Lemma 4

Let $A$ be a noetherian ring, $F$ a closed subset of $\mathrm{Spec}(A)$, $U$ the complementary open set, and $u : M \to N$ a homomorphism of finitely generated $A$-modules.

a) Suppose that $u_p : M_p \to N_p$ is injective for every $p \in U$ and that one has $\mathrm{prof}_F(M) \geqslant 1$. Then $u$ is injective.

b) Suppose that $u_p : M_p \to N_p$ is bijective for every $p \in U$ and that one has $\mathrm{prof}_F(M) \geqslant 2$ and $\mathrm{prof}_F(N) \geqslant 1$. Then $u$ is bijective.

a) The hypotheses imply $\mathrm{Supp}(\mathrm{Ker}\,u) \subset F$, hence $\mathrm{Hom}_A(\mathrm{Ker}\,u, M) = 0$ (No. 5, Remark 1); therefore one has $\mathrm{Ker}\,u = 0$.

b) One already knows that $u$ is injective, and one has $\mathrm{Supp}(\mathrm{Coker}\,u) \subset F$. By loc. cit., one has $\mathrm{Hom}_A(\mathrm{Coker}\,u, N) = 0$ and $\mathrm{Ext}_A^1(\mathrm{Coker}\,u, M) = 0$. From the exact sequence of extension modules

one deduces $\mathrm{Hom}_A(\mathrm{Coker}\,u, \mathrm{Coker}\,u) = 0$, which implies $\mathrm{Coker}\,u = 0$.