Passing vector space structure via surjective map Let $V$ be a $K$-vector space, and let $W$ be a set. Let further $+ : W \times W \to W , \cdot : K \times W \to W$ be maps, and let 
$ A : V \to W$ be a surjective map, such that
$$ \forall x ,y \in V \forall \lambda \in K : A(\lambda x + y) = \lambda A(x) + A(y). $$

I am asked to show that this induces a $K$-vector space structure on $W$.

However, I can't seem to prove either of the following equivalent statements $$ \begin{align*} (1) &amp;\text{  } A(0) \text{ is right-neutral}, \\ (2)&amp;\text{  } 1\cdot w = w \text{ for any }w \in W. \end{align*} $$

I am now very uncertain, whether the statement actually holds (without assuming (1) or (2)) - but I can't find a counter example either. Thus my question: is the statement true?
For $1)$ Let $w = A(v)$, which we can do since $A$ is surjective. $w + A(0) = A(v) + A(0) = A(v+0) = A(v) = w$, so $A(0)$ is right-neutral

For $2)$ let $w = A(v)$, which we can do since A is surjective. Then $1\cdot w = 1\cdot A(v) = A(1\cdot v) = A(v) = w$, as needed.

Edit: Okay. Currently, I think it's possible that we don't necessarily get a vector space structure out of the above definition without at least a few extra structural demands on + and $\cdot$ on W. However, the proper rebuke would come in the form of a counterexample structure... which is too much to ask.
For $1)$ Let $w = A(v)$, which we can do since $A$ is surjective. $w + A(0) = A(v) + A(0) = A(v+0) = A(v) = w$, so $A(0)$ is right-neutral

For $2)$ let $w = A(v)$, which we can do since A is surjective. Then $1\cdot w = 1\cdot A(v) = A(1\cdot v) = A(v) = w$, as needed.

Edit: Okay. Currently, I think it's possible that we don't necessarily get a vector space structure out of the above definition without at least a few extra structural demands on + and $\cdot$ on W. However, the proper rebuke would come in the form of a counterexample structure... which is too much to ask.