# Does tensor functor preserve exact sequences in modules?

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.

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 $$0 \rightarrow M‘ \otimes F \rightarrow M \otimes F \rightarrow M‘‘ \otimes F$$