(i) amounts to the fact that $\varphi_{D}\circ p_{\mathcal{X}}=\varphi_{D}$ for all $D\in\Div_{0}(\mathcal{Y})$, which is a special case of Lemma 3.4.

Let us now prove (ii). The map $\emb^{\prime}_{\mathcal{Y}}:=p_{\mathcal{X}}\circ\emb_{\mathcal{Y}}:\Delta_{\mathcal{Y}}\to X$ is continuous, and the previous identity implies that $\ev_{\mathcal{Y}}\circ\emb^{\prime}_{\mathcal{Y}}=\ev_{\mathcal{Y}}\circ\emb_{\mathcal{Y}}=\id$. By the uniqueness part of Theorem 3.1 it suffices to prove that $c_{\mathcal{Y}}\circ\emb^{\prime}_{\mathcal{Y}}=c_{\mathcal{Y}}\circ\emb_{\mathcal{Y}}$ on $\Delta_{\mathcal{Y}}$. Pick $s\in\Delta_{\mathcal{Y}}$ and set $x:=\emb_{\mathcal{Y}}(s)$, $x^{\prime}:=\emb^{\prime}_{\mathcal{Y}}(s)$. On the one hand (i) shows that

so $c_{\mathcal{Y}}(x^{\prime})\in\overline{\{c_{\mathcal{Y}}(x)\}}$ by (i) of Lemma 3.4. On the other hand $p_{\mathcal{X}}(x)=x^{\prime}$ by definition, so $c_{\mathcal{X}}(x)\in\overline{\{c_{\mathcal{X}}(x^{\prime})\}}$ and hence $c_{\mathcal{Y}}(x)\in\overline{\{c_{\mathcal{Y}}(x^{\prime})\}}$ by continuity of the map $\mathcal{X}\to\mathcal{Y}$ for the Zariski topology. ∎