What is the fiber of an etale map over a point? Given a field $K$, $K$-schemes $X$ and $Y$, an étale map $f:X\to Y$, $x\in X$ a point of the topological space to $X$ and $y:=f(x)\in Y$ a point of the topological space to $Y$. What is the basechange $X\times_Y Spec(k(y))$ of $f$ along $g:Spec(k(y))\to Y$?
I think it is a finite disjoint union $\coprod_n Spec(K_n)$ for fields $K_n$, all (algebraic?) extensions of $K$. The union is indexed by the number of preimages of $y$ for the topological space map $f$. Is this true?
If $g$ is a geometric point ($k(y)=\bar K$) of $Y$, is it true that I can find $g':Spec(k(y))\to X$ with $g=f\circ g'$?
 A: A scheme $S$ over a field $k$ is etale if and only if it is unramified if and only if it is a disjoint union of spectra of finite separable extensions of $k$. 
So, if $X\rightarrow Y$ is an etale morphism, then because this is a property preserved by base change, for any $y\in Y$, $X_y\rightarrow\mathrm{Spec}(k(y))$ is etale, so $X_y$ is a disjoint union of spectra of finite separable extensions of $k(y)$ (the residue fields of the various points in the fiber).
Your question about geometric points amounts to the following: given an algebraically closed field $K$, a point of $y\in Y$, and an injection $k(y)\hookrightarrow K$, does there exist a point $x\in X$ with $f(x)=y$ and an injection $k(x)\hookrightarrow K$ such that $k(y)\rightarrow k(x)\hookrightarrow K$ is the given injection? Assuming that $y$ is actually of the form $f(x)$, as you have, $f$ gives you an injection $k(y)\rightarrow k(x)$, and you're asking whether there exists a $k(y)$-injection $k(x)\hookrightarrow K$. Since $k(x)/k(y)$ is finite separable and $K$ is algebraically closed, the answer is yes. 
