Let $f:X\dashrightarrow Y$ be a birational map between projective varieties not contracting any divisor. Assume that $X$ is smooth, and that $Y$ has at most ordinary singularities at finitely many points. 


Is it true then that $K\_X = f^{\*}K\_Y$ ? 



Pull-back of the canonical divisor via a rational map
There are several issues with this question.


One issue is that if $f$ is a rational map and not a morphism, then you have to say what you mean by $f^\*$. Another issue is that if $K\_Y$ is not at least $\mathbb Q$-Cartier, then $f^\*K\_Y$ doesn't make sense even if $f$ is a morphism. (More precisely, there is no good way to define $f^\*K\_Y$. In other words, $f^\*$ can really only be defined for $\mathbb Q$-Cartier divisors).


I am also not entirely sure what "ordinary singularities" mean, but let me assume that it includes at least "normal". You kind of need that to talk about divisors anyway. I guess it might also imply that $K\_Y$ is $\mathbb Q$-Cartier, in which case the second issue is resolved. 


I guess you could do the following: If $X$ is normal, then $f$ is defined in codimension $1$, so let $U\subseteq X$ be a dense open set where $f$ is defined and $X\setminus U$ is at least codimension $2$ in $X$. If $f$ does not contract a divisor, then the exceptional set is also at least of codimension $2$, so we might as well assume that $f|\_U$ is an isomorphism onto its image. So, you could define "$f^\*K\_Y$" as the "closure" of $K\_U$ on $X$ which should indeed be $K\_X$, but this feels more as a push-forward from $Y$ to $X$ than a pull-back, so I'd rather denote it (as it is done in the literature) by $f^{-1}\_\*K\_Y$. But with that it is true that $f^{-1}\_\*K\_Y\sim K\_X$ and you don't even have to worry about $K\_Y$ being $\mathbb Q$-Cartier.


