Fullness of pullback functor in algebraic geometry

Question

Given $f:X\to Y$ a morphism of schemes (or stacks if it's not harder), I am interested in a geometric reformulation of the condition that the functor $f^*:D^b(Coh(Y))\to D^b(Coh(X))$ is full. I can only find full and faithful appearing together in the literature, and I need to extricate the two conditions. Does anyone know a simple formulation, or a good reference?

Intuitively, asking for $f^*$ to be full seems alot like asking that anytime you have a sheaf $F$ on $Y$, and a section of it defined only on $X$ (i.e. a section of $f^*F$), it can be extended to a section on all of $Y$. And so that would seem to indicate that the image of $f$ should have codimension-two complement. However, that intuition only really applies to underived $f^*$, and maybe deriving $f^*$ eliminates the codimension 2 requirement? Also, this intution is assuming that f is mono, so that $f^*$ is just restriction, which I don't think is true a priori.

As a side-note: I'd be interested in the same question (geometric characterization of fullness) for $f_*$ and $f^!$.`

Answer 1

In case $X, Y$ are smooth, $ f^* $ is full if and only if it is full and faithful. This is explained in the introduction of arXiv:1101.5931 (by Canonaco-Orlov-Stellari), which also studies when this implication holds more generally. Thus the pull-back is full and faithful if and only if $Rf_* \mathcal{O}_X = \mathcal{O}_Y$; for example, when $f$ is birational.

Here is a short explanation of their argument for the case of pull-back functors: $f^*$ is full and faithful if and only if $\mathrm{Ext}^i_{Y}(f^* O_{x_1}, f^* O_{x_2})$ is

1. $\mathbb C$ for $x_1 = x_2$, $i = 0$,
2. 0 for $i \notin [0, \mathrm{dim} Y]$, and
3. 0 for $x_1 \neq x_2$.

(This is due to Bondal-Orlov and Bridgeland.)

Since $f^* $ is full, the 2nd and 3rd condition are automatically satisfied. Thus it will be full and faithful if and only if $f^* O_x \neq 0$ for all $x \in X$.

Pick $x \in X$ such that $f^* O_x$ is non-zero. Note that $\mathrm{Hom}^{\bullet}_Y(f^* O_x, f^* O_x) = \mathrm{Hom}^{\bullet}_X(O_x, f_* f^* O_x)$ is a quotient of $\mathrm{Hom}^{\bullet}_X(O_x, O_x)$, and $f_* f^* O_x$ is supported at $x$. From this one can show that $O_x$ is a sheaf, and in fact isomorphic to $O_x$; all we need is that it's Chern character (in cohomology) is non-zero. But the Chern character of $f_* f^* O_x$ is independent of $x$, so $f^* O_x \neq 0$ for all $x \in X$.