Suppose we have a fiber square (pullback diagram) of schemes.
$$\begin{matrix}
 \hspace{.7cm} X' \stackrel{v}{\longrightarrow} & X \cr
 \downarrow^{g} & \downarrow^f \cr 
 \hspace{.7cm} Y' \stackrel{u}{\longrightarrow} & Y 
 \end{matrix}$$
It is well known that if $u$ is flat then $\mathbf{L}u^\ast \circ \mathbf{R}f\_\ast \cong \mathbf{R} g\_\ast \circ \mathbf{L}v^\ast$. 
I am interested in the following question. When there is the isomorphism $$u^\ast\circ f\_\ast\cong g\_\ast\circ v^\ast$$ 
where these functors considered just as functors between categoties of quasi-coherent sheaves (without derived categories and derived functors)? It's easy to see that it's true when all the schemes are affine, without any conditions for morphisms. Could it be true in general case without any conditions for schemes and morphisms?



pullback-pushforward isomorphism without derived functors
I don't think this is true in general. 


For instance, 


* Set $f : X \to Y = \mathbb{A}^2$ be the blowup of the origin.
* Set $Y' = \text{Spec} k$ and then fix $u : Y' \to Y$ to be the inclusion of the origin (note that $u$ is not flat).


It follows that $X'$ is the reduced exceptional divisor of $f$, in other words, $X'$ is a copy of $\mathbb{P}^1$. 


Consider a sheaf $G = O\_X(n X')$ for some integer $n > 0$ (here I am viewing $X'$ as a divisor in $X$). We pullback and pushforward in two ways.


1. Now $f\_\* G = O\_Y$ since we are just allowing a pole of some order over a point. Therefore, $u^\* f\_\* G = u^\* O\_Y = O\_{Y'} = k$.
2. On the other hand, $v^\* G = v^\* O\_X(nX') = O\_{X'}(-n)$ since $X'$ has self-intersection -1 on $X$. Thus 
$$
g\_\* v^\* O\_X(nX') = g\_\* O\_{X'}(-n) = H^0(X', O\_{X'}(-n)) = 0.
$$


So since $0 \neq k$ it's not true.


