Geometric points in fibre of finite étale morphism $\phi : Y \rightarrow X$ is independent of fibre I am reading the following notes http://www.math.toronto.edu/~jacobt/Lecture6.pdf and trying to understand the conclusion of Lemma 2.1.
We are trying to show that the number of geometric points above $\bar{x}$ is equal to $n$, where locally the étale morphism is induced on affines by $A \rightarrow B$ for $B$ a rank $n$ free $A$-module. The proof shows that $B \otimes_A \bar{k} = \bar{k}^n$, where the morphism $A \rightarrow \bar{k}$ is induced by $\bar{x}$. The proof then suggests that we have $n$ geometric points lying above $\bar{x}$. 
Why is this true?
On the level of rings, we want to find morphisms $f: B \rightarrow \bar{k}$ such that $A \rightarrow \bar{k}$ is the composition of $f \circ\phi^{\#}$. Tensoring this commutative triangle with $\bar{k}$ we get maps $A \otimes \bar{k} \rightarrow \bar{k}$, $A \otimes \bar{k} \rightarrow \bar{k}^n$ and $\bar{k}^n \rightarrow \bar{k}$ such that the analogous commutativity conditions hold. There are $n$ ''natural" maps $\bar{k}^n \rightarrow \bar{k}$, namely the projections to each factor. But why would any of these projections cause the triangle to commute, and even then given a such a map how do we get an induced map on rings pre-tensoring i.e. $B \rightarrow \bar{k}$?
Any help is appreciated.

This may be a little more complicated than needed but I haven't taken the time to simplify the argument. Please propose simplifications (or corrections if it is wrong) in the comments.
Consider the cartesian diagram
$$
\require{AMScd}
\begin{CD}
A @>{\phi}>> B\\
@VVV @VVV \\
\bar{k} @>>> B\otimes_{A} \bar{k}\cong \bar{k}^n
\end{CD}
$$
Then a geometric point above $\bar{x}$ which, as you describe, would be a morphism $B\rightarrow \bar{k}$. The key is that, by the universal property of the tensor product, the maps $B\rightarrow \bar{k}, \bar{k}\xrightarrow[]{id}\bar{k}$ should induce a map $g:B\otimes_{A} \bar{k}\cong \bar{k}^n \rightarrow \bar{k}$ such that the composition $\bar{k}\rightarrow \bar{k}^n\rightarrow\bar{k}$ is the identity. There is an obvious way to go in the opposite direction, so we are left with finding this kind of maps.
Well, now you need to understand more explicitly what these maps are. Since $B$ is a free $A$-module, more explicitly it must have a basis $\{ b_1,\ldots,b_n\}$ and obviously $b_i$ will map to $(0,0,\ldots,1,0,\ldots,0)\in \bar{k}^n.$ On the other hand, $\bar{k}$ is embedded diagonally in $\bar{k}^n$, ie $\lambda\rightarrow (\lambda,\ldots,\lambda)$. You can see this by considering how should the module structures work. 
Then, there you have it. There are exactly $n$ maps which will "invert" the diagonal embedding, and these are the projections.
