**Proof.**
We have a commutative diagram

\[ \xymatrix{ X' \times _ X X' \ar[d]^{f''} \ar@<1ex>[r]^-a \ar@<-1ex>[r]_-b & X' \ar[d]^{f'} \ar[r]_ h & X \ar[d]^ f \\ Y' \times _ Y Y' \ar@<1ex>[r]^-p \ar@<-1ex>[r]_-q & Y' \ar[r]^ g & Y } \]

The element $res_1(c')$ is the restriction (see Remark 42.33.5) of $c'$ for the cartesian square with morphisms $a, f', p, f''$ and the element $res_2(c')$ is the restriction of $c'$ for the cartesian square with morphisms $b, f', q, f''$. Assume $res_1(c') = res_2(c')$ and let $\beta \in \mathop{\mathrm{CH}}\nolimits _ k(Y)$. By Lemma 42.22.4 we can find a $\beta ' \in \mathop{\mathrm{CH}}\nolimits _ k(Y')$ with $g_*\beta ' = \beta $. Then we set

\[ c \cap \beta = h_*(c' \cap \beta ') \]

To see that this is independent of the choice of $\beta '$ it suffices to show that $h_*(c' \cap (p_*\gamma - q_*\gamma ))$ is zero for $\gamma \in \mathop{\mathrm{CH}}\nolimits _ k(Y' \times _ Y Y')$. Since $c'$ is a bivariant class we have

\[ h_*(c' \cap (p_*\gamma - q_*\gamma )) = h_*(a_*(c' \cap \gamma ) - b_*(c' \cap \gamma )) = 0 \]

the last equality since $h_* \circ a_* = h_* \circ b_*$ as $h \circ a = h \circ b$.

Observe that our choice for $c \cap \beta $ is forced by the requirement that $res(c) = c'$ and the compatibility of bivariant classes with proper pushforward.

Of course, in order to define the bivariant class $c$ we need to construct maps $c \cap -: \mathop{\mathrm{CH}}\nolimits _ k(Y_1) \to \mathop{\mathrm{CH}}\nolimits _{k + p}(Y_1 \times _ Y X)$ for any morphism $Y_1 \to Y$ locally of finite type satisfying the conditions listed in Definition 42.33.1. Denote $Y'_1 = Y' \times _ Y Y_1$, $X_1 = X \times _ Y Y_1$. The morphism $Y'_1 \to Y_1$ is an envelope by Lemma 42.22.3. Hence we can use the base changed diagram

\[ \xymatrix{ X'_1 \times _{X_1} X'_1 \ar[d]^{f''_1} \ar@<1ex>[r]^-{a_1} \ar@<-1ex>[r]_-{b_1} & X'_1 \ar[d]^{f'_1} \ar[r]_{h_1} & X_1 \ar[d]^{f_1} \\ Y'_1 \times _{Y_1} Y'_1 \ar@<1ex>[r]^-{p_1} \ar@<-1ex>[r]_-{q_1} & Y'_1 \ar[r]^{g_1} & Y_1 } \]

and the same arguments to get a well defined map $c \cap - : \mathop{\mathrm{CH}}\nolimits _ k(Y_1) \to \mathop{\mathrm{CH}}\nolimits _{k + p}(X_1)$ as before.

Next, we have to check conditions (1), (2), and (3) of Definition 42.33.1 for $c$. For example, suppose that $t : Y_2 \to Y_1$ is a proper morphism of schemes locally of finite type over $Y$. Denote as above the base changes of the first diagram to $Y_1$, resp. $Y_2$, by subscripts ${}_1$, resp. ${}_2$. Denote $t' : Y'_2 \to Y'_1$, $s : X_2 \to X_1$, and $s' : X'_2 \to X'_1$ the base changes of $t$ to $Y'$, $X$, and $X'$. We have to show that

\[ s_*(c \cap \beta _2) = c \cap t_*\beta _2 \]

for $\beta _2 \in \mathop{\mathrm{CH}}\nolimits _ k(Y_2)$. Choose $\beta '_2 \in \mathop{\mathrm{CH}}\nolimits _ k(Y'_2)$ with $g_{2, *}\beta '_2 = \beta _2$. Since $c'$ is a bivariant class and the diagrams

\[ \vcenter { \xymatrix{ X'_2 \ar[d]_{s'} \ar[r]_{h_2} & X_2 \ar[d]^ s \\ X'_1 \ar[r]^{h_1} & X_1 } } \quad \text{and}\quad \vcenter { \xymatrix{ X'_2 \ar[d]_{s'} \ar[r]_{f'_2} & Y'_2 \ar[d]^{t'} \\ X'_2 \ar[r]^{f'_1} & Y'_1 } } \]

are cartesian we have

\[ s_*(c \cap \beta _2) = s_*(h_{2, *}(c' \cap \beta '_2)) = h_{1, *}s'_*(c' \cap \beta '_2) = h_{1, *}(c' \cap (t'_*\beta '_2)) \]

and the final expression computes $c \cap t_*\beta _2$ by construction: $t'_*\beta '_2 \in \mathop{\mathrm{CH}}\nolimits _ k(Y'_1)$ is a class whose image by $g_{1, *}$ is $t_*\beta _2$. This proves condition (1). The other conditions are proved in the same manner and we omit the detailed arguments.
$\square$

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