I have the following problem.
Let $M$ be a faithfully flat module. I need to show that if sequence
$$0\longrightarrow N'\otimes M \stackrel{\varphi\otimes{id_{M}}}\longrightarrow N\otimes M\stackrel{\psi\otimes id_{M}}\longrightarrow N''\otimes M\longrightarrow 0$$
is exact, then sequence $$0\longrightarrow N'\stackrel{\varphi}\longrightarrow N\stackrel{\psi}\longrightarrow N''\longrightarrow 0$$ is exact. The only thing I have left to show is that the $\psi$ is surjective but I don't know how.
As mentioned by Asvin, this is often taken as definition of faithful flatness. To develop on this, $M$ is a faithfully flat module if it is flat and the functor $- \otimes M$ is faithful, namely, $f \otimes id_M = 0$ implies $f = 0$ for all $R$-module homomorphisms $f$. In particular we can show that $\psi \otimes id_M$ surjective implies $\psi$ surjective using only faithfulness (no flatness required).
It is convenient to use the category-theoretic generalisation of surjectiveness called epimorphism. Assume $f, g : N'' \rightarrow X$ are morphisms such that $f \circ \psi = g \circ \psi$. Then $$(f \otimes id_M)\circ(\psi\otimes id_M) = (g \otimes id_M)\circ(\psi\otimes id_M)$$ and since $\psi$ is surjective by assumption, we have $f \otimes id_M = g \otimes id_M$. Using the group structure of module homomorphisms, we obtain $(f-g)\otimes id_M$ and hence $f-g = 0$ by faithfulness of the $- \otimes M$ functor. Hence $f=g$ as required, so $\psi$ is surjective.