For any morphism of sheaves \(\varphi: \mcf \to \mcg\), show that for each point \(P\) one has \((\ker \varphi)_P = \ker(\varphi_P)\) and \((\im \varphi)_P = \im(\varphi_P)\).
Show that \(\varphi\) is injective (respectively surjective) if and only if the induced map on stalks \(\varphi_P\) is injective (respectively surjective) for all \(P\).
Show that a sequence \[ \cdots \to \mcf^{i-1} \mapsvia{\phi^{i-1}} \mcf^{i} \mapsvia{\phi^{i}} \mcf^{i+1} \to \cdots \] of sheaves and morphisms is exact if and only if for each \(P \in X\) the corresponding sequence of stalks is exact as a sequence of abelian groups.
Solution
Part a. The clean argument is formal: taking the stalk at \(P\) is a filtered colimit over the neighbourhoods of \(P\), a kernel is a finite limit, and filtered colimits commute with finite limits. This gives \((\ker \phi)_P = \ker(\phi_P)\) at once. For the image, write \(\im \phi \da \ker(\mcg \to \coker \phi)\), so the image is again a kernel, and the same commutation applies.
A direct diagram chase gives the same conclusion.
\((\ker \phi)_P \subseteq \ker(\phi_P)\): let \(s \in (\ker \phi)_P\) and represent it by a pair \((U, \tilde s)\) with \(\tilde s \in (\ker \phi)(U)\), so that \(\ro{\tilde s}{P} = s\) and \(\phi(U)(\tilde s) = 0\) in \(\mcg(U)\). Passing to germs at \(P\) and using that the square relating sections over \(U\) to stalks at \(P\) commutes, we get \(\phi_P(s) = \ro{0}{P} = 0\).
\(\ker(\phi_P) \subseteq (\ker \phi)_P\): let \(s_P \in \ker(\phi_P)\) and lift it to a pair \((U, s)\) with \(s \in \mcf(U)\) and \(\ro{s}{P} = s_P\). Put \(t \da \phi(U)(s) \in \mcg(U)\). Commutativity gives \(\ro{t}{P} = \phi_P(s_P) = 0\) in \(\mcg_P\). Since the germ of \(t\) at \(P\) vanishes, there is an open \(V\) with \(P \in V \subseteq U\) and \(\ro{t}{V} = 0\) in \(\mcg(V)\). Because \(\phi\) is a morphism of sheaves the restriction squares commute, so \(\phi(V)(\ro{s}{V}) = \ro{t}{V} = 0\), which says \(\ro{s}{V} \in (\ker \phi)(V)\). Taking germs at \(P\) exhibits \(s_P = \qty{\ro{s}{V}}_P\) as an element of \((\ker \phi)_P\).
The diagram chase for images is the same argument run in the other direction.
Part b. Injectivity. (\(\Rightarrow\)) If \(\phi\) is injective then \(\ker \phi = 0\) as a sheaf, so by part (a) we get \(\ker(\phi_P) = (\ker \phi)_P = 0\) for every \(P\), and \(\phi_P\) is injective.
(\(\Leftarrow\)) Suppose \(\ker(\phi_P) = 0\) for all \(P\), and set \(K \da \ker \phi\), a sheaf. By part (a), \(K_P = 0\) for all \(P\). Given a section \(s \in K(U)\), each \(P \in U\) has a neighbourhood \(U_P \subseteq U\) with \(\ro{s}{U_P} = 0\), since the germ \(s_P\) vanishes. The sets \(\ts{U_P \st P \in U}\) cover \(U\), and \(K\) is a sheaf, so the identity axiom forces \(s = 0\). Hence \(K = 0\) and \(\phi\) is injective.
Surjectivity. The same argument applied to \(\coker \phi\), whose stalks are the cokernels of the \(\phi_P\) by part (a), shows \(\coker \phi = 0\) if and only if every \(\phi_P\) is surjective.
Part c. Exactness of the sequence of sheaves means \(\ker \phi^i = \im \phi^{i-1}\) as subsheaves of \(\mcf^i\).
(\(\Rightarrow\)) If \(\ker \phi^i = \im \phi^{i-1}\) then these sheaves have the same stalks, so by part (a) \[ \ker(\phi^i_P) = (\ker \phi^i)_P = (\im \phi^{i-1})_P = \im(\phi^{i-1}_P), \] which is exactness at \(P\).
(\(\Leftarrow\)) Conversely, if the stalk sequences are exact then the same chain of identities read in the other order gives \((\ker \phi^i)_P = (\im \phi^{i-1})_P\) for every \(P\). Two subsheaves of \(\mcf^i\) with the same stalks coincide, so \(\ker \phi^i = \im \phi^{i-1}\).
Compactly, using part (a) throughout, \[ \ker \phi^i = \im \phi^{i-1} \iff (\ker \phi^i)_P = (\im \phi^{i-1})_P \ \forall P \iff \ker(\phi^i_P) = \im(\phi^{i-1}_P) \ \forall P. \]