hcju/stacksqa huggingface dataset:
Since the $\delta$-dimension of a closed subset is the supremum of the $\delta$-dimensions of the irreducible components, it suffices to prove the final statement. We may replace $Y$ by the integral closed subscheme $Z$ and $X$ by $f^{-1}(Z) = Z \times_Y X$. Hence we may assume $Z = Y$ is integral and $f$ is a flat morphism of relative dimension $r$. Since $Y$ is locally Noetherian the morphism $f$ which is locally of finite type, is actually locally of finite presentation. Hence Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-fppf-open} applies and we see that $f$ is open. Let $\xi \in X$ be a generic point of an irreducible component of $X$. By the openness of $f$ we see that $f(\xi)$ is the generic point $\eta$ of $Z = Y$. Thus $Z'$ dominates $Z = Y$. Finally, we see  that $\xi$ and $\eta$ are in the schematic locus of $X$ and $Y$ by Properties of Spaces, Proposition \ref{spaces-properties-proposition-locally-quasi-separated-open-dense-scheme}. Since $\xi$ is a generic point of $X$ we see that $\mathcal{O}_{X, \xi} = \mathcal{O}_{X_\eta, \xi}$ has only one prime ideal and hence has dimension $0$ (we may use usual local rings as $\xi$ and $\eta$ are in the schematic loci of $X$ and $Y$). Thus by Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-dimension-fibre-at-a-point} (and the definition of morphisms of given relative dimension) we conclude that the transcendence degree of $\kappa(\xi)$ over $\kappa(\eta)$ is $r$. In other words, $\delta(\xi) = \delta(\eta) + r$ as desired.
The composition is flat of relative dimension $r + s$ by Morphisms of Spaces, Lemmas \ref{spaces-morphisms-lemma-dimension-fibre-at-a-point-additive} and \ref{spaces-morphisms-lemma-composition-flat}. Suppose that \begin{enumerate} \item $A \subset Z$ is a closed integral subspace of $\delta$-dimension $k$, \item $A' \subset Y$ is a closed integral subspace of $\delta$-dimension $k + s$ with $A' \subset g^{-1}(A)$, and \item $A'' \subset Y$ is a closed integral subspace of $\delta$-dimension $k + s + r$ with $A'' \subset f^{-1}(W')$. \end{enumerate} We have to show that the coefficient $n$ of $[A'']$ in $(g \circ f)^*[A]$ agrees with the coefficient $m$ of $[A'']$ in $f^*(g^*[A])$. We may choose a commutative diagram $$ \xymatrix{ U \ar[d] \ar[r] & V \ar[d] \ar[r] & W \ar[d] \\ X \ar[r] & Y \ar[r] & Z } $$ where $U, V, W$ are schemes, the vertical arrows are \'etale, and there exist points $u \in U$, $v \in V$, $w \in W$ such that $u \mapsto v \mapsto w$ and such that $u, v, w$ map to the generic points of $A'', A', A$. (Details omitted.) Then we have flat local ring homorphisms $\mathcal{O}_{W, w} \to \mathcal{O}_{V, v}$, $\mathcal{O}_{V, v} \to \mathcal{O}_{U, u}$, and repeatedly using Lemma \ref{lemma-length} we find $$ n = \text{length}_{\mathcal{O}_{U, u}}( \mathcal{O}_{U, u}/\mathfrak m_w\mathcal{O}_{U, u}) $$ and $$ m = \text{length}_{\mathcal{O}_{V, v}}( \mathcal{O}_{V, v}/\mathfrak m_w\mathcal{O}_{V, v}) \text{length}_{\mathcal{O}_{U, u}}( \mathcal{O}_{U, u}/\mathfrak m_v\mathcal{O}_{U, u}) $$ Hence the equality follows from Algebra, Lemma \ref{algebra-lemma-pullback-transitive}.
Part (1) follows from part (2) by Lemma \ref{lemma-cycle-closed-coherent} and the fact that $f^*\mathcal{O}_Z = \mathcal{O}_{f^{-1}(Z)}$. \medskip\noindent Proof of (2). As $X$, $Y$ are locally Noetherian we may apply Cohomology of Spaces, Lemma \ref{spaces-cohomology-lemma-coherent-Noetherian} to see that $\mathcal{F}$ is of finite type, hence $f^*\mathcal{F}$ is of finite type (Modules on Sites, Lemma \ref{sites-modules-lemma-local-pullback}), hence $f^*\mathcal{F}$ is coherent (Cohomology of Spaces, Lemma \ref{spaces-cohomology-lemma-coherent-Noetherian} again). Thus the lemma makes sense. Let $W \subset Y$ be an integral closed subspace of $\delta$-dimension $k$, and let $W' \subset X$ be an integral closed subspace of dimension $k + r$ mapping into $W$ under $f$. We have to show that the coefficient $n$ of $[W']$ in $f^*[{\mathcal F}]_k$ agrees with the coefficient $m$ of $[W']$ in $[f^*{\mathcal F}]_{k+r}$. We may choose a commutative diagram $$ \xymatrix{ U \ar[d] \ar[r] & V \ar[d] \\ X \ar[r] & Y } $$ where $U, V$ are schemes, the vertical arrows are \'etale, and there exist points $u \in U$, $v \in V$ such that $u \mapsto v$ and such that $u, v$ map to the generic points of $W', W$. (Details omitted.) Consider the stalk $M = (\mathcal{F}|_V)_v$ as an $\mathcal{O}_{V, v}$-module. (Note that $M$ has finite length by our dimension assumptions, but we actually do not need to verify this. See Lemma \ref{lemma-length-finite}.) We have $(f^*\mathcal{F}|_U)_u = \mathcal{O}_{U, u} \otimes_{\mathcal{O}_{V, v}} M$. Thus we see that $$ n = \text{length}_{\mathcal{O}_{U, u}} (\mathcal{O}_{U, u} \otimes_{\mathcal{O}_{V, v}} M) \quad \text{and} \quad m = \text{length}_{\mathcal{O}_{V, v}}(M) \text{length}_{\mathcal{O}_{V, v}}( \mathcal{O}_{U, u}/\mathfrak m_v \mathcal{O}_{U, u}) $$ Thus the equality follows from Algebra, Lemma \ref{algebra-lemma-pullback-module}.
The assertion that $f'$ is proper follows from Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-base-change-proper}. The assertion that $g'$ is flat of relative dimension $r$ follows from Morphisms of Spaces, Lemmas \ref{spaces-morphisms-lemma-dimension-fibre-after-base-change} and \ref{spaces-morphisms-lemma-base-change-flat}. It suffices to prove the equality of cycles when $\alpha = [W]$ for some integral closed subspace $W \subset X$ of $\delta$-dimension $k$. Note that in this case we have $\alpha = [\mathcal{O}_W]_k$, see Lemma \ref{lemma-cycle-closed-coherent}. By Lemmas \ref{lemma-cycle-push-sheaf} and \ref{lemma-pullback-coherent} it therefore suffices to show that $f'_*(g')^*\mathcal{O}_W$ is isomorphic to $g^*f_*\mathcal{O}_W$. This follows from cohomology and base change, see Cohomology of Spaces, Lemma \ref{spaces-cohomology-lemma-flat-base-change-cohomology}.
A finite locally free morphism is flat and finite by Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-finite-flat}, and a finite morphism is proper by Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-finite-proper}. We omit showing that a finite morphism has relative dimension $0$. Thus the formula makes sense. To prove it, let $Z \subset Y$ be an integral closed subscheme of $\delta$-dimension $k$. It suffices to prove the formula for $\alpha = [Z]$. Since the base change of a finite locally free morphism is finite locally free (Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-base-change-finite-locally-free}) we see that $f_*f^*\mathcal{O}_Z$ is a finite locally free sheaf of rank $d$ on $Z$. Thus clearly $f_*f^*\mathcal{O}_Z$ has length $d$ at the generic point of $Z$. Hence $$ f_*f^*[Z] = f_*f^*[\mathcal{O}_Z]_k = [f_*f^*\mathcal{O}_Z]_k = d[Z] $$ where we have used Lemmas \ref{lemma-pullback-coherent} and \ref{lemma-cycle-push-sheaf}.
Part (1) follows from the definition of a prime divisor (Spaces over Fields, Definition \ref{spaces-over-fields-definition-Weil-divisor}), Decent Spaces, Lemma \ref{decent-spaces-lemma-codimension-local-ring}, and the definition of a dimension function (Topology, Definition \ref{topology-definition-dimension-function}). \medskip\noindent Let $D \subset X$ be an effective Cartier divisor. Let $Z \subset D$ be an irreducible component and let $\xi \in |Z|$ be the generic point. Choose an \'etale neighbourhood $(U, u) \to (X, \xi)$ where $U = \Spec(A)$ and $D \times_X U$ is cut out by a nonzerodivisor $f \in A$, see Divisors on Spaces, Lemma \ref{spaces-divisors-lemma-characterize-effective-Cartier-divisor}. Then $u$ is a generic point of $V(f)$ by Decent Spaces, Lemma \ref{decent-spaces-lemma-decent-generic-points}. Hence $\mathcal{O}_{U, u}$ has dimension $1$ by Krull's Hauptidealsatz (Algebra, Lemma \ref{algebra-lemma-minimal-over-1}). Thus $\xi$ is a codimension $1$ point on $X$ and $Z$ is a prime divisor as desired.
We are going to deduce this from the case of schemes by \'etale localization. Let $Z \subset Y$ be an integral closed subspace of $\delta$-dimension $n - 1$. We want to show that the coefficient of $[Z]$ in $p_*\text{div}(f)$ and $\text{div}(g)$ are equal. Apply Spaces over Fields, Lemma \ref{spaces-over-fields-lemma-finite-in-codim-1} to the morphism $p : X \to Y$ and the generic point $\xi \in |Z|$. We find that we may replace $Y$ by an open subspace containing $\xi$ and assume that $p : X \to Y$ is finite. Pick an \'etale neighbourhood $(V, v) \to (Y, \xi)$ where $V$ is an affine scheme. By Lemma \ref{lemma-etale-pullback} it suffices to prove the equality of cycles after pulling back to $V$. Set $U = V \times_Y X$ and consider the commutative diagram $$ \xymatrix{ U \ar[r]_a \ar[d]_{p'} & X \ar[d]^p \\ V \ar[r]^b & Y } $$ Let $V_j \subset V$, $j = 1, \ldots, m$ be the irreducible components of $V$. For each $i$, let $U_{j, i}$, $i = 1, \ldots, n_j$ be the irreducible components of $U$ dominating $V_j$. Denote $p'_{j, i} : U_{j, i} \to V_j$ the restriction of $p' : U \to V$. By the case of schemes (Chow Homology, Lemma \ref{chow-lemma-proper-pushforward-alteration}) we see that $$ p'_{j, i, *}\text{div}_{U_{j, i}}(f_{j, i}) = \text{div}_{V_j}(g_{j, i}) $$ where $f_{j, i}$ is the restriction of $f$ to $U_{j, i}$ and $g_{j, i}$ is the norm of $f_{j, i}$ along the finite extension $R(U_{j, i})/R(V_j)$. We have \begin{align*} b^* p_*\text{div}_X(f) & = p'_* a^* \text{div}_X(f) \\ & = p'_*\left(\sum\nolimits_{j, i} (U_{j, i} \to U)_*\text{div}_{U_{j, i}}(f_{j, i})\right) \\ & = \sum\nolimits_{j, i} (V_j \to V)_*p'_{j, i, *}\text{div}_{U_{j, i}}(f_{j, i}) \\ & = \sum\nolimits_j (V_j \to V)_*\left(\sum\nolimits_i \text{div}_{V_j}(g_{j, i})\right) \\ & = \sum\nolimits_j (V_j \to V)_*\text{div}_{V_j}(\prod\nolimits_i g_{j, i}) \end{align*} by Lemmas \ref{lemma-flat-pullback-proper-pushforward}, \ref{lemma-etale-pullback-principal-divisor}, and \ref{lemma-compose-pushforward}. To finish the proof, using Lemma \ref{lemma-etale-pullback-principal-divisor} again, it suffices to show that $$ g \circ b|_{V_j} = \prod\nolimits_i g_{j, i} $$ as elements of the function field of $V_j$. In terms of fields this is the following statement: let $L/K$ be a finite extension. Let $M/K$ be a finite separable extension. Write $M \otimes_K L = \prod M_i$. Then for $t \in L$ with images $t_i \in M_i$ the image of $\text{Norm}_{L/K}(t)$ in $M$ is $\prod \text{Norm}_{M_i/M}(t_i)$. We omit the proof.
Let $\{W_j\}_{j \in J}$ be a locally finite collection of integral closed subspaces of $U$ of $\delta$-dimension $k + 1$, and let $f_j \in R(W_j)^*$ be elements such that $(\alpha - \beta)|_U = \sum (i_j)_*\text{div}(f_j)$ as in the definition. Let $W_j' \subset X$ be the corresponding integral closed subspace of $X$, i.e., having the same generic point as $W_j$. Suppose that $V \subset X$ is a quasi-compact open. Then also $V \cap U$ is quasi-compact open in $U$ as $V$ is Noetherian. Hence the set $\{j \in J \mid W_j \cap V \not = \emptyset\} = \{j \in J \mid W'_j \cap V \not = \emptyset\}$ is finite since $\{W_j\}$ is locally finite. In other words we see that $\{W'_j\}$ is also locally finite. Since $R(W_j) = R(W'_j)$ we see that $$ \alpha - \beta - \sum (i'_j)_*\text{div}(f_j) $$ is a cycle on $X$ whose restriction to $U$ is zero. The lemma follows by applying Lemma \ref{lemma-exact-sequence-open}.
What do we have to show? Well, suppose we are given a collection $$ i_j : W_j \longrightarrow Y $$ of closed immersions, with each $W_j$ integral of $\delta$-dimension $k + 1$ and rational functions $g_j \in R(W_j)^*$. Moreover, assume that the collection $\{|i_j|(|W_j|)\}_{j \in J}$ is locally finite in $|Y|$. Then we have to show that $$ f^*(\sum i_{j, *}\text{div}(g_j)) = \sum f^*i_{j, *}\text{div}(g_j) $$ is rationally equivalent to zero on $X$. The sum on the right makes sense by Lemma \ref{lemma-inverse-image-locally-finite}. \medskip\noindent Consider the fibre products $$ i'_j : W'_j = W_j \times_Y X \longrightarrow X. $$ and denote $f_j : W'_j \to W_j$ the first projection. By Lemma \ref{lemma-flat-pullback-proper-pushforward} we can write the sum above as $$ \sum i'_{j, *}(f_j^*\text{div}(g_j)) $$ By Lemma \ref{lemma-prepare-flat-pullback-rational-equivalence} we see that each $f_j^*\text{div}(g_j)$ is rationally equivalent to zero on $W'_j$. Hence each $i'_{j, *}(f_j^*\text{div}(g_j))$ is rationally equivalent to zero. Then the same is true for the displayed sum by the discussion in Remark \ref{remark-infinite-sums-rational-equivalences}.
Let $Z \subset X$ be an integral closed subspace of $\delta$-dimension $n - 1$. Let $\xi \in |Z|$ be its generic point. To prove the first equality we compare the coefficients of $Z$ on both sides. Choose an elementary \'etale neighbourhood $(U, u) \to (X, \xi)$, see Decent Spaces, Section \ref{decent-spaces-section-residue-fields-henselian-local-rings} and recall that $\mathcal{O}_{X, \xi}^h = \mathcal{O}_{U, u}^h$ in this case. After replacing $U$ by an open neighbourhood of $u$ we may assume there is a trivializing section $s_U$ of $\mathcal{L}|_U$. Write $s|_U = f s_U$ for some $f \in \Gamma(U, \mathcal{O}_U)$. Then $Z \times_X U$ is equal to $V(f)$ as a closed subscheme of $U$, see Divisors on Spaces, Definition \ref{spaces-divisors-definition-zero-scheme-s}. As in Spaces over Fields, Section \ref{spaces-over-fields-section-c1} denote $\mathcal{L}_\xi$ the pullback of $\mathcal{L}$ under the canonical morphism $c_\xi : \Spec(\mathcal{O}_{X, \xi}^h) \to X$. Denote $s_\xi$ the pullback of $s_U$; it is a trivialization of $\mathcal{L}_\xi$. Then we see that $c_\xi^*(s) = fs_\xi$. The coefficient of $Z$ in $[Z(s)]_{n - 1}$ is by definition $$ \text{length}_{\mathcal{O}_{U, u}}(\mathcal{O}_{U, u}/f\mathcal{O}_{U, u}) $$ Since $\mathcal{O}_{U, u} \to \mathcal{O}_{X, \xi}^h$ is flat and identifies residue fields this is equal to $$ \text{length}_{\mathcal{O}_{X, \xi}^h} (\mathcal{O}_{X, \xi}^h/f\mathcal{O}_{X, \xi}^h) $$ by Algebra, Lemma \ref{algebra-lemma-pullback-module}. This final quantity is equal to $\text{ord}_{Z, \mathcal{L}}(s)$ by Spaces over Fields, Definition \ref{spaces-over-fields-definition-order-vanishing-meromorphic}, i.e., to the coefficient of $Z$ in $\text{div}_\mathcal{L}(s)$ as desired.
Let $U \to X$ be an \'etale morphism such that $\mathcal{L}|_U$ is trivial. Then $T \times_X U \to U$ is isomorphic to the projection morphism $\mathbf{G}_m \times U \to U$, where $\mathbf{G}_m$ is the multipliciative group scheme, see Groupoids, Example \ref{groupoids-example-multiplicative-group}. Thus (1) is clear. \medskip\noindent To see (2) observe that $q_*q^*\mathcal{L} = \bigoplus_{n \in \mathbf{Z}} \mathcal{L}^{\otimes n + 1}$. Thus there is an obvious isomorphism $q_*q^*\mathcal{L} \to q_*\mathcal{O}_T$ of $q_*\mathcal{O}_T$-modules. By Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-affine-equivalence-modules} this determines an isomorphism $q^*\mathcal{L} \to \mathcal{O}_T$. \medskip\noindent Part (3) holds because forming the relative spectrum commutes with arbitrary base change and the same thing is clearly true for the isomorphism $\alpha$. \medskip\noindent Part (4) follows immediately from (1) and the definitions. \medskip\noindent Part (5) follows from the fact that if $Z$ is an integral algebraic space, then $\mathbf{G}_m \times Z$ is an integral algebraic space. \medskip\noindent Part (6) follows from the fact that lengths are preserved: if $(A, \mathfrak m)$ is a local ring and $B = A[x]_{\mathfrak m A[x]}$ and if $M$ is an $A$-module, then $\text{length}_A(M) = \text{length}_B(M \otimes_A B)$. This implies that if $\mathcal{F}$ is a coherent $\mathcal{O}_X$-module and $\xi \in |X|$ with $\xi' \in |T|$ the generic point of the fibre over $\xi$, then the length of $\mathcal{F}$ at $\xi$ is the same as the length of $q^*\mathcal{F}$ at $\xi'$. Tracing through the definitions this gives (6) and more. \medskip\noindent The map in part (7) comes from Decent Spaces, Remark \ref{decent-spaces-remark-functoriality-henselian-local-ring}. However, in our case we have $$ \Spec(\mathcal{O}_{X, \xi}^h) \times_X T = \mathbf{G}_m \times \Spec(\mathcal{O}_{X, \xi}^h) = \Spec(\mathcal{O}_{X, \xi}^h[t, t^{-1}]) $$ and $\xi'$ corresponds to the generic point of the special fibre of this over $\Spec(\mathcal{O}_{X, \xi}^h)$. Thus $\mathcal{O}_{T, \xi'}^h$ is the henselization of the localization of $\mathcal{O}_{X, \xi}^h[t, t^{-1}]$ at the corresponding prime. Part (7) follows from this and some commutative algebra; details omitted.
Observe that $\text{div}_T(q^*(s)) = \text{div}_{\mathcal{O}_T}(q^*(s))$ by the compatibility between the constructions given in Spaces over Fields, Sections \ref{spaces-over-fields-section-Weil-divisors} and \ref{spaces-over-fields-section-c1}. We will show the agreement with $\text{div}_{\mathcal{O}_T}(q^*(s))$ in this proof. We will use all the properties of $q : T \to X$ stated in Lemma \ref{lemma-Gm-torsor} without further mention. Let $Z \subset T$ be a prime divisor. Then either $Z \to X$ is dominant or $Z = q^{-1}(Z')$ for some prime divisor $Z' \subset X$. If $Z \to X$ is dominant, then the coefficient of $Z$ in either side of the equality of the lemma is zero. Thus we may assume $Z = q^{-1}(Z')$ where $Z' \subset X$ is a prime divisor. Let $\xi' \in |Z'|$ and $\xi \in |Z|$ be the generic points. Then we obtain a commutative diagram $$ \xymatrix{ \Spec(\mathcal{O}_{T, \xi}^h) \ar[r]_-{c_\xi} \ar[d]_h & T \ar[d]^q \\ \Spec(\mathcal{O}_{X, \xi'}^h) \ar[r]^-{c_{\xi'}} & X } $$ see Decent Spaces, Remark \ref{decent-spaces-remark-functoriality-henselian-local-ring}. Choose a trivialization $s_{\xi'}$ of $\mathcal{L}_{\xi'} = c_{\xi'}^*\mathcal{L}$. Then we can use the pullback $s_\xi$ of $s_{\xi'}$ via $h$ as our trivialization of $\mathcal{L}_\xi = c_\xi^* q^*\mathcal{L}$. Write $s/s_{\xi'} = a/b$ for $a, b \in \mathcal{O}_{X, \xi'}$ nonzerodivisors. By definition the coefficient of $Z'$ in $\text{div}_\mathcal{L}(s)$ is $$ \text{length}_{\mathcal{O}_{X, \xi'}^h}( \mathcal{O}_{X, \xi'}^h/a \mathcal{O}_{X, \xi'}^h) - \text{length}_{\mathcal{O}_{X, \xi'}^h}( \mathcal{O}_{X, \xi'}^h/b \mathcal{O}_{X, \xi'}^h) $$ Since $Z = q^{-1}(Z')$, this is also the coefficient of $Z$ in $q^*\text{div}_\mathcal{L}(s)$. Since $\mathcal{O}_{X, \xi'}^h \to \mathcal{O}_{T, \xi}^h$ is flat the elements $a, b$ map to nonzerodivisors in $\mathcal{O}_{T, \xi}^h$. Thus $q^*(s)/s_\xi = a/b$ in the total quotient ring of $\mathcal{O}_{T, \xi}^h$. By definition the coefficient of $Z$ in $\text{div}_T(q^*(s))$ is $$ \text{length}_{\mathcal{O}_{T, \xi}^h}( \mathcal{O}_{T, \xi}^h/a \mathcal{O}_{T, \xi}^h) - \text{length}_{\mathcal{O}_{T, \xi}^h}( \mathcal{O}_{T, \xi}^h/b \mathcal{O}_{T, \xi}^h) $$ The proof is finished because these lengths are the same as before by Algebra, Lemma \ref{algebra-lemma-pullback-module} and the fact that $\mathfrak m_\xi^h = \mathfrak m_{\xi'}^h\mathcal{O}_{T, \xi}^h$ shown in Lemma \ref{lemma-Gm-torsor}.
The additivity follows directly from Spaces over Fields, Lemma \ref{spaces-over-fields-lemma-c1-additive} and the definitions. To see that $c_1(\mathcal{O}_X) \cap \alpha = 0$ consider the section $1 \in \Gamma(X, \mathcal{O}_X)$. This restricts to an everywhere nonzero section on any integral closed subspace $W \subset X$. Hence $c_1(\mathcal{O}_X) \cap [W] = 0$ as desired.
Let $\varphi : V \to Y$ be a surjective \'etale morphism where $V$ is a scheme. It suffices to prove the equality after pulling back by $\varphi$, see Lemma \ref{lemma-etale-pullback}. That same lemma tells us that $\varphi^*[Y_i] = [\varphi^{-1}(Y_i)] = \sum [V_{i, j}]$ where $V_{i, j}$ are the irreducible components of $V$ lying over $Y_i$. Hence if we first apply the case of schemes (Chow Homology, Lemma \ref{chow-lemma-prepare-geometric-cap}) to $\varphi^*s_i$ on $Y_i \times_Y V$ we find that $\varphi^*[Z(s_i)]_k = [Z(\varphi^*s_i)] = \sum [Z(s_{i, j})]_k$ where $s_{i, j}$ is the pullback of $s$ to $V_{i, j}$. Applying the case of schemes to $\varphi^*s$ we get $$ \varphi^*[Z(s)]_k = [Z(\varphi^*s)]_k = \sum n_i[Z(s_{i, j})]_k $$ by our remark on multiplicities above. Combining all of the above the proof is complete.
Using sleight of hand we will deduce this from Lemma \ref{lemma-prepare-flat-pullback-rational-equivalence}. (An alternative is to redo the proof of that lemma in the setting of meromorphic sections of invertible modules.) Namely, let $q : T \to Y$ be the morphism of Lemma \ref{lemma-Gm-torsor} constructed using $\mathcal{L}$ on $Y$. We will use all the properties of $T$ stated in this lemma. Consider the fibre product diagram $$ \xymatrix{ T' \ar[r]_{q'} \ar[d]_h & X \ar[d]^f \\ T \ar[r]^q & Y } $$ Then $q' : T' \to X$ is the morphism constructed using $f^*\mathcal{L}$ on $X$. Then it suffices to prove $$ (q')^*f^*\text{div}_\mathcal{L}(s) = \sum n_i (q')^*\text{div}_{f^*\mathcal{L}|_{X_i}}(s_i) $$ Observe that $T'_i = q^{-1}(X_i)$ are the irreducible components of $T'$ and that $n_i$ is the multiplicity of $T'_i$ in $T'$. The left hand side is equal to $$ h^*q^*\text{div}_\mathcal{L}(s) = h^*\text{div}_T(q^*(s)) $$ by Lemma \ref{lemma-Gm-torsor-divisor-meromorphic-section} (and Lemma \ref{lemma-compose-flat-pullback}). On the other hand, denoting $q'_i : T'_i \to X_i$ the restriction of $q'$ we find that Lemma \ref{lemma-Gm-torsor-divisor-meromorphic-section} also tells us the right hand side is equal to $$ \sum n_i \text{div}_{T_i}((q'_i)^*(s_i)) $$ In these two formulas the expressions $q^*(s)$ and $(q'_i)^*(s_i)$ represent the rational functions corresponding to the pulled back meromorphic sections of $q^*\mathcal{L}$ and $(q'_i)^*f^*\mathcal{L}|_{X_i}$ via the isomorphism $\alpha : q^*\mathcal{L} \to \mathcal{O}_T$ and its pullbacks to spaces over $T$. With this convention it is clear that $(q'_i)^*(s_i)$ is the composition of the rational function $q^*(s)$ on $T$ and the morphism $h|_{T'_i} : T'_i \to T$. Thus Lemma \ref{lemma-prepare-flat-pullback-rational-equivalence} exactly says that $$ h^*\text{div}_T(q^*(s)) = \sum n_i \text{div}_{T_i}((q'_i)^*(s_i)) $$ as desired.
Write $\alpha = \sum n_i[W_i]$. We will show that $$ f^*(c_1(\mathcal{L}) \cap [W_i]) = c_1(f^*\mathcal{L}) \cap f^*[W_i] $$ in $\CH_{k + r - 1}(X)$ by producing a rational equivalence on the closed subspace $f^{-1}(W_i)$ of $X$. By the discussion in Remark \ref{remark-infinite-sums-rational-equivalences} this will prove the equality of the lemma is true. \medskip\noindent Let $W \subset Y$ be an integral closed subspace of $\delta$-dimension $k$. Consider the closed subspace $W' = f^{-1}(W) = W \times_Y X$ so that we have the fibre product diagram $$ \xymatrix{ W' \ar[r] \ar[d]_h & X \ar[d]^f \\ W \ar[r] & Y } $$ We have to show that $f^*(c_1(\mathcal{L}) \cap [W]) = c_1(f^*\mathcal{L}) \cap f^*[W]$. Choose a nonzero meromorphic section $s$ of $\mathcal{L}|_W$. Let $W'_i \subset W'$ be the irreducible components of $\delta$-dimension $k + r$. Write $[W']_{k + r} = \sum n_i[W'_i]$ with $n_i$ the multiplicity of $W'_i$ in $W'$ as per definition. So $f^*[W] = \sum n_i[W'_i]$ in $Z_{k + r}(X)$. Since each $W'_i \to W$ is dominant we see that $s_i = s|_{W'_i}$ is a nonzero meromorphic section for each $i$. By Lemma \ref{lemma-prepare-flat-pullback-cap-c1} we have the following equality of cycles $$ h^*\text{div}_{\mathcal{L}|_W}(s) = \sum n_i\text{div}_{f^*\mathcal{L}|_{W'_i}}(s_i) $$ in $Z_{k + r - 1}(W')$. This finishes the proof since the left hand side is a cycle on $W'$ which pushes to $f^*(c_1(\mathcal{L}) \cap [W])$ in $\CH_{k + r - 1}(X)$ and the right hand side is a cycle on $W'$ which pushes to $c_1(f^*\mathcal{L}) \cap f^*[W]$ in $\CH_{k + r - 1}(X)$.