Let \(\mcf'\) be a subsheaf of a sheaf \(\mcf\). Show that the natural map of \(\mcf\) to the quotient sheaf \(\mcf/\mcf'\) is surjective and has kernel \(\mcf'\). Thus there is an exact sequence \[ 0 \to \mcf' \to \mcf \to \mcf/\mcf' \to 0. \]
Conversely, if \[ 0 \to \mcf' \to \mcf \to \mcf'' \to 0 \] is an exact sequence, show that \(\mcf'\) is isomorphic to a subsheaf of \(\mcf\), and that \(\mcf''\) is isomorphic to the quotient of \(\mcf\) by this subsheaf.
Solution
Let \[ q:\mcf\longrightarrow\mcf/\mcf' \] denote the natural morphism to the quotient sheaf, defined as the sheafification of the presheaf quotient \[ U\longmapsto\mcf(U)/\mcf'(U). \]
1. For every \(x\in X\), the stalk of the quotient sheaf is canonically\[ (\mcf/\mcf')_x\cong\mcf_x/\mcf'_x, \]and the induced stalk map \(q_x\) is the ordinary quotient homomorphism.
Taking a stalk is a filtered colimit over neighborhoods of \(x\). Filtered colimits of abelian groups commute with cokernels, so \[ \left(\varinjlim_{x\in U}\mcf(U)\right) \Big/ \left(\varinjlim_{x\in U}\mcf'(U)\right) \cong \varinjlim_{x\in U}\bigl(\mcf(U)/\mcf'(U)\bigr). \] Sheafification does not change stalks. Hence the stalk of the quotient sheaf is exactly \(\mcf_x/\mcf'_x\), and \(q_x\) is induced by the quotient maps on sections.
2. The natural morphism \(q:\mcf\to\mcf/\mcf'\) is surjective and has kernel \(\mcf'\).
By step 1, for every \(x\in X\) the map \[ q_x:\mcf_x\longrightarrow\mcf_x/\mcf'_x \] is surjective and has kernel \(\mcf'_x\). A morphism of sheaves is surjective exactly when all its stalk maps are surjective, so \(q\) is surjective.
The kernel sheaf of \(q\) has stalk \[ (\ker q)_x=\ker(q_x)=\mcf'_x. \] The inclusion \(\mcf'\hookrightarrow\mcf\) therefore identifies \(\mcf'\) with \(\ker q\), since a morphism of sheaves inducing an isomorphism on every stalk is an isomorphism. Thus \[ \boxed{0\longrightarrow\mcf'\longrightarrow\mcf\longrightarrow\mcf/\mcf'\longrightarrow0} \] is exact, proving part (a).
3. In an exact sequence\[ 0\longrightarrow\mcf'\xrightarrow{i}\mcf\xrightarrow{p}\mcf''\longrightarrow0, \]the morphism \(i\) identifies \(\mcf'\) with a subsheaf of \(\mcf\).
Exactness at \(\mcf'\) says that \(i\) is injective. Equivalently, every stalk map \[ i_x:\mcf'_x\longrightarrow\mcf_x \] is injective. The image sheaf \(\operatorname{im}i\) is, by definition, a subsheaf of \(\mcf\). The induced morphism \[ \mcf'\longrightarrow\operatorname{im}i \] is an isomorphism on every stalk, hence an isomorphism of sheaves. Thus we may identify \(\mcf'\) with the subsheaf \(\operatorname{im}i\subseteq\mcf\).
4. With the identification of step 3, there is a canonical isomorphism\[ \boxed{\mcf/\mcf'\cong\mcf''}. \]
Exactness at \(\mcf\) gives \[ \ker p=\operatorname{im}i=\mcf'. \] Therefore \(p\) kills the subsheaf \(\mcf'\) and induces a unique morphism \[ \bar p:\mcf/\mcf'\longrightarrow\mcf''. \] On the stalk at \(x\), this is the homomorphism \[ \bar p_x:\mcf_x/\mcf'_x\longrightarrow\mcf''_x \] induced by \(p_x\). The stalk sequence \[ 0\longrightarrow\mcf'_x\longrightarrow\mcf_x\longrightarrow\mcf''_x\longrightarrow0 \] is exact, so the ordinary first isomorphism theorem gives \[ \mcf_x/\mcf'_x\cong\mcf''_x. \] Thus \(\bar p\) is an isomorphism on every stalk and hence an isomorphism of sheaves. This proves part (b).
5. Q.E.D.
Steps <1>1–<1>2 prove part (a), and steps <1>3–<1>4 prove part (b).