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]. 



Base change for Borel-Moore homology
$\require{AMScd}$
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 $g^\*f\_\* = f'\_\*\tilde g^\*:H\_\*(Y)\to H\_\*(X')$. 


