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 $f^!$.`



Fullness of pullback functor in algebraic geometry
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](http://arxiv.org/abs/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 $x \in X$.


