Let $X$ be an algebraic variety over $\mathbb C$. Let $X^{an}\to Y$ be a finite etale morphism with $Y$ a complex analytic space.

I read somewhere that $Y$ algebraizes, ie, $Y=V^{an}$ for some algebraic variety $V$ over $\mathbb C$.

Why is this, and what is an "easy" proof for this? (Consequently, the morphism $X^{an}\to Y = V^{an}$ also algebraizes by Riemann's existence theorem.)

Let $X$ be an algebraic variety over $\mathbb C$. Let $X^{an}\to Y$ be a finite etale morphism with $Y$ a complex analytic space.

I read somewhere that $Y$ algebraizes, ie, $Y=V^{an}$ for some algebraic space $V$ over $\mathbb C$. (Edit: I previously wrote $Y$ algebraic variety. For my purposes, I just want $Y$ to be an algebraic space.)

Why is this, and what is an "easy" proof for this? (Consequently, the morphism $X^{an}\to Y = V^{an}$ also algebraizes by Riemann's existence theorem. Actually, this is not right. It algebraizes in the sense that it comes from an algebraic morphism $W\to V$, but $W$ might not be isomorphic to $X$.  )