Is this a isomorphism $(\mathbb Z,+) \to (Z,+)$ where $\varphi(n) = 2n$? Why not? This is what I did :
a) $\varphi(x+y) = \varphi(x)\varphi(y) $:
 $2(x+y)= 2x + 2y$ so $\varphi$ is a homomorphism.
b) $\varphi(x) = \varphi(y)$:
   $2(1) = 2(1)$  so $\varphi$ is one-to-one
c) $\varphi(x) = 2n$
$\varphi(1) = 2(1) = 2$ so is it onto? I say yes, but the answer says "$\varphi(n) \neq 1$ for all $n \in \mathbb Z$" so not isomorphic.
Why I say yes: for every element in $\mathbb Z$, the first group, there is a element $2n$ in the second group. What am I missing here?  

As pointed out in the comments, this question results from confusion over the definition of onto.
In order for $\phi$ to be onto, we need there to be an $x$ such that, for every $y$, we have $y=\phi(x)$.  A simple counterexample is $y=1$; since $y$ is odd, there is no integer $x$ such that $y=2x$.
