I'm trying to show that for a flat $R$-module $M$, condition 1 implies condition 2, where the conditions are the following:
- $f: N_1\rightarrow N_2$ is an isomorphism of $R$-modules iff $f\otimes id: N_1\otimes M\rightarrow N_2\otimes M$ is.
- A sequence $P_1\stackrel{g}\to P_2\stackrel{h}\to P_3$ of $R$-modules is exact iff the sequence $P_1\otimes M \stackrel{g\otimes id}\to P_2\otimes M \stackrel{h\otimes id}\to P_3\otimes M$ has this property.
Assume the hypothesis and suppose we have an exact sequence $P_1\stackrel{g}\to P_2\stackrel{h}\to P_3$. This means that $\operatorname{im}g=\ker h$ impliying $\operatorname{im}g\otimes M=\ker h\otimes M$. We need to show that $\operatorname{im}(g\otimes id)=\ker(h\otimes id)$. Now $\ker h\otimes M=\ker(h\otimes id)$, but it is not necessarily true that $\operatorname{im}g\otimes M=\operatorname{im}(g\otimes id)$. How to show that this is in fact true?
Conversely, assume the hypothesis and suppose we have an exact sequence $P_1\otimes M \stackrel{g\otimes id}\to P_2\otimes M \stackrel{h\otimes id}\to P_3\otimes M$, which means that $\operatorname{im}(g\otimes id)=\ker(h\otimes id)$, and we want to show that $\operatorname{im}g=\ker h$. Here is the same problem with $\operatorname{im}$, but in fact I even don't understand how to proceed if this problem wouldn't take place.
Update:
In fact, now I think that even my claim $\ker h\otimes M=\ker(h\otimes id)$ isn't true, because it holds whenever $h$ is a surjective homomorphism, which is again not necessarily true...