compact and $A$ is open, this is called compact-open topology. 
However in your case, since the spaces are finite dimensional (hence $C(V, W)$ is finite dimensional), so the normables topologies are the same, since they're normables and the norms are equivalents.
If your space is not locally compact, which is, in general, the case for infinite dimensional topological vector spaces, but it's Hausdorff and second countable, it's metrizable , so you have your topology. 
A: There are many, many, many ways to define a topology of any given space. 
If $(V,t)$ and $(W,\tau)$ are two topological spaces then $(V \times W, t \times \tau)$ is a topological space, where the open sets of $V\times W$ are given by the union of products of the open sets in $V$ and $W$.
Such a topology is called a product topology. 
A: You are right, you just identify the space of homomorphisms with a space of matrices and that via the entries with $\mathbb C^{\dim_{\mathbb C}V\times\dim_{\mathbb C}W}$, which you give the usual topology.
Isomorphisms correspond to matrices with determinant not equal to $0$, the determinant is a continuous function in the entries of the matrix (a polynomial even), hence the set of all isomorphisms is open.
