Prove vector space structure, and linearity Suppose that $S$ is a set and $V$ is a vector space. Show that the set $V^s$ of all maps from S to V have a unique structure of vector space such that for any s in S the evaluation map ev$_s: V^s -> V$ is linear.
Ok so i know that for something to have a structure of a vector space it has to meet certain requirements, which for the sake of the questions clarity i will list here:
Let u,v,w be in V. c,d are any scalar. (Vector addition and scalar multiplication  are defined)
1) u + v is in V
2) u + v = v + u
3) u + (v + w) = (u + v) + w
4) u + 0 = u
5) u - u = 0
6) cu is in V 
7) c(u + v) = cu + cv
8) (c + d)u = cu + du
9) c(du) = (cd)u
10) 1(u) = u
Now, in theory all i have to do now is take all elements of $V^s$  and run them by all these tests, that should do it to prove that it has vector space structure.
A) but how do i do this with maps?
For the second part of the question,
What the evaluation does is take an element from $V^s$ and asign it to an element in V so, in a way it is a function:
  $V^s$ -> $ V$
   $v^s$ -> $f(v^s)$
Now to prove that  the map evaluation is linear i suppose all that is to be done is as follows:
1) $f(v^s_1 + v^s_2) = f(v^s_1) + f(v^s_2)$ 
2) $f(av^s) = a f(v^s)$
Have i finished? I dont know how to continue from here, because every time i turn to a book for whatever reason i find the following phrase " it is easy to prove that ..." and then it is never proven.
I never really had a propper training in math, so i find this really abstract and i would really appreciate it if you forgive any mathematical inaccuracy.
Thank you in advance.
 A: You are confused about the evaluation map. It is defined by $\operatorname{ev}_s(f) = f(s)$. (Remember that an element of $V^S$ is a function $f:S \to V$.)
You also haven't actually proven anything. In order to prove that $\operatorname{ev}_s$ is linear, one has to know what the vector space structure on $V^S$ is. To figure out what that is, let us suppose that the evaluation map is linear and see what we find. If the evaluation map were to be linear we should have
$$(\color{red}{f+g})(s) = \operatorname{ev}_s(\color{red}{f + g}) = \operatorname{ev}_s(f) + \operatorname{ev}_s(g) = f(s) + g(s)$$
and
$$(\color{red}{\alpha f})(s) = \operatorname{ev}_s(\color{red}{\alpha f}) = \alpha \operatorname{ev}_s(f) = \alpha f(s)$$
for every $f,g \in V^S$ and $\alpha \in F$ (where $F$ is the scalar field of $V$). Note that the parts in red are not actually defined yet and are thus meaningless! But we see that the only way for the evaluation map to even stand a chance to be linear is to define the vector space structure
$$(f+g)(s) := f(s) + g(s)$$ and $$(\alpha f)(s) := \alpha f(s)$$
on $V^S$. Conversely, this assignment makes $\operatorname{ev}_s$ into a linear map by the first computations above.
Now all that remains to prove is that $V^S$, with these operations, actually forms a vector space over $F$. This is routine and I'll leave it to you. You'll see that the axioms follow directly from the corresponding axioms for $V$.
