Uniqueness of vector space structure Let's suppose that $V$ is a vector space, $S$ is a set and $\phi : V \rightarrow S$ is a surjective map. 
The problem is asking me to show that there is at most one structure of a vector space on $S$ such that $\phi$ is linear with respect to that structure. In the context of the problem, "at most one" means that there may be none at all, but if there are any, they must be the same.
My problem is that I don't know if I have to suppose that the structure exists and show that any other coincide (also, how do I suppose another structure to prove uniqueness?), or I have to prove the existence of it, and then prove its uniqueness.
Thanks. 
 A: A vector space structure on $S$ is a tuple $(S,+,\cdot)$ where $+$ is a rule for adding elements of $S$ and $\cdot$ is a rule for multiplying elements of $S$ by a scalar. Of course, $(S,+\cdot)$ must satisfy the vector space axioms.
To prove your claim, suppose $(S,+,\cdot)$ and $(S,\oplus,\odot)$ are two vector space structures on $S$ and that $\phi:V\to S$ is a linear surjection with respect to both.
To prove that $+$ and $\oplus$ agree, let $s_1,s_2\in S$. Since $\phi$ is surjective, there exist $v_1,v_2\in V$ such that $\phi(v_i)=s_i$. It follows that
$$
s_1+s_1=\phi(v_1)+\phi(v_2)=\phi(v_1+v_2)=\phi(v_1)\oplus \phi(v_2)=s_1\oplus s_2
$$
This proves that $+=\oplus$.
To prove that $\cdot$ and $\odot$ agree, let $s\in S$ and let $\lambda\in \Bbb R$. Since $\phi$ is surjective, there exists a $v\in V$ such that $\phi(v)=s$. It follows that
$$
\lambda\cdot s=\lambda\cdot \phi(v)=\phi(\lambda\cdot v)=\lambda\odot \phi(v)=\lambda\odot s
$$
This proves that $\cdot=\odot$.
This proves that if a linear structure exists on $S$ making $\phi$ a linear surjection, then that structure is unique. 
