# pullback-pushforward isomorphism without derived functors

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?

• I think you have to assume $f$ quasi-compact and quasi-separated and $Y$ quasi-compact for the derived formula to be true (if you assume that $u$ is flat). See SGA 4 XVII 4. – Damian Rössler Apr 29 '12 at 7:50

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.

• In general (in you example in particular) even the formula with derived functors is not true. The reason is that the original fiber square is not Tor-independent. To make everything true one has to replace $X'$ with the DERIVED fiber product of $X$ and $Y'$ over $Y$. – Sasha Apr 29 '12 at 3:20
• Sasha, thanks. That makes sense, I don't really feel like I understand the derived fiber product though (although I've seen talks which include it occasionally). Is there a nice survey out there? – Karl Schwede Apr 29 '12 at 4:28
• I guess you can find a machinery in appropriate version of derived algebraic geometry (Toen, Lurie). But morally the problem is that when you define the fiber product you take tensor product and forget higher Tors. But if you wish the base change to hold you should take Tors into account. A reasonable thing to do is to include them into the structure sheaf, thus it should be a DG-algebra (or a simplicial algebra). – Sasha Apr 29 '12 at 4:40

If $u$ is flat then so is $v$ and hence $Lu^* = u^*$, $Lv^* = v^*$. Therefore for each sheaf $F$ one has $$u^*(Rf_*F) \cong Rg_*(v^*F).$$ Taking the cohomology sheaves we get $$u^*(R^if_*F) \cong R^ig_*(v^*F)$$ for all $i$ (here we also use the fact that $u^*$ and $v^*$ are exact). In particular, for $i = 0$ one obtains $u^*(f_*F) \cong g_*(v^*F)$.