$Lf^*$ is fully faithful

Question

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!

Answer 1

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.