I don't understand the smoothness condition in the following theorem,


Let $f: X\longrightarrow Y$ be a projective morphism of $\underline{smooth}$ projective varieties such that $Rf\_\*\mathcal{O}\_X=\mathcal{O}\_Y$. Then the functor


\begin{equation}
Lf^\* : D^b(Y)\longrightarrow D^b(X)
\end{equation}
is fully faithful.


The proof is easy,
\begin{equation}
Hom\_{D(X)}(Lf^\*\mathcal{F}^{\bullet},Lf^\*\mathcal{F}^{\bullet}) \simeq Hom\_{D(Y)} (\mathcal{F}^{\bullet},Rf\_\*Lf^\*\mathcal{F}^{\bullet}) \simeq Hom\_{D(Y)}(\mathcal{F}^{\bullet},\mathcal{F}^{\bullet}\otimes Rf\_\*\mathcal{O}\_X)\simeq Hom\_{D(Y)}(\mathcal{F}^{\bullet},\mathcal{F}^{\bullet}).
\end{equation}


As far as I know, the projection formula works for any proper morphism, so why we need both varieties to be smooth? 


Can we generalize it to the case which $X$ is smooth and $Y$ is singular?
(Sasha explained why this is impossible, but what about the other case, i.e. $X$ singular, and $Y$ is smooth)


Thanks! 



$Lf^*$ is fully faithful
Smoothness is not necessary. What is important is that $f$ has finite $Tor$-dimension (otherwise $Lf^\*$ does not preserve boundedness); a sufficient (but not necessary) condition for this is smoothness of $Y$.


On the other hand, it is impossible to have $X$ smooth and $Y$ singular. Indeed, in this case one can find objects $F,G \in D^b(Y)$ with $\dim Ext^\bullet(F,G) = \infty$ (e.g. $F = G = O\_y$ for $y \in Sing(Y)$), while for any objects of $D^b(X)$ the analogous dimension is finite.


