"Let $F$ be a functor from $A-$modules to $A-$modules that preserves cokernels. Then $F$ take exact short sequences into right exact short sequences."
I found a mistake in my proof. I also can't find it anywhere because everybody seems to find this too obvious.
Here's my proof: Let $0 \rightarrow L \rightarrow^{\alpha} M \rightarrow^{\beta} N \rightarrow 0$ be a short exact sequence and let $F$ be a functor that preserves cokernels.
We want to show that $F(L) \rightarrow^{F(\alpha)} F(M) \rightarrow^{F(\beta)} F(N) \rightarrow 0$ is exact
Since $\beta$ is surjective, then $coker\beta = 0$. Then we have,
$0 = F(0) = F(coker \beta) = cokerF(\beta)$. So $F(\beta)$ is surjective.
Now we have to show that $ImF(\alpha) = KerF(\beta)$.
$ImF(\alpha) \subset KerF(\beta)$ is easy because $\beta \circ \alpha = 0 \Rightarrow F(\beta) \circ F(\alpha) = 0$
Since the original sequence is exact, we have $Im\alpha = Ker\beta$. Then,
$\frac{F(M)}{KerF(\beta)} = F(N) = F(\frac{M}{Ker\beta}) = F(\frac{M}{Im\alpha}) = F(coker\alpha) = cokerF(\alpha) = \frac{F(M)}{ImF(\alpha)} $
My mistake: $ImF(\alpha) \subset KerF(\beta)$ and $\frac{F(M)}{KerF(\beta)} = \frac{F(M)}{ImF(\alpha)} \Rightarrow KerF(\beta) = ImF(\alpha)$.
I know this last claim is not true.
I realy appreciate any help whatsoever.
Your argument is almost right. If you knew that $\ker{F(\beta)}$ contained ${\rm im}\, F(\alpha)$ you would be able to conclude that (why?).
Consider the epimorphism $L\stackrel{\alpha'}\to \ker\beta\to 0$. Then we know $\iota\alpha'=\alpha$ where $\iota$ is the inclusion $\ker\beta\to M$. By hypothesis $\rm coker\,\alpha'=0$ so that the map $F(\alpha'):FL\to F(\ker\beta)$ is an epimorphism.
But the image of $F(\alpha')$ is the same of that of $F(\alpha)$ so ${\rm im}\,F(\alpha)= F(\ker\beta)=F({\rm im}\,\alpha)$.
Thus means that ${\rm im}\,F(\alpha)=F({\rm im}\,\alpha)$ is contained in $\ker F(\beta)$, and now you argument goes through: if $A\subset B$ and $M/B=M/A$ then necessarily $A=B$.