Let $X$ be a normal algebraic surface (over $\mathbb{C}$) and $Y$ its smooth locus, i.e., the complement of the singularities of $X$. Suppose $Z\subset Y$ is a Zariski open subset of $X$. Then is it true that $\pi_1(Y)$ is a quotient of $\pi_1(Z)$? (For computing the fundamental group, I'm considering not the Zariski topology but the usual topology.)
I've read a speical case of the above situation in https://projecteuclid.org/journals/nihonkai-mathematical-journal/volume-29/issue-2/Log-del-Pezzo-surfaces-of-rank-one-containing-the-affine/nihmj/1565048776.full, above Corollary 1.1: "If a normal algebraic surface contains the affine plane $\mathbb{C}^2$ as a Zariski open subset, then its smooth locus is simply-connected." I've asked for a proof of this statement in https://math.stackexchange.com/questions/4970345/complex-surface-containing-bbb-c2-as-a-zariski-open-subset, but I didn't get an answer.
Also, I've seen that for a normal algebraic surface containing $\mathbb{C}\times \mathbb{C}^*$ as a Zariski open subset, where $\mathbb{C}^*=\mathbb{C}-\{\text{point}\}$, the fundamental group of its smooth locus is a quotient group of $\pi_1(\mathbb{C}\times \mathbb{C}^*)=\mathbb{Z}$ (but I cannot remember where I've read this).
So it seems that the statement in the first paragraph is true, but I cannot see why. Can this be deduced from a van Kampen theorem argument?