Good ol' Diagram Chasing fun - is my kernel finitely generated? I'm in a bit of a pickle with a homework problem.
Let $R$ be a commutative ring, and $M$ a finitely presented module (i.e. there exist $F_1$, $F_2$ free so that there is an exact sequence $A \to B \to M \to 0$). Prove that any epimorphism $f$ from a free module $F$ onto $M$ has a finitely generated kernel. 
I managed to prove this using the following commutative diagram
$\beta$ and $\alpha$ were constructed by me">
the snake lemma, and a third short exact sequence which wedged $\ker f$ between two finitely generated modules (Im $\alpha$ and $\frac{F}{\textrm{Im} \beta}$ respectively). $\beta$ is defined using the fact that each generator of $F_2$ can be mapped to at least one element of $F$ which has the same image in $M$.
I am however stuck at the following loosening of the problem conditions:
"Does the statement still hold if $F$ is not free, but solely finitely generated?"
I managed to figure out that my construction of $\beta$ hinges on the fact that $F_2$ is projective (as it is free), but I cannot seem to find any other issue which would stop my proof from holding. 
 A: Suppose we have an epimorphism $F'\twoheadrightarrow M$ where $F'$ is finitely generated. Take an epimorphism $F\twoheadrightarrow F'$ where $F$ is finitely generated free (or just projective). We obtain an exact commutative diagram
$$\require{AMScd} \begin{CD}
0 @>>> K @>>> F @>>> M @>>> 0\\
@. @VVV @VVV @|\\
0 @>>> K' @>>> F' @>>> M @>>> 0
\end{CD}$$
By the first part, we know that $K$ is finitely generated. By the Snake Lemma we know that the map $K\to K'$ is onto. Thus $K'$ is finitely generated (by the image of a finite set of generators for $K$).
Also, the first part is basically Schanuel's Lemma: Given two projectives $P,P'$ mapping onto $M$, with kernels $K$ and $K'$ respectively, then we can form the pullback diagram
$$\begin{CD}
@.@.0 @. 0\\
@.@. @VVV @VVV\\
@.@. K @= K\\
@.@. @VVV @VVV\\
0 @>>> K' @>>> X @>>> P @>>> 0\\
@. @| @VVV @VVV\\
0 @>>> K' @>>> P' @>>> M @>>> 0\\
@.@. @VVV @VVV\\
@.@. 0 @. 0
\end{CD}$$
where we have used that parallel maps in a pullback diagram have isomorphism kernels. Since $P$ and $P'$ are projective, we see that
$$ P\oplus K' \cong X \cong P'\oplus K. $$
Thus if three of $P,P',K,K'$ are finitely generated, then so too is the fourth.
