Let $X$ be a toric variety, and let $\pi: \mathbb A^n-V(B) \to X = (\mathbb A^n-V(B))/(\mathbb C^\*)^\rho$ be the quotient map defining $X$ in the Cox construction. A subvariety $Y\subset X$ is called *quasismooth* if $\pi^{-1}(Y)$ is smooth. Clearly, $Y$ smooth implies $Y$ quasismooth.


If we assume now that $X$ is smooth, is $Y$ quasismooth if and only if it is smooth?



Quasismooth vs smooth in a smooth toric variety
Yes. Morphism $\pi$ is smooth if $X$ is smooth. Smoothness is stable under base change, so $\pi$ is smooth implies $\pi^{-1}(Y) \to Y$ is smooth. By [Lemma 35.17.4 in "Stack Project"](https://stacks.math.columbia.edu/tag/034D) "being regular" (which is equivalent to "being smooth" when the base field is $\mathbb C$) is a property local in the smooth topology, hence, $\pi^{-1}(Y)$ smooth, $\pi^{-1}(Y) \to Y$ smooth and surjective implies $Y$ is smooth.


Why $\pi$ is smooth if $X$ is smooth. $X$ is smooth iff defining fan $\Sigma$ is smooth. Consider fan $\tilde{\Sigma}$ in $\mathbb R^{\Sigma(1)}$ corresponding to toric variety $\mathbb C^{\Sigma(1)} - Z(\Sigma)$ which for every cone $\sigma \in \Sigma$ have cone $\tilde{\sigma} = \operatorname{cone}(e\_{\rho} | \rho \in \sigma(1)) \in \tilde{\Sigma}$ and all corresponding faces (see Cox Little Schenck "Toric varieties" §5.1 for more accurate construction of $\tilde{\Sigma}$). We have to show that morphism $\pi : X\_{\tilde{\Sigma}} \to X\_{\Sigma}$ induced by canonical morphism of fans $\tilde{\Sigma} \to \Sigma$ is smooth. Since smoothness is a property local on source, it is enough to check smoothness of restriction $\pi|\_{U\_{\tilde{\sigma}}}$ for every affine open set $U\_{\tilde{\sigma}}, \sigma \in \Sigma$. Computation in coordinates shows that morphism $\pi|\_{U\_{\tilde{\sigma}}} : U\_{\tilde{\sigma}} \to U\_{\sigma}$ is isomorphic to standard projection $\pi : \mathbb C^{k} \times (\mathbb C^\*)^{\dim X-k} \times (\mathbb C^\*)^{r} \to \mathbb C^{k} \times (\mathbb C^\*)^{\dim X-k}$ where $k=\dim \sigma$, which is definitely smooth.


