I am reading the following proof in the Red Book of Varieties and Schemes by Mumford saying that a surjective morphism of schemes is stable under base change and I came across some things that I didn't understand.
"The maps $r^*$ and $s^*$ define inclusions of fields"
The morphisms of schemes $r$ induces a map on stalks $\mathscr{O}_{S,r(x)}\to \mathscr{O}_{X,x}$, which is a local homomorphism, so we get a map $k(r(x))\to k(x)$. Why should this be an inclusion ? This might be obvious, but my intuition about this $k(x)$ is still very vague.
Edit: This is clear now, morphisms of fields are always injective.
"defining $\alpha^*$ and $\beta^*$ to be the compositions"
Here, he is defining what the map on stalks should be. Does this then immediately define the morphism of sheaves?
Any insight would be very much appreciated.
Your first question has already been handled by Daniel Hast and Severin Schraven in the comments: all morphisms of fields are injective because the kernel must be a proper ideal.
To answer your second question, the map $\mathcal{O}_X\to\alpha_*\mathcal{O}_{\operatorname{Spec} \Omega}$ is specified by a compatible collection of maps $\mathcal{O}_X(U)\to (\alpha_*\mathcal{O}_{\operatorname{Spec} \Omega})(U)$ for $U\subset X$ open. But $(\alpha_*\mathcal{O}_{\operatorname{Spec} \Omega})(U)$ is equal to $0$ when $x\notin U$ and $\Omega$ when $x\in U$, so we can define $\mathcal{O}_X(U)\to (\alpha_*\mathcal{O}_{\operatorname{Spec} \Omega})(U)$ as the zero map when $x\notin U$ and the map $\mathcal{O}_X(U)\to\mathcal{O}_{X,x}\to k(x) \to \Omega$ when $x\in U$. These maps are compatible because the map $\mathcal{O}_X(U)\to\mathcal{O}_{X,x}$ factors as $\mathcal{O}_X(U)\to\mathcal{O}_X(V)\to\mathcal{O}_{X,x}$ for $x\in V\subset U$. So this really is enough to define the morphism you want.