Let $X\subset\mathbb P^4$ a projective hypersurface with an ordinary double point at $o\in X$.

Blow-up $\mathbb P^4$ at $o$ and let $E\simeq\mathbb P^3$ the exceptional divisor of this blow-up. Consider the strict transform $\widetilde X$ of $X$. Then, it is easy to verify that $Y=\widetilde X\cap E$ is a quadric hypersurface in $E\simeq\mathbb P^3$, thus isomorphic to $\mathbb P^1\times\mathbb P^1$.

Now, take an hyperplane (if you want, generic) section $C$ of $Y$ in $E$, that is $C=Y\cap H(\simeq\mathbb P^1)$, where $H$ is a hyperplane in $E\simeq\mathbb P^3$. Obviously, the curve $C$ meets each line of the two rulings of $Y$ in exactly one point. So, choose one ruling and define the obvious morphism $\varphi\colon Y\to C$ obtained by collapsing any line of the choosen ruling to its intersection point with $C$.

**Here is my question.**

Does there exist a *smooth* projective variety $\overline X$, containing a curve isomorphic to $C$ (which I still call $C$), and a regular morphism $\Phi\colon\widetilde X\to\overline X$ such that:

- $\Phi|_{\widetilde X\setminus Y}\colon\widetilde X\setminus Y\quad\overset{\simeq}\to\quad\overline X\setminus C$
- $\Phi|_{Y}\colon Y\to C$ is the $\varphi$ above.

I guess this is very classical, but I wasn't able to find it.

Thanks in advance!