Base change for Borel-Moore homology

Question

For a seperated scheme of finite type $X$ over $\mathbf{C}$, let $H_*(X)$ denote its Borel-Moore homology, which is defined by $$ H_k(X) = R^{-k}\Gamma(X, \omega_X) $$ where $\omega\in D_c(X, \mathbf{C})$ is a dualising object in the derived category of constructible $\mathbf{C}$-sheaves on $X$. It is immediate from the definition and the six operations for constructible sheaves that $H_*$ is covariant with respect to proper morphisms and contravariant with respect to smooth morphisms. Let $f_*$ and $f^*$ denote respectively the direct and inverse image maps.

Question

If we are given a cartesian diagramme of schemes over $\mathbf{C}$ $\require{AMScd}$ \begin{CD} Y' @>\tilde g>> Y \\ @VVf'V @VVfV \\ X' @>g>> X \end{CD} where $f$ is proper and $g$ is smooth, do we have an equality of the maps between homology groups $f'_*\tilde g^* = g^*f_*: H_*(Y)\to H_*(X')$?

In the book [N. Chriss & V. Ginzburg, Representation theory and complex geometry, 8.3.34] , this is shown in the case where $g$ is locally a trivial fibration, by reducing it to the commutativity of the following diagramme of derived functors

\begin{CD} f_*f^! @>>> \mathrm{id}_X @>>> g_*g^* \\ @VVV @. @AAA \\ f_*\tilde g_*\tilde g^*f^! @. = @. g_* f'_*{f'}^! g^* \end{CD}

in which all the morphisms are adjunction morphisms except for the base change morphism $\tilde g^*f^! \cong {f'}^! g^*$ in the bottom line, whose definition can be found in [SGA4, Exposé 18, 3.1.14.2]. One gets the result by apply the diagramme to $\omega_X$ and then take global sections.

The commutativity of the diagramme above can be easily checked in the case where $g: X'\to X$ is locally a trivial fibration. While I don't know how to do it if $g$ is only assumed to be smooth, I still suspect the commutativity to remain true.

Let me also add that there is such an equality $f'_*\tilde g^* = g^*f_*: A_*(Y)\to A_*(X')$ for Chow groups and for $f$ proper and $g$ flat, see for example [Fulton, Intersection theory, Prop. 1.7].

Answer 1

$\require{AMScd}$ After the struggle of a whole day due to my unfamiliarity with the category theory, I found the answer.

The commutativity is due essentially to the following facts:

1. Given any morphism of schemes $f: X \to Y$ we have the adjunction $(f^*, f_*, \epsilon_f, \eta_f)$: $$ f^*: D_c(Y, \mathbf{C})\to D_c(X, \mathbf{C}),\; f_*: D_c(X, \mathbf{C})\to D_c(Y, \mathbf{C}) $$ $$ \epsilon_f : \mathrm{id} \to f_*f^*, \; \eta_f : f^*f_* \to \mathrm{id} $$ Then, given any two morphisms $X\xrightarrow{f}Y\xrightarrow{g} Z$, we then have $(g_*\epsilon_fg^*)\epsilon_g = \epsilon_{gf}$, as indicated below \begin{CD} \mathrm{id}@>\epsilon_{gf}>> (gf)_*(gf)^* \\ @V\epsilon_{g}VV @| \\ g_*g^*@>g_*\epsilon_fg^*>> g_*f_*f^*g^* \end{CD}
2. Given any morphism of schemes $f: X \to Y$ we have the adjunction $(f_!, f^!, \sigma_f, \tau_f)$: $$ f_!: D_c(X, \mathbf{C})\to D_c(Y, \mathbf{C}),\; f^!: D_c(Y, \mathbf{C})\to D_c(X, \mathbf{C}) $$ $$ \sigma_f : \mathrm{id} \to f^!f_!, \; \tau_f : f_!f^! \to \mathrm{id} $$ Then, given any two morphisms $X\xrightarrow{f}Y\xrightarrow{g} Z$, we then have $\tau_g(g_!\tau_fg^!) = \tau_{gf}$, as indicated below \begin{CD} g_!f_!f^!g^!@>g_!\tau_{f}g^!>> f_!f^! \\ @| @V\tau_{f}VV \\ (gf)_!(gf)^!@>\tau_{gf}>> \mathrm{id} \end{CD}

The first one is obvious, while the second one is a sheaf-theoretic version of the Fubini theorem.

Now we turn to the diagramme in question \begin{CD} Y' @>\tilde g>> Y \\ @Vf'VV @VfVV\\ X' @>g>> X \end{CD} where $f$ is proper and $g$ is smooth and equidimensional.

Suppose we are given two objects $A, B\in D_c(X, \mathbf{C})$ and a morphism $\varphi\in \mathrm{Hom}(f^*A, f^!B)\cong \mathrm{Hom}(A, f_*f^!B)$. Form the following diagramme \begin{CD} g^* A @>g^*\epsilon_f>> g^*f_*f^* A @>g^*f_*\varphi>> g^*f_*f^! B @>g^*\tau_f>> g^* B \\ @V\epsilon_{f'}g^*VV @V\mathrm{BC}VV @V\mathrm{BC}VV @A\tau_{f'}g^*AA \\ f'_*{f'}^*g^*A @= f'_*\tilde g^*f^*A @>f'_*\tilde g^*\varphi>> f'_*\tilde g^*f^!B @= f'_*{f'}^!g^*B \end{CD} This diagramme is commutative:

• The square in the middle is commutative, by applying the base change isomorphism (BC) $g^*f_*\cong f'_*\tilde g^*$ to $\varphi$.
• The square in the left is equivalent to the following adjoint \begin{CD} A @>\epsilon_f>> f_*f^* A \\ @V(g_*\epsilon_{f'}g^*)\epsilon_gVV @Vf_*\epsilon_{\tilde g}f^*VV \\ g_*f'_*{f'}^*g^*A @= f_*\tilde g_*\tilde g^*f^*A \end{CD} This is commutative, as we have mentionned in the above fact 1. that $(g_*\epsilon_{f'}g^*)\epsilon_g = \epsilon_{gf'}= \epsilon_{f\tilde g} = (f_*\epsilon_{\tilde g}f^*)\epsilon_f$.
• Similarly, the square in the right is commutative in regard of the fact 2 and the canonical isomorphisms $g^* \cong g^![2d], \; \tilde g^* \cong \tilde g^![2d]$, where $d = \dim X'/X$.

Now, if we set $A = \mathbf{C}$ and $B = \omega_X[-k]$ in $D_c(X, \mathbf{C})$ for $k\in \mathbf{Z}$, then every homology class $\alpha\in H_k(Y)\cong \mathrm{Hom}_{D_c(Y)}(f^*A, f^! B)$ corresponds to a morphism $\varphi_{\alpha}: f^*A \to f^! B$. Then the composite $$(g^*\tau_f)(g^*f_*\varphi_{\alpha})(g^*\epsilon_f): g^*A \to g^*B$$ corresponds exactly to the class $g^*f_*\alpha\in H_{k+2d}(X')$, whereas the composite $$ (\tau_{f'}g^*)(f'_*\tilde g^*\varphi_{\alpha})(\epsilon_{f'}g^*): g^*A \to g^*B $$ corresponds to the class $f'_*\tilde g^*\alpha\in H_{k+2d}(X')$.

The commutativity of the above 8-term diagramme applying to $\varphi = \varphi_{\alpha}$ then tells us that $$ (g^*\tau_f)(g^*f_*\varphi_{\alpha})(g^*\epsilon_f) = (\tau_{f'}g^*)(f'_*\tilde g^*\varphi_{\alpha})(\epsilon_{f'}g^*) $$ and consequently $g^*f_*\alpha = f'_*\tilde g^*\alpha$. Thus $g^*f_* = f'_*\tilde g^*:H_*(Y)\to H_*(X')$.