Is the function $f(z)=z^2$ injective?

Question

I know this may sound silly, but I'm not so sure if the method of constructing Riemann surfaces for the domain also applies to the codomain. For example, we can make the function $f(z)=\sqrt(z)$ injective by adding to the domain two more complex planes (thus making the codomain equal to $\mathbb{C}$). Is the same thing necessary, but or the codomain, of the function $f(z)=z^2$? If so, when we speak of the function $f(z)=z^2$, is it usually taken as understood that the codomain actually consists of two complex planes (Riemann sheets), just as we usually assume that the domain of the function $f(z)=\sqrt{z}$ is really two complex planes (Riemann sheets)?