TITLE: commutativity of restriction and Gysin morphisms in a cartesian square
QUESTION [5 upvotes]: Let $X, Y, Z$ be compact topological manifolds $f: Y \to X, g: Z \to X$ be embeddings of submanifolds meeting transversely and let $W = Y \times_X Z$:
$$
\begin{array}{ccc}
    Y & \to^f & X \\
    \uparrow^G & & \uparrow^g\\
    W & \to^F & Z \\
\end{array}
$$
My question is: How does one show that the two morphisms $F_! \circ G^*$ and $g^*
\circ f_!$ from $H^*(Y, \mathbb Q)$ to $H^*(Z, \mathbb Q)$ coincide?
Here $F_!, f_!$ are Gysin morphisms; let us not concern ourselves with degree shift for brevity.
Using the natural morphisms ${\mathbb Q}_Y \to G_*{\mathbb Q}_W$ and ${\mathbb Q}_W \to RF^!{\mathbb Q}_Z$ and adjunction the first morphism can be represented as the composition of the following morphisms
$$
RHom_Y({\mathbb Q}_Y, {\mathbb Q}_Y) \to RHom_Y({\mathbb Q}_Y, G_* {\mathbb Q}_W) \cong RHom_W(G^*{\mathbb Q}_Y,{\mathbb Q}_W) \cong RHom_W({\mathbb Q}_W,{\mathbb Q}_W) \to RHom_W({\mathbb Q}_W, RF^!{\mathbb Q}_Z) \cong RHom_Z(RF_! {\mathbb Q}_W, {\mathbb Q}_Z) \cong RHom_Z({\mathbb Q}_Z, {\mathbb Q}_Z) 
$$
and using the natural morphisms  ${\mathbb Q}_Y \to Rf^!{\mathbb Q}_X$ and  ${\mathbb Q}_X \to g_*{\mathbb Q}_Z$  the second morphism can be represented as the composition of
$$
RHom_Y({\mathbb Q}_Y, {\mathbb Q}_Y) \to RHom_Y({\mathbb Q}_Y, Rf^!{\mathbb Q}_X) \cong RHom_X(Rf_!{\mathbb Q}_Y, {\mathbb Q}_X) \cong RHom_X({\mathbb Q}_X, {\mathbb Q}_X) \to RHom_X({\mathbb Q}_X, g_*{\mathbb Q}_Z) \cong RHom_Z(g^*{\mathbb Q}_X, {\mathbb Q}_Z) \cong RHom_Z({\mathbb Q}_Z, {\mathbb Q}_Z)
$$
The base change theorem (Iversen, "Cohomology of sheaves" VII.2.6)
gives an isomorphism of the corresponding functors, i.e. $RF_! \circ
G^* \mathcal F \cong g^* \circ Rf_!  \mathcal F$ functorially in the object $\mathcal F$ in the derived category. I am not sure how one aplies this statement about functors to deduce the statement about morphisms on cohomology.

REPLY [3 votes]: Let me sketch how I would do it in the six functor formalism. Let us consider slightly more generally a cartesian square of spaces
$$
\begin{array}{ccc}
    Y & \to^f & X \\
    \uparrow^G & & \uparrow^g\\
    W & \to^F & Z \\
\end{array}
$$
in which:

$f$ is proper
we have fixed an isomorphism $f^! \mathbf Z = \mathbf Z[d]$ for some $d$ (i.e. $f$ is oriented)
the map $G^\ast f^! \to F^! g^\ast$, obtained as the mate of the inverse of $g^\ast f_!\stackrel \sim\to F_!G^\ast $, is an isomorphism (transversality). In particular $F$ is also oriented.

We have a commutative diagram of natural transformations between endofunctors of the derived category $D(X)$ (all functors derived from now on)
$$
\begin{array}{ccc}
    f_!f^! & \to & \mathrm{id} \\
    \downarrow & & \downarrow\\
    g_\ast g^\ast f_!f^! & \to & g_\ast g^\ast \\
\end{array}
$$
and I claim that if we apply this diagram to the unit object $\mathbf Z$ and take global sections then we obtain a commuting diagram
$$
\begin{array}{ccc}
    H(Y)[d] & \to & H(X) \\
    \downarrow & & \downarrow\\
    H(W)[d] & \to & H(Z) \\
\end{array}
$$
where the vertical maps are pullback in cohomology and horizontal are Gysin maps. The point is that using base change ($g^\ast f_! = F_! G^\ast$), transversality ($G^\ast f^! = F^! g^\ast$), and properness+commutativity (in the form $f_!G_\ast = g_\ast F_!$) it follows that
$$ f_! G_\ast G^\ast f^! \cong g_\ast g^\ast f_!f^! \cong g_\ast F_!F^! g^\ast $$
where the first expression makes clear that the left vertical map is pullback in cohomology, the second makes clear commutativity of the square above, third makes clear that the lower horizontal map is the Gysin map.
What is needed to finish the proof is to check that the isomorphisms above actually make the diagrams commute. So, first of all, both sides of
$$f_!G_\ast G^\ast \cong g_\ast g^\ast f_!$$
receive a natural transformation from $f_!$, and one must check that the obvious triangle commutes. This means that we need a stronger statement than just base change in the form of the existence of some isomorphism $F_!G^\ast \cong g^\ast f_!$, we want more specifically that the map $g^\ast f_! \to F_! G^\ast$ obtained as the mate of $ f_! G_\ast\stackrel \sim \to g_\ast F_! $ is an isomorphism.
Similarly both sides of
$$ g^\ast f_!f^! \cong F_!F^! g^\ast $$
admit a natural transformation to $g^\ast$, and again one should check that the triangle commutes, and again it is because the transversality and base change isomorphisms are obtained as mates.