In Atiyah Macodnald, ones defines a $A$ module $N$ to be flat $A$ module if for all $A$ modules and homomorphisms $M' \to ^f M \to ^g M''$ that are exact, if you tensor with $N$ (and take the maps $f \otimes 1$, $g \otimes 1$) it remains exact.
A moment before defining, we prove a nice theorem that if $M' \to ^f M \to ^g M'' \to 0$ is exact (of $A$ modules,homomorphisms), and $N$ is any $A$ module, then the sequence tensored with $N$ (And the same maps mentioned above) remains exact.
Finally to my question: We then claim that the following are equivalent for an $A$ module $N$
$N$ is flat
If $0 \to M' \to M \to M'' \to 0$ is any exact sequences of $A$ modules, the tensored sequence (with the same maps we mentioned a couple times $f \otimes 1$..) is exact.
If $f:M' \to M$ is injective, then $f \otimes 1:M' \otimes N \to M \otimes N$ is injective.
1 $\to$ 2 $\to$ 3 is clear. $3 \to 1$ is also clear using the proposition. I have a proof for 3->1 that says that $\frac {M}{Im_f(M')} \otimes N$ is naturally isomorphic to $\frac {M \otimes N}{Im_{f \otimes 1} (M \otimes N)}$, so that the injectiveness passes from one to another and we get 1.
Atiyh takes 2 $\to $ 1 for granted- probably something easier I'm missing? I'm guessing he means given $M' \to ^f M \to ^g M''$, think of it instead as $0 \to \frac {M'}{Ker} \to ^{\bar{f}} M \to ^g Im_{g}(M) \to 0$ and then he knows this tensored with $N$ is exact, and prove in a similiar I showed $3 \to 1$ it means the original tensored with $N$ is exact, but it seems even more work.
Am I missing something silly?