Every nonzero vector space can be viewed as a space of functions

[Buri]

Homework Statement

Let V be a nonzero vector space over a field F, and suppose that S is a bases for V. Let C(S,F) denote the vector space of all functions f ∈ Ω(S,F) (i.e. the set of functions from S to a field F) such that f(s) = 0 for all but a finite number of vectors in S. Let Ψ: C(S,F) → V be defined by Ψ(f) = 0 if f is the zero function, and Ψ(f) = Σ {s ∈ S, f(s) ≠ 0} f(s)s, otherwise. Prove that Ψ is an isomorphism.

The attempt at a solution

Okay, I've already proved that Ψ is linear, that it is 1-1, but I'm having troubles proving that it is onto. Here's what I've done:

I'd like to be able to show that for any v ∈ V there is a f ∈ Ω(S,F) such that Ψ(f) = v. So v = (a1)s1 + (a2)s2 + ... where S = {s1, s2, ...} is a basis for V (I haven't been told whether V is finite or infinite dimensional). However Ψ(f), when f is nonzero, is a linear combination of FINITE number of elements of the basis. I do realize we could write Ψ(f) = f(s1)s1 + f(s2)s2 + ... where some of these will be zero as f is only nonzero at a finite number of them. See if v = (a1)s1 + a2(s2) + ... + (an)sn then I could define f to be f(si) = ai and I would be done. But the problem is that I can't see and can't show that Ψ(f) could ever equal a vector in v which is a linear combination of an infinite number of elements of the basis.

Any help? This isn't homework, I'm taking a look at linear algebra on my own this summer. Thanks a lot!

EDIT: I've figured it out! :)

The vectors in V will be written as a linear combination of FINITE number of vectors of S - this is by definition! I hadn't read a section on Maximal linearly independent sets (the course I'll be taking in the fall skips it) so I hadn't seen this result.

 

[Bacle]

Well, S here is the basis for both vector spaces, i.e., the free vector space on
S has S as its basis, so the two should be isomorphic.