The following is taken from: $\textit{Partial Differential Control Theory Vol 1: Mathematical tools}$ by J F. Pommaret
$\color{Green}{Background:}$
$\textbf{Definition 1.50.}$ $M$ is call a $\textit{free module}$ if it has a $\textit{basis},$ that is a system of generators linearly independent over $A.$ When $M$ is free, the number of generators in a basis is called the $\textit{rank}$ of $M$ and does not depend on the basis if it is finite. Indeed, if $\{x_i\}$ and $\{y_j\}$ are two bases and $y_j=\sum {a^i}_jx_i,$ then $\text{det}({a^i}_j)$ must be invertible in $A.$ If $M$ is free of rank $n,$ then $M\cong A^n.$
$\textbf{Lemma 1.54.}$ In a short exact sequence of modules:
$$0\to M'\xrightarrow{f}M\xrightarrow{g}M''\to 0$$
the module $M$ is noetherian if and only if the momdules $M'$ and $M''$ are noetherian.
$\textbf{Proposition 1.55.}$ If $A$ is a noetherian ring and $M$ is a finitely generated $A-$module, then $M$ is a noetherian module.
$\textit{Proof.}$ Applying the preceding lemma (Lemma 1.54) to the short exact sequence $0\to A^{r-1}\to A^r\to A\to 0$ where the right epimorphism is the projection onto one factor, we deduce by induction that $A''$ is a noetherian module. Now, if $M$ is generated by $\{x_1,\ldots,x_n\}$ then there is an epimorphism $A^n\to M:(1,0,\ldots,0)\to x_1,\ldots,(0,\ldots,0,1)\to x_n.$ According to the preceding lemma, $M$ is a noetherian module too.
$\color{Red}{Questions:}$
Can someone help me with the beginning of the above proof, I am not clear on which right morphism the author is referring to in the sentence: "Applying the preceding lemma (Lemma 1.54) to the short exact sequence $0\to A^{r-1}\to A^r\to A\to 0$ where the right epimorphism is the projection onto one factor..."
Thank you in advance