Given a flat and projective morphism of noetherian schemes, $f: X \rightarrow Y$ and $F$, $G$ two coherent $O\_X$-modules, flat over $Y$. Furthermore given a morphism $u: Y' \rightarrow Y$ of noetherian schemes and a quasi coherent $O\_{Y'}$-module $\mathcal{M}$. Then we have $X':=X\times\_YY'$ with $p\_i$ the i-th projection.


1) Take a locally free resolution $P\_{\cdot} \rightarrow F \rightarrow 0$. Now the article says: "Since $f$ and $F$ are flat over $Y$, $p\_1^{\\*}P\_{\cdot}$ is a resolution of $p\_1^{\\*}F$". Don't we need $u$ to be flat for this? Why is this implied by the flatness of $f$ and $F$?


2) Assume $Y$ and $Y'$ are affine, say $Y=Spec(A)$ and $Y'=Spec(A')$. Given an $A'$-module $M$. For a descending induction, the article says there is an exact sequence:


$0\rightarrow p\_1^{\\*}G'(n)\otimes\_{A'} M \rightarrow \bigoplus\limits\_{i=1}^r O\_{X'}(-k+n)\otimes\_{A'} M \rightarrow p\_1^{\\*}G(n)\otimes\_{A'} M\rightarrow 0$.


I see that we have $0\rightarrow G'(n) \rightarrow \bigoplus\limits\_{i=1}^r O\_{X}(-k+n)\rightarrow G(n)\rightarrow 0$ on X, now if $p\_1$ would be flat (see (1)), we'd have $0\rightarrow p\_1^{\\*}G'(n) \rightarrow \bigoplus\limits\_{i=1}^r O\_{X'}(-k+n) \rightarrow p\_1^{\\*}G(n)\rightarrow 0$ on $X'$, but why is $\otimes\_{A'} M$ exact in this case? Or is there any other way to construct such a sequence?


Background: I'm reading the article "Universal families of extensions" by Herbert Lange (J. Algebra 83 (1983), 101–112), where he constructs base change homomorphisms for relative $Ext$-sheaves on schemes. I get the basic idea of the construction, but now i'm trying to understand the details, and these two questions stayed open.



Subtleties in the construction of base change morphisms
It is easy, $p\_1^\*G = G\otimes\_A A'$ is flat over $A'$ since $G$ is flat over $A$. Hence $Tor\_1^{A'}(p\_1^\*G,M) = 0$, hence your sequence is exact after tensoring with $M$. The same argument works for the first question as well.


