Does taking $\mathbb{C}$-points of a scheme preserve pullbacks or pushouts?

Question

Let $K$ be the field $\mathbb{C}$ of complex numbers and let $X$ be a scheme of finite type over $S=\operatorname{Spec(K)}$.

The set $X(K)=\hom_{Sch/S}(S, X)$ of $K$-rational points of $X$ carries a natural topology coming from the topology of $K$. E.g. if $X=\operatorname{Spec}(A)\subseteq \mathbb{A}_K^d$ is affine, then the set $\mathbb{A}_K^d(K)$ is given the topology of $K^d$ and the set $X(K)$ is given the the subspace topology.

Suppose $X$, $Y$, $Z$, $W$ are schemes of finite type over $S$ and $D=$ $$ \begin{array}{rcl} X &\to &Y\\ \downarrow && \downarrow\\ Z &\to& W \end{array} $$ is a pushout (resp. pullback) diagram in $Sch/S$. Is then $D(K)$ a pushout (resp. pullback) diagram of topological spaces?

If this is too hard to answer in general, I am also interested in the statement for affine schemes $Aff/S$ over $S$ instead of all schemes $Sch/S$ over $S$.

Answer 1

Yes, this is true for fiber products. See Proposition 3.1 of Brian Conrad's article "Weil and Grothendieck approaches to adelic points."

Answer 2

This is not true for pushouts. Let $i:U\rightarrow \mathbb{A}^1_{\mathbb{C}}$ be an open immersion with $U\neq \mathbb{A}^1_{\mathbb{C}}$ and $U\neq \emptyset$ and $f:U\rightarrow \mathrm{Spec}(\mathbb{C})$ be a unique map to a point. Then scheme-theoretical pushout $\mathbb{A}^1_{\mathbb{C}}\cup_{f}\mathrm{Spec}(\mathbb{C})$ is just $\mathrm{Spec}(\mathbb{C})$ which is a point. So even set-theoretical pushout of sets of $\mathrm{Spec}(\mathbb{C})$-rational points and pushout in the category of schemes do not agree.