Let $f:X\to Y$ be a morphism of (nice) topological spaces. consider the basechange along $g:Y'\to Y$. We write $F:X'\to Y'$ and $G:X'\to X$. Consider some class of sheaves on those spaces, e.g. abelian sheaves or sheaves of $k$-vector spaces for a fixed field. We can take this to the unbounded derived categories. Then Spaltenstein showed that there is a natural morphism $Lg^* \circ Rf_!\Rightarrow RF_! \circ LG^*$ and that it is an isomorphism.
However $g$ (and hence $G$) is required to be flat, hence $g^*$ (resp. $G^*$) is exact and we basically only have to right derive, instead of dealing with a left and a right derived functor at the same time.
First question: Is there a way to do this without the flatness condition? If not can somebody provide a counterexample?
Next I would like to bring this together with the formula \begin{equation} Rf_!(-\otimes^L Lf^*-)\cong (Rf_! - )\otimes^L - \end{equation}
To bring them into a single "framework" consider a multimorphism $g:Y'\to Y_1,\ldots Y_n$ consisting of $g_i:Y'\to Y_i$ and define the pullback as $$ g^*(-,\ldots, -):=\bigotimes g_i^*- $$ We get the formula above as base change of
$\require{AMScd}$ \begin{CD} X @>f>> Y\\ @V f,1 V V @VV 1,1 V\\ Y, X@>>1,f> Y,Y \end{CD}
Now I want to prove basechange in the general case, where we have morphisms spaces $X,Y_1,\ldots, Y_n, Y'$ and morphisms $f:X\to Y_j$ for fixed $j$ and $g_i:Y'\to Y_i$ for each $i$ in a cartesian square
\begin{CD} X\times_{Y_j} Y' @>F>> Y'\\ @V G V V @VV g V\\ Y_1,\ldots, X, \ldots, Y_n @>>1,f,1> Y_1,\ldots, Y_j,\ldots, Y_n \end{CD}
where $F=\text{pr}_{Y'}$ and $G_i=\begin{cases} g_i\circ \text{pr}_{Y'} & i\neq j\\ \text{pr}_X & i=j\end{cases}$
and preferably I want to do this without using the other two "base changes", i.e. in one go.
Since $g$ contains a tensor product, $g^*$ will not be exact, even if $g$ is flat.
Second question: How would one construct a natural transformation $Rf_!\circ Lg^*\to LG^*\circ RF_!$ in this generalized case?
