Let $f: X\rightarrow Y$ be a birational morphism between normal varieties. Let $V\subset X$ be a prime divisor of $X$, i.e. irreducible closed subvariety of codimension 1.
Choosing an open subset $U$ of $X$ for which $f$ is regular on such that $V\cap U \neq \emptyset$, we define the birational transform of $V$ to the closure of $f(U\cap V)$ in $Y$, and we denote it as $\tilde{V}$. My question is, is $\tilde{V}$ still of codimension 1 in $Y$?
We know that $\dim X = \dim Y$ since they are birational to one another. Suppose $f$ maps $U$ isomorphically to an open subset $U’ \subset Y$. Then $1=\operatorname{codim}(U\cap V , U) = \operatorname{codim}(f(U\cap V), U’)$ (please correct me if I’m wrong.)
Can we further assert that $\operatorname{codim}(\tilde{V},Y) =1$? This appears to be the case in many of the resources I have encountered online but I can’t seem to see why. This question may be obvious to many but I just wish to untangle some technical baggages.. any help given would be greatly appreciated!