Does tensor functor preserve exact sequences in modules?

Question

We know that the tensor functor in modules preserves right exactness, that is, if $$M' \rightarrow M \overset{f}{\rightarrow} M''\rightarrow 0$$ exact then $$F\otimes M' \rightarrow F\otimes M \rightarrow F\otimes M''\rightarrow 0$$ is exact. Moreover tensoring by flat modules preserves also left exactness. In all proofs for this result (that I've ever seen!), the surjectivity of $f$ is used. What we can say about the exact sequences in which the right map is not surjective? For example if $$0\rightarrow M' \rightarrow M \overset{f}{\rightarrow} M''$$ exact sequence of $R-$modules ($M, M'$ and $M''$ are not zero and $f$ is not epimorphism) and $F$ is a flat $R-$module, then is the induced sequence $$0\rightarrow F\otimes M' \rightarrow F\otimes M \rightarrow F\otimes M''$$ exact? If it is not true, what conditions on $R$ or $F$ are needed?

I tried to find a direct proof, but it seems difficult, because of different representations of any element in tensor products. I know that we can always change $M''$ to $f(M)$ to have $f$ as an epimorphism. But I was wondering if doing this for $M''$ makes difficulties in some problems.

I will be appreciated if anyone can help.

Answer 1

Assume that $f$ is not surjective. We thus have a nontrivial image factorization $M \rightarrow \operatorname{img}(f) \rightarrow M‘‘$.

As you said tensoring with $F$ results in a short exact sequence $$0\rightarrow M‘ \otimes F \rightarrow M \otimes F \rightarrow \operatorname{img}(f)\otimes F \rightarrow 0$$ Meanwhile you also have the short exact sequence $$0\rightarrow \operatorname{img}(f) \rightarrow M‘‘ \rightarrow M‘‘/\operatorname{img}(f) \rightarrow 0$$ so tensoring with $F$ yields the short exact sequence $$0 \rightarrow \operatorname{img}\otimes F \rightarrow M‘‘\otimes F \rightarrow M‘‘/\operatorname{img}(f) \otimes F \rightarrow 0$$ But this means that $M \otimes F \rightarrow \operatorname{img}(f) \otimes F \rightarrow M‘‘ \otimes F$ is a factorization of $M \otimes F \rightarrow M‘‘ \otimes F$ into a surjection followed by an injection, hence $$\operatorname{ker}(M \otimes F \rightarrow M‘‘ \otimes F) = \operatorname{ker}(M \otimes F \rightarrow \operatorname{img}(f) \otimes F)$$ So exactness of the first sequence implies exactness of $$0 \rightarrow M‘ \otimes F \rightarrow M \otimes F \rightarrow M‘‘ \otimes F$$