The following question is taken from Arrows, Structures and Functors the categorical imperative by Arbib and Manes
Let $X$ be a vector space. If $A$ is a subset of $X$, show that there is at most one vector space structure on $A$ such that the inclusion map $i:A\rightarrow X$ is linear, and that this occurs iff $A$ is a subspace of $X$. If $E$ is an equivalence relation on $X,$ show that there is at most one vector space structure on $X/E$ such that $\eta_E:X\rightarrow X/E$ is linear and that this occurs iff $[0]$ is a subspace of $X$ (in which case, $E=\{(x,x')\mid x-x'\in [0]\})$.)
I am having trouble showing that "show that there is at most one vector space structure on $X/E$ such that $\eta_E:X\rightarrow X/E$ is linear and that this occurs iff $[0]$ is a subspace of $X$ (in which case, $E=\{(x,x')\mid x-x'\in [0]\})$.)". I am having difficulties because I am not clear on what the notation for $E=\{(x,x')\mid x-x'\in [0]\})$ means. Does it mean I am quotioning out all the vectors of the form $x=x'$ also, I know that $[x]\in X/E,$ so what is modding out $E$ the same as modding out $[0]$. Finally, can I assume that $\eta_E(x)=x+E$?
Furthermore, I know that $X/X$ is the zero space, which means it consists of a single element. For in $X/X$, $X+X=(0+X)+(0+X)=(0+0)+X=0+X=X.$ So $X$ is the null space of $X$.
Also, $x\sim x'$ iff $x-x'\in E,$ then $x\sim 0$ iff $x-0\in E$ which implies $x\in E$. Then the coset $x+E$ is the zero element in the quotient iff $x\in E$. This means $[0]=E$.
As to the proof that there is at most one vector space structure on $X/E$, here is the proof.
Proof: To prove that $+$ and $\oplus$ agree, let $[y_1],[y_2]\in X/E$. Since $\eta_E$ is surjective, there exist $x_1,x_2\in X$ such that $\eta_E(x_i)=[y_i],$ for $i=1,2$ It follows that $$[y_1]+[y_2]=\eta_E(x_1)+\eta_E(x_2)=\eta_E(x_1+x_2)=\eta_E(x_1)\oplus \eta_E(x_2)=[y_1]\oplus [y_2]$$ Hence $+=\oplus$.
To prove that $\cdot$ and $\odot$ agree, let $[y]\in X/E$ and let $\lambda\in \Bbb F$. Again because of surjectivity of $\eta_E$, there exists a $x\in X$ such that $\eta_E(x)=[y]$. It follows that $$ \lambda\cdot [y]=\lambda\cdot \eta_E(x)=\eta_E(\lambda\cdot x)=\lambda\odot \eta_E(x)=\lambda\odot [y] $$
This proves that $\cdot=\odot$.
I think that you’re a bit confused, maybe because you already know what a quotient vector space is.
The first thing to note is that $E$ is not a subspace of $X$! $E$ is an equivalence relation on $X$, i.e. a subset of $X \times X$ with certain properties. In particular, it makes no sense to write $x + E$ for an element $x \in X$. This is despite the fact that we use the same notation $X/E$ that we would use for quotienting out a subspace.
I will only give hints to the actual exercise for now.
For the first part (there is at most one vector space structure), it is probably easiest to only use the fact that $\eta_E$ is surjective (you might know this as an onto map). That is: If $\eta : X \to Y$ is a surjective map and $X$ is a vector space, show that there is at most one vector space structure on $Y$ that makes $\eta$ into a linear map.
The “iff” in the second part is just wrong. That $[0]$ is a subspace of $X$ is not sufficient for there to be such a vector space structure on $X/E$. (Can you find a counter-example?)
However, if there is such a vector space structure, $[0]$ will be a subspace and $E$ will be of the form given in the exercise. As a first step, show that $[0]$ is the kernel of $\eta_E$.