Infinite chinese remainder theorem for $\Bbb{Z}/p, p \text{ prime}$?

Question

Let $X = \prod_{p=2 \\ \text{prime}}^{\infty}\Bbb{Z}/p$. Let $\phi : \Bbb{Z} \to X$ be the natural projection in each component.

Then for any $x \in X$, there exists an integer $y$ such that $\phi(y) = x$ by infinite Chinese remaindering?

Answer 1

No: Take $x = \prod_p x(p)$ with $x(2) = 1$ and all other $x(p) = 0$. If $x = \phi(y)$, then $y$ would have to be nonzero but divisible by every prime $p > 2$.

Answer 2

No:

$X$ is obviously uncountable. How are you going to map onto it from $\mathbb Z$?