We may assume that $B$ is the localization of a finitely presented $A$-algebra $B_0$ and that $N$ is the localization of a finitely presented $B_0$-module $M_0$, see Lemma \ref{lemma-reduce-finite-type-injective-into-flat-mod-m}. By Lemma \ref{lemma-generic-flatness-stratification} there exists a ``generic flatness stratification'' for $\widetilde{M_0}$ on $\Spec(B_0)$ over $\Spec(A)$. Translating back to $N$ we find a sequence of closed subschemes $$ S = \Spec(A) \supset S_0 \supset S_1 \supset \ldots \supset S_t = \emptyset $$ with $S_i \subset S$ cut out by a finitely generated ideal of $A$ such that the pullback of $\widetilde{N}$ to $\Spec(B) \times_S (S_i \setminus S_{i + 1})$ is flat over $S_i \setminus S_{i + 1}$. We will prove the proposition by induction on $t$ (the base case $t = 1$ will be proved in parallel with the other steps). Let $\Spec(A/J_i)$ be the scheme theoretic closure of $S_i \setminus S_{i + 1}$. \medskip\noindent {\bf Claim 1.} $N/J_iN$ is flat over $A/J_i$. This is immediate for $i = t - 1$ and follows from the induction hypothesis for $i > 0$. Thus we may assume $t > 1$, $S_{t - 1} \not = \emptyset$, and $J_0 = 0$ and we have to prove that $N$ is flat. Let $J \subset A$ be the ideal defining $S_1$. By induction on $t$ again, we also have flatness modulo powers of $J$. Let $A^h$ be the henselization of $A$ and let $B'$ be the localization of $B \otimes_A A^h$ at the maximal ideal $\mathfrak m_B \otimes A^h + B \otimes \mathfrak m_{A^h}$. Then $B \to B'$ is faithfully flat. Set $N' = N \otimes_B B'$. Note that $N'$ is $A^h$-flat if and only if $N$ is $A$-flat. By Theorem \ref{theorem-flattening-local} there is a smallest ideal $I \subset A^h$ such that $N'/IN'$ is flat over $A^h/I$, and $I$ is finitely generated. By the above $I \subset J^nA^h$ for all $n \geq 1$. Let $S_i^h \subset \Spec(A^h)$ be the inverse image of $S_i \subset \Spec(A)$. By Lemma \ref{lemma-closed-points-complement} we see that $V(I)$ contains the closed points of $U = \Spec(A^h) - S_1^h$. By construction $N'$ is $A^h$-flat over $U$. By Lemma \ref{lemma-make-smaller-flatness-ideal} we see that $N'/I_2N'$ is flat over $A/I_2$, where $I_2 = \Ker(I \to \Gamma(U, I/I^2))$. Hence $I = I_2$ by minimality of $I$. This implies that $I = I^2$ locally on $U$, i.e., we have $I\mathcal{O}_{U, u} = (0)$ or $I\mathcal{O}_{U, u} = (1)$ for all $u \in U$. Since $V(I)$ contains the closed points of $U$ we see that $I = 0$ on $U$. Since $U \subset \Spec(A^h)$ is scheme theoretically dense (because replaced $A$ by $A/J_0$ in the beginning of this paragraph), we see that $I = 0$. Thus $N'$ is $A^h$-flat and hence Claim 1 holds. \medskip\noindent We return to the situation as laid out before Claim 1. With $A^h$ the henselization of $A$, with $B'$ the localization of $B \otimes_A A^h$ at the maximal ideal $\mathfrak m_B \otimes A^h + B \otimes \mathfrak m_{A^h}$, and with $N' = N \otimes_B B'$ we now see that the flattening ideal $I \subset A^h$ of Theorem \ref{theorem-flattening-local} is nilpotent. If $nil(A^h)$ denotes the ideal of nilpotent elements, then $nil(A^h) = nil(A) A^h$ (More on Algebra, Lemma \ref{more-algebra-lemma-henselization-nil}). Hence there exists a finitely generated nilpotent ideal $I_0 \subset A$ such that $N/I_0N$ is flat over $A/I_0$. \medskip\noindent {\bf Claim 2.} For every prime ideal $\mathfrak p \subset A$ the map $\kappa(\mathfrak p) \otimes_A N \to \kappa(\mathfrak p) \otimes_A M$ is injective. We say $\mathfrak p$ is bad it this is false. Suppose that $C$ is a nonempty chain of bad primes and set $\mathfrak p^* = \bigcup_{\mathfrak p \in C} \mathfrak p$. By Lemma \ref{lemma-find-pure-spreadout} there is a finitely generated ideal $\mathfrak a \subset \mathfrak p^*A_{\mathfrak p^*}$ such that there is a pure spreadout over $V(\mathfrak a)$. If $\mathfrak p^*$ were good, then it would follow from Lemma \ref{lemma-properties-pure-spreadout} that the points of $V(\mathfrak a)$ are good. However, since $\mathfrak a$ is finitely generated and since $\mathfrak p^*A_{\mathfrak p^*} = \bigcup_{\mathfrak p \in C}A_{\mathfrak p^*}$ we see that $V(\mathfrak a)$ contains a $\mathfrak p \in C$, contradiction. Hence $\mathfrak p^*$ is bad. By Zorn's lemma, if there exists a bad prime, there exists a maximal one, say $\mathfrak p$. In other words, we may assume every $\mathfrak p' \supset \mathfrak p$, $\mathfrak p' \not = \mathfrak p$ is good. In this case we see that for every $f \in A$, $f \not \in \mathfrak p$ the map $u \otimes \text{id}_{A/(\mathfrak p + f)}$ is universally injective, see Lemma \ref{lemma-universally-injective-if-flat}. Thus it suffices to show that $N/\mathfrak p N$ is separated for the topology defined by the submodules $f(N/\mathfrak pN)$. Since $B \to B'$ is faithfully flat, it is enough to prove the same for the module $N'/\mathfrak p N'$. By Lemma \ref{lemma-flat-finite-type-local-colimit-free} and More on Algebra, Lemma \ref{more-algebra-lemma-content-exists-flat-Mittag-Leffler} elements of $N'/\mathfrak pN'$ have content ideals in $A^h/\mathfrak pA^h$. Thus it suffices to show that $\bigcap_{f \in A, f \not \in \mathfrak p} f(A^h/\mathfrak p A^h) = 0$. Then it suffices to show the same for $A^h/\mathfrak q A^h$ for every prime $\mathfrak q \subset A^h$ minimal over $\mathfrak p A^h$. Because $A \to A^h$ is the henselization, every $\mathfrak q$ contracts to $\mathfrak p$ and every $\mathfrak q' \supset \mathfrak q$, $\mathfrak q' \not = \mathfrak q$ contracts to a prime $\mathfrak p'$ which strictly contains $\mathfrak p$. Thus we get the vanishing of the intersections from Lemma \ref{lemma-big-intersection-is-zero}. \medskip\noindent At this point we can put everything together. Namely, using Claim 1 and Claim 2 we see that $N/I_0 N \to M/I_0M$ is $A/I_0$-universally injective by Lemma \ref{lemma-universally-injective-if-flat}. Then the diagrams $$ \xymatrix{ N \otimes_A (I_0^n/I_0^{n + 1}) \ar[r] \ar[d] & M \otimes_A (I_0^n/I_0^{n + 1}) \ar@{=}[d] \\ I_0^n N /I_0^{n + 1} N \ar[r] & I_0^n M /I_0^{n + 1} M } $$ show that the left vertical arrows are injective. Hence by Algebra, Lemma \ref{algebra-lemma-what-does-it-mean-again} we see that $N$ is flat. In a similar way the universal injectivity of $u$ can be reduced (even without proving flatness of $N$ first) to the one modulo $I_0$. This finishes the proof.