The Stacks project

Proof. Let $Z \subset X$ be an integral closed subscheme dominating $R$. By the construction of $sp_{X/R}$ we have $sp_{X/R}([Z_ K]) = [Z_\kappa ]_ r$ and this characterizes the specialization map. Set $Z' = f(Z) \subset Y$. Then $Z'$ is an integral closed subscheme of $Y$ dominating $R$. Thus $sp_{Y/R}([Z'_ K]) = [Z'_\kappa ]_ r$.

We can think of $[Z]$ as an element of $Z_{r + 1}(X)$. By definition we have $f_*[Z] = 0$ if $\dim (Z') < r + 1$ and $f_*[Z] = d[Z']$ if $Z \to Z'$ is generically finite of degree $d$. Since proper pushforward commutes with flat pullback by $Y_ K \to Y$ (Chow Homology, Lemma 42.15.1) we see that correspondingly $f_{K, *}[Z_ K] = 0$ or $f_{K, *}[Z_ K] = d[Z'_ K]$. Let us apply Chow Homology, Lemma 42.29.8 to the commutative diagram

We obtain that $f_{\kappa , *}[Z_\kappa ]_ r = 0$ or $f_{\kappa , *}[Z_\kappa ] = d[Z'_\kappa ]_ r$ because clearly $i^*[Z] = [Z_ k]_ r$ and $j^*[Z'] = [Z'_\kappa ]_ r$. Putting everything together we conclude. $\square$