Basic Property of Finite Presentation of Modules I'm struggling to prove some rudimentary facts about finite presentations of a module $A$ over a ring $\Lambda$, which is defined in Hilton/Stammbach's A Course in Homological Algebra as a short exact sequence $0 \to N \to P \to A \to 0$ with $N$ finitely-generated and $P$ finitely-generated projective.
I'd like to show that if $A$ has a finite presentation, then for all short exact sequences $0 \to R \to S \to A \to 0$ with $S$ finitely-generated, $R$ is also finitely-generated. Here's my progress so far: If we denote the surjective maps in the two exact sequences above by $\epsilon : P \to A$ and $\epsilon' : S \to A$, then since $P$ is projective we have a map $\alpha : P \to S$ such that $\epsilon = \epsilon'\alpha$. We can identify $N$ with $\operatorname{Ker}{\epsilon}$ and $R$ with $\operatorname{Ker}{\epsilon'}$. We then have $\alpha(\operatorname{Ker}{\epsilon}) = \alpha(P) \cap \operatorname{Ker}{\epsilon'}$, and here I get stuck. We know that $\alpha(\operatorname{Ker}{\epsilon})$ and $\alpha(P)$ are finitely-generated, but how do we use this to show that $\operatorname{Ker}{\epsilon'}$ is finitely-generated? A hint would be much appreciated.
 A: The simplest way is to make a pullback diagram
$$\require{AMScd}
\begin{CD}
{} @. {} @. 0 @. 0 \\
@. @. @VVV @VVV \\
{} @. {} @. N @= N \\
@. @. @VVV @VVV \\
0 @>>> R @>>> Q @>>> P @>>> 0 \\
@. @| @VVV @VVV \\
0 @>>> R @>>> S @>>> A @>>> 0 \\
@. @. @VVV @VVV \\
{} @. {} @. 0 @. 0 \\
\end{CD}
$$
The rows and columns are exact; the arrows marked with equality are easily seen to be isomorphisms when $Q$ is the pullback of the maps $P\to A$ and $S\to A$.
Since $N$ and $S$ are finitely generated by assumption, $Q$ is finitely generated as well. Since $P$ is projective, the sequence $0\to R\to Q\to P\to 0$ splits and therefore $R$ is finitely generated.
Another way is to use Schanuel's lemma. In order to do this, take a finitely generated projective module $S'$ and an epimorphism $S'\to S$; let $K$ be the kernel of the composition $S'\to A$. Then, by Schanuel's lemma, $K\oplus P\cong N\oplus S'$, which implies $K$ is finitely generated because $N\oplus S'$ is.
The homomorphism theorem provides a surjective homomorphism $K\to R$.
