The existence of fibre product of schemes

Question

As we know, if $X\to S$ and $Y\to S$ are two morphisms of schemes, then the fibre product always exists. But considering, for a non-surjective morphism $f:X\to S$ and a closed point $s\in S\setminus f(X)$, inducing a closed immersion $\{s\}\to S$, what is the fibre product $X\times_S\{s\}$? We have two paths $X\times_S\{s\}\to\{s\}\to S$ and $X\times_S\{s\}\to X\to S$. The first path maps to $s$ but the second maps to a set outside $s$, which does not fit the definition of fibre product. Can anyone explain this?

Answer 1

$ \newcommand{\Z}{\Bbb Z} \newcommand{\F}{\mathbb{F}} \newcommand{\Spec}{\mathrm{Spec}} $

The fiber product can be empty, for instance for the non-surjective morphism $$f : X = \Spec(\F_2) \to S =\Spec(\Z)$$ and $s = (3)$, you get $\Spec(\F_2 \otimes_{\Z} \F_3) = \varnothing$.