Draft lesson.

Godement resolutions and flabby sheaf cohomology

Written by GPT-6.1 Sol (OpenAI), October 2026. Public domain (CC0).

A sheaf is determined by its germs, but an arbitrary family of germs need not come from a section. Allowing every family produces a useful larger sheaf. Repeating that operation gives a canonical resolution, and the difference between a stalkwise splitting and a sheaf splitting explains why its global sections can have positive cohomology.

Throughout, \(X\) is a topological space, \(\mathcal O\) is a sheaf of unital rings, and modules are left modules. Rings need not be commutative. All sets and products lie in fixed universes, and we use choice. No separation, countability, compactness or paracompactness hypothesis is imposed on \(X\).

Prerequisites are Injective objects in locally module stacks, Sections 1–5, for the abelian sheaf category, exact stalks and enough injectives; Injective replacements and right derived functors, Sections 2–6, for ordinary injective resolutions and their long exact sequences; and Universal cohomology and acyclic resolutions, Sections 3–4, for dimension shifting and natural comparison with an acyclic resolution. The finite-space exercises use the sheaf–diagram equivalence and point adjoints proved in Finite-poset sheaves and canonical homological resolutions, Sections 1–3 and 6.

1. From stalks back to sections

Write \(\mathcal A=\operatorname{Mod}(\mathcal O)\) and \(\mathcal D=\prod_{x\in X}\operatorname{Mod}(\mathcal O_x)\). Define

S\mathcal F=(\mathcal F_x)_{x\in X},\qquad R(V)(U)=\prod_{x\in U}V_x. \tag{1.1}

An element of \(\mathcal O(U)\) acts in the \(x\)-coordinate through its germ in \(\mathcal O_x\); restriction forgets the coordinates outside the smaller open. These formulas define an \(\mathcal O\)-module sheaf: compatible tuples on a cover give a well-defined entry at each point, and those entries give the unique glued tuple. A family of module maps defines \(R\) on arrows coordinatewise.

There is a natural adjunction \(S\dashv R\). Given \(b:\mathcal F\to R(V)\), form \(b_x:\mathcal F_x\to (R(V))_x\), then evaluate a tuple germ at its indexing point \(x\). This gives a map \(\ell_x:\mathcal F_x\to V_x\). Evaluation is well-defined since representatives of the same germ agree on a neighborhood containing \(x\), and it is \(\mathcal O_x\)-linear. Conversely, from \((\ell_x)_x\) define

b_U(s)=(\ell_x(s_x))_{x\in U}. \tag{1.2}

This is linear and commutes with restriction. Applying the first operation to (1.2) recovers \(\ell_x\), since every germ has a section representative near \(x\). Starting with \(b\), the \(x\)-entry of (1.2) is the \(x\)-entry of \(b_U(s)\), so the second composite recovers \(b\). Composition with arrows on either side commutes with these formulas, proving naturality.

The unit and counit therefore are

\eta_{\mathcal F,U}(s)=(s_x)_{x\in U},\qquad \varepsilon_{V,x}:(R(V))_x\longrightarrow V_x, \quad [v]_x\longmapsto v_x. \tag{1.3}

The equality \(\varepsilon_{S\mathcal F}S\eta_{\mathcal F}=1_{S\mathcal F}\) says that the germ of the family of germs evaluates to the original germ. The equality \(R\varepsilon_V\eta_{RV}=1_{RV}\) holds on sections because each tuple entry is recovered by evaluation at that entry's point. Thus both triangle identities hold, as actual module maps.

It is generally false that \((RV)_x=V_x\). The counit is an evaluation map, not an identification. For example, take a family with a single nonzero entry \(V_x\). Its sheaf is the point direct image \(i_{x*}V_x\): sections on \(U\) are \(V_x\) if \(x\in U\), and zero otherwise. Its stalk at \(y\) is \(V_x\) precisely when every neighborhood of \(y\) contains \(x\), equivalently \(y\in\overline{\{x\}}\). In that case all transition maps in the neighborhood system are identities, and the action uses the germ map \(\mathcal O_y\to\mathcal O_x\). If some neighborhood excludes \(x\), neighborhoods contained in it are cofinal and the stalk is zero. This describes all stalks, even on a non-\(T_1\) space.

2. Exactness, flabbiness and injectivity

The functor \(S\) is exact: stalks are filtered colimits of section modules, and a locally exact sequence gives an exact sequence of germs. It is faithful, since a section whose every germ is zero is locally zero and hence zero. Exactness of a sheaf complex is equivalent to exactness at every stalk: a kernel germ whose image in the next stalk is zero is represented on a neighborhood by a kernel section, and belonging locally to the image is exactly the image sheaf condition. In particular, \(S\) reflects exactness and isomorphisms. The unit \(\eta\) is monic by the same local-zero argument.

The functor \(R\) is exact as well. For a short exact sequence of families, the section sequence on any \(U\) is a product of short exact module sequences. It is exact, including surjectivity by choosing coordinate lifts. A sectionwise exact sequence of sheaves is exact, so this proves the assertion. Consequently \(T=RS\) is an exact endofunctor of \(\mathcal A\).

A sheaf \(\mathcal G\) is called flabby (or flasque) if every restriction \(\mathcal G(V)\to\mathcal G(U)\), for opens \(U\subseteq V\), is onto. Every \(RV\) is flabby: extend a tuple by zero in the missing coordinates. This argument concerns the section product in (1.1); it makes no assertion that a stalk commutes with that product.

An exact left adjoint makes its right adjoint preserve injectives. Here is the extension argument in this case. For an injective family \(J\), a map \(A\to RJ\) along a monomorphism \(A\hookrightarrow B\) corresponds to \(SA\to J\). Exactness makes \(SA\hookrightarrow SB\) monic, so injectivity extends it to \(SB\to J\). The inverse adjunction gives the desired extension to \(B\). A family is injective exactly when each component is: extensions can be chosen componentwise, and one-component test families prove necessity.

Thus if every stalk ring \(\mathcal O_x\) is a field, every \(RV\) is injective: a linear map on a subspace extends by completing a basis. Over general rings, the same conclusion holds when every \(V_x\) is injective. Together with the monic unit and injective embeddings of each stalk module, this recovers the enough-injective construction from the first prerequisite. Arbitrary \(RV\) remains flabby even when these injectivity hypotheses fail.

3. Multiplication of the germ-family operation

The adjunction supplies a multiplication \(\mu_{\mathcal F}=R\varepsilon_{S\mathcal F}:T^2\mathcal F\to T\mathcal F\). A section of \(T^2\mathcal F\) on \(U\) is a family, indexed by \(x\in U\), of germs at \(x\) of sections of \(T\mathcal F\). Evaluate each germ at its own indexing point to get a family in \(T\mathcal F(U)\). This is the section formula for \(\mu\).

The two triangles in Section 1 give the two unit laws

\mu_{\mathcal F}\eta_{T\mathcal F}=1_{T\mathcal F},\qquad \mu_{\mathcal F}T\eta_{\mathcal F}=1_{T\mathcal F}. \tag{3.1}

For associativity apply counit naturality to the family map \(\varepsilon_{S\mathcal F}:SRS\mathcal F\to S\mathcal F\). It gives \(\varepsilon_{S\mathcal F}SR\varepsilon_{S\mathcal F}=\varepsilon_{S\mathcal F}\varepsilon_{SRS\mathcal F}\). Applying \(R\) gives

\mu_{\mathcal F}T\mu_{\mathcal F} =\mu_{\mathcal F}\mu_{T\mathcal F}:T^3\mathcal F\to T\mathcal F. \tag{3.2}

These computations prove the monad laws directly. In particular, the two ways of evaluating a family of iterated germ families agree; no exchange between infinite products and stalks is involved.

4. Every differential in the Godement complex

Put \(J^n(\mathcal F)=T^{n+1}\mathcal F\) for \(n\ge0\), and augment by \(d^{-1}=\eta_{\mathcal F}\). For \(0\le i\le n+1\), define

\delta_i^n=T^i\eta_{T^{n+1-i}\mathcal F}:J^n(\mathcal F)\to J^{n+1}(\mathcal F), \qquad d^n=\sum_{i=0}^{n+1}(-1)^i\delta_i^n. \tag{4.1}

For example, \(d^0=\eta_{T\mathcal F}-T\eta_{\mathcal F}\), and \(d^1=\eta_{T^2\mathcal F}-T\eta_{T\mathcal F}+T^2\eta_{\mathcal F}\). Every displayed map is a sheaf module map. If \(i<j\), naturality of \(\eta\), applied between the two insertion positions, gives

\delta_j^{n+1}\delta_i^n=\delta_i^{n+1}\delta_{j-1}^n. \tag{4.2}

More explicitly, insert the unit at the later position and move the earlier unit past it using \(\eta_B a=T(a)\eta_A\); applying the unchanged outer \(T^i\) gives exactly the two sides of (4.2). In \(d^{n+1}d^n\), the term indexed by \((j,i)\) with \(i<j\) has coefficient \((-1)^{j+i}\). Equation (4.2) pairs it with the term indexed by \((i,j-1)\), whose coefficient is its negative. These pairs exhaust the summands, so \(d^{n+1}d^n=0\). Unit naturality also gives \(d^0\eta_{\mathcal F}=0\).

For a sheaf map \(a:\mathcal F\to\mathcal G\), take \(J^n(a)=T^{n+1}a\). Unit naturality makes this commute with each coface, hence each differential, and with the augmentation. This defines the entire augmented complex functorially, including its maps on morphisms.

5. The contraction is a family of stalk maps

Apply \(S\) and define, for \(n\ge0\),

h^n=\varepsilon_{S T^n\mathcal F}: S J^n(\mathcal F)\longrightarrow S T^n\mathcal F. \tag{5.1}

In degree zero the target is \(S\mathcal F\). A component of \(h^n\) evaluates the outermost tuple germ at the point in question. There is no degree-dependent sign in (5.1). The first triangle says \(h^0 S d^{-1}=1\). The same triangle, on \(T^{n+1}\mathcal F\), gives \(h^{n+1}S\delta_0^n=1\). For \(i\ge1\), write \(\delta_i^n=T\delta_{i-1}^{n-1}\), interpreting \(\delta_0^{-1}=\eta_{\mathcal F}\) when \(n=0\). Counit naturality then gives

h^{n+1}S\delta_i^n=S\delta_{i-1}^{n-1}h^n. \tag{5.2}

Sum (5.2) with the signs of (4.1). The \(i=0\) term is the identity, and setting \(j=i-1\) makes every other sign the negative of the sign in \(Sd^{n-1}h^n\). Hence

h^{n+1}S d^n+S d^{n-1}h^n=1_{S J^n(\mathcal F)} \quad(n\ge0). \tag{5.3}

This contracts the augmented complex of families. Each stalk complex is therefore exact, and exactness is reflected by \(S\). We have proved the exact resolution

0\longrightarrow\mathcal F\xrightarrow{\eta}T\mathcal F \xrightarrow{d^0}T^2\mathcal F\xrightarrow{d^1}T^3\mathcal F\longrightarrow\cdots. \tag{5.4}

All its terms are flabby by Section 2. If every stalk ring is a field, they are injective as well. Counit naturality also makes \(h^n\) natural in \(\mathcal F\) as a family map. It does not make it a sheaf map: there need be no sheaf morphism whose stalks are these evaluations. In particular, (5.3) gives no contraction after applying global sections. Exercises 3 and 4 separate the two issues.

6. Why injective sheaves are flabby

We spell out the open-set extension argument. For an open inclusion \(j:U\hookrightarrow X\), a section of the extension by zero \(j_!M\) on \(V\) is a section \(s\in M(V\cap U)\) with the following property: for each \(y\in V\setminus U\), there is an open neighborhood \(W\subseteq V\) of \(y\) such that \(s|_{W\cap U}=0\). Equivalently its support is closed in \(V\) and contained in \(U\). This condition is local on \(V\), so gluing proves the sheaf condition. Its stalk is \(M_x\) inside \(U\) and zero outside. Thus \(j_!\) is exact, as one can check on these stalks.

A map \(M\to N|_U\) extends to \(j_!M\to N\) by using the given map in \(U\) and the zero section near the other points; the vanishing condition makes them agree on overlaps. Restriction recovers the original map. Conversely a map out of \(j_!M\) is determined on these same neighborhoods by its restriction to \(U\). These constructions prove \(j_!\dashv j^*\), with both inverse Hom maps. Consequently restriction \(j^*\) preserves injectives by the exact-left-adjoint argument of Section 2.

For \(M=\mathcal O|_U\), extension of sections by zero defines a monomorphism \(j_!M\hookrightarrow\mathcal O\), since its stalks are an identity or a zero inclusion. If \(I\) is injective, extending maps across this monomorphism proves surjectivity of

\operatorname{Hom}_{\mathcal O}(\mathcal O,I)=I(X) \longrightarrow \operatorname{Hom}_{\mathcal O}(j_!\mathcal O|_U,I)=I(U). \tag{6.1}

The identifications evaluate a module map at the unit section, and the inverse multiplies a section by a local scalar. Thus (6.1) is the actual restriction map. Lifting a section from \(U\) to \(X\), then restricting to any larger open \(V\), proves that \(I\) is flabby. Also an injective resolution on \(X\) restricts to an injective resolution on \(U\). It follows directly that the derived functors of \(\mathcal F\mapsto\mathcal F(U)\) on \(X\) are the ordinary sheaf cohomology groups on \(U\).

7. Lifting sections across a flabby kernel

Lemma 7.1. If \(0\to A\to B\xrightarrow{p}C\to0\) is an exact sequence of sheaf modules and \(A\) is flabby, then \(B(U)\to C(U)\) is onto for every open \(U\). If \(B\) is also flabby, so is \(C\).

Proof. Fix \(c\in C(U)\). A partial lift is a pair \((V,b)\), where \(V\subseteq U\) is open and \(p(b)=c|_V\). Order these pairs by extension of the section. The empty open supplies an initial partial lift. A chain has an upper bound: its sections agree on nested domains and glue uniquely on their union. Zorn's lemma gives a maximal partial lift \((V,b)\).

If \(x\in U\setminus V\), stalkwise surjectivity of \(p\) provides a neighborhood \(W\subseteq U\) of \(x\) and a lift \(b'\in B(W)\) of \(c|_W\). Indeed lift the germ of \(c\) at \(x\), choose a representative, then shrink until its image equals the section \(c\). On \(W\cap V\), the difference \(b'-b\) lies in the section kernel, which is \(A(W\cap V)\). Extend that section to \(a\in A(W)\) using flabbiness. Then \(b'-a\) agrees with \(b\) on the intersection. Gluing gives a partial lift on \(V\cup W\), strictly extending the maximal one. Therefore \(V=U\), proving surjectivity. Left exactness of sections supplies the other two exactness assertions.

If \(A\) and \(B\) are flabby, lift a section of \(C(U)\) to \(B(U)\), extend it to \(B(V)\) for any open \(V\supseteq U\), and project to \(C(V)\). Its restriction is the original section. Thus \(C\) is flabby. Every operation used here commutes with maps of short exact sequences and restriction when a lift is chosen; existence alone is needed. \(\square\)

8. Flabby acyclicity on every open

Theorem 8.1. For a flabby \(\mathcal F\), one has \(H^q(U,\mathcal F)=0\) for every open \(U\) and every \(q>0\).

Proof. Choose an injective embedding \(\mathcal F\hookrightarrow I^0\), which exists by the first prerequisite. The cokernel \(K^0\) is flabby by Section 6 and Lemma 7.1. Continue by embeddings \(K^n\hookrightarrow I^{n+1}\) and cokernels \(K^{n+1}\). All the \(K^n\) are flabby by induction. With \(K^{-1}=\mathcal F\), every sequence

0\longrightarrow K^{n-1}(U)\longrightarrow I^n(U) \longrightarrow K^n(U)\longrightarrow0 \tag{8.1}

is exact by Lemma 7.1. The sheaf differentials are the composites \(I^n\twoheadrightarrow K^n\hookrightarrow I^{n+1}\). Equation (8.1) shows that the resulting section complex is exact in each positive degree. It is an injective resolution of \(\mathcal F\) before taking sections, so its cohomology computes the derived groups and proves the vanishing. Section 6 identifies those groups with cohomology on \(U\). The same proof includes the empty open, where every section module is zero. \(\square\)

No injectivity of the stalks of \(\mathcal F\) is asserted here. Acyclicity is a consequence of the global extension property and the complete partial-lift argument.

9. Computing cohomology and its maps

Combining (5.4) and Theorem 8.1 gives a functorial acyclic resolution for the left exact section functor on every open. The ordinary acyclic-resolution comparison in the prerequisites therefore gives

H^q(U,\mathcal F)\simeq H^q\bigl(\Gamma(U,J^\bullet(\mathcal F))\bigr), \qquad q\ge0. \tag{9.1}

Here are explicit dimension-shifting maps realizing the comparison. Set \(Z_{-1}=\mathcal F\) and \(Z_n=\operatorname{coker}(Z_{n-1}\hookrightarrow J^n)\); exactness identifies \(Z_n\) with \(\ker(J^{n+1}\to J^{n+2})\). Thus \(0\to Z_{n-1}\to J^n\to Z_n\to0\) is exact. A degree-\(q\) cocycle is a section of \(Z_{q-1}\) inside \(J^q\); its boundaries are the images of \(J^{q-1}(U)\). For \(q\ge1\), the connecting homomorphism for the short exact sequence with \(n=q-1\) induces

\frac{Z_{q-1}(U)}{\operatorname{im}(J^{q-1}(U))} \xrightarrow{\ \sim\ } H^1(U,Z_{q-2}) \xrightarrow{\ \sim\ } H^2(U,Z_{q-3}) \longrightarrow\cdots\xrightarrow{\ \sim\ }H^q(U,\mathcal F). \tag{9.2}

The first arrow is an isomorphism by the long exact sequence and \(H^1(U,J^{q-1})=0\); each following arrow is its higher-degree connecting homomorphism, an isomorphism since both adjacent positive cohomology groups of the intervening \(J^n\) vanish. When \(q=1\) there are no further arrows. In degree zero, the inclusion identifies the section kernel with \(\mathcal F(U)\). Thus all degrees, including their actual comparison maps, are specified. Connecting homomorphisms are those of the ordinary injective-resolution long exact sequence; their functorial construction and sign convention were proved in the second prerequisite. Naturality of that sequence and of every \(T^{n+1}(a)\) proves naturality of (9.1) in \(\mathcal F\).

For an open \(V\subseteq X\), neighborhoods of \(x\in V\) contained in \(V\) are cofinal. The stalk families for restriction are therefore canonically the restricted families, and for any open \(W\subseteq V\) both \((T_X\mathcal F)|_V(W)\) and \(T_V(\mathcal F|_V)(W)\) equal \(\prod_{x\in W}\mathcal F_x\). These identifications preserve the germ unit and evaluation counit, hence every coface, differential and stalk contraction. The restriction map on the right of (9.1) is induced by the actual section restriction in every degree. Under (9.2) it agrees with the usual cohomology restriction: one may restrict an injective resolution by Section 6, and its connecting maps commute with restriction of section complexes. This proves compatibility for every inclusion of opens, not only for global sections.

10. Four exercises with full solutions

10.1 A graph on the Sierpinski space

Exercise 1 (basic). Give \(X=\{0<1\}\) its upper-set topology and use the constant field \(k\) as the ring sheaf. A sheaf corresponds to \(f:V_0\to V_1\). Compute its first Godement term and its cokernel, their global sections, and all cohomology. Determine whether \((0\to k)\) is injective, and compute the stalk support of the two point direct images.

Solution. The minimal opens are \(X\) and \(\{1\}\), so \(\Gamma(X,\mathcal F)=V_0\). Formula (1.1) gives \(T\mathcal F=(V_0\oplus V_1\xrightarrow{\mathrm{pr}_2}V_1)\). The unit is \(v\mapsto(v,fv)\) at zero and the identity at one. The quotient is \(Q=(V_1\to0)\), with quotient map \((a,b)\mapsto b-f a\) at zero. Kernel and surjectivity are immediate at both vertices, giving an exact sequence \(0\to\mathcal F\to T\mathcal F\to Q\to0\).

Both larger terms are injective: \(T\mathcal F\) is covered by Section 2 and \(Q=i_{0*}V_1\) is \(R\) of a one-component vector family. Applying global sections gives \(V_0\to V_0\oplus V_1\to V_1\), with the same graph and difference formulas. The last arrow is onto by \(b\mapsto(0,b)\). Thus \(H^0(X,\mathcal F)=V_0\) and \(H^q(X,\mathcal F)=0\) for \(q>0\). This two-term injective resolution computes the same groups as the full complex by (9.1).

For \(f:(0\to k)\), its embedding in \(T\mathcal F=(k\xrightarrow{1}k)\) cannot split. A retraction would have zero component at zero and identity component at one; commutation with the arrow would require the identity of \(k\) to be zero. An injective subobject would have a retraction, so \(\mathcal F\) is not injective. Finally, \(i_{1*}k=(k\xrightarrow{1}k)\) has nonzero stalks at both points, whereas \(i_{0*}k=(k\to0)\) has only its zero-point stalk nonzero. These are exactly the closures of \(\{1\}\) and \(\{0\}\).

10.2 Three points and a nonzero first group

Exercise 2 (intermediate). Let the poset have relations \(0<2\) and \(1<2\) and no relation between \(0\) and \(1\). Put \(\mathcal F_0=\mathcal F_1=0\), \(\mathcal F_2=k\), with zero structure arrows. Use the germ-family embedding to compute every \(H^q(X,\mathcal F)\).

Solution. The minimal-open formula gives \(T\mathcal F\) equal to \(k\) at each vertex, with both arrows the identity. The unit is zero at the two minimal vertices and identity at vertex two. Its cokernel is \(Q=(k,k,0)\), with zero arrows, and quotient identity maps at the minimal vertices. It is the finite direct sum \(i_{0*}k\oplus i_{1*}k\), hence is injective by the point-family adjunction. The same holds for \(T\mathcal F\), so again this is a length-one injective resolution.

Compatible section triples of \(\mathcal F\) must have zero last component, giving \(\Gamma\mathcal F=0\). Compatible triples of \(T\mathcal F\) are \((a,a,a)\), while \(\Gamma Q=k^2\). The induced differential is \(k\to k^2\), \(a\mapsto(a,a)\). Its kernel is zero and its cokernel is identified with \(k\) by \((b,c)\mapsto c-b\). Therefore \(H^0=0\), \(H^1=k\), and \(H^q=0\) for \(q\ge2\). This also shows concretely why the stalk contraction of the full Godement complex cannot force global positive cohomology to vanish.

10.3 A germ embedding that cannot split

Exercise 3 (advanced). Let \(X\) be connected and have two disjoint nonempty open subsets \(U_0,U_1\). Let \(\mathcal O=k_X\), the sheaf of locally constant functions to a field, and \(\mathcal F=k_X\). Show that \(\mathcal F\) is not flabby and its Godement embedding does not split as a sheaf map. Explain why this does not contradict (5.3).

Solution. Every global locally constant function is constant: each nonempty fiber is open and its complement is open, so two nonempty fibers would disconnect \(X\). On \(U_0\cup U_1\), the function equal to zero on \(U_0\) and one on \(U_1\) is locally constant. Both opens are closed relative to their union, so this prescription is locally constant at every point of the union. It cannot extend globally by connectedness. Thus restriction is not onto, and \(\mathcal F\) is not flabby. By Section 6, it is not injective.

Each stalk of \(k_X\) is canonically \(k\), and \(T\mathcal F(U)=\prod_{x\in U}k\), the module of all functions without a local-constancy condition. It is injective by Section 2. If a retraction \(T\mathcal F\to\mathcal F\) existed, \(\mathcal F\) would be injective: compose any map into it with the embedding, extend into \(T\mathcal F\), and follow by the retraction. This contradicts the preceding paragraph. Although \(h^0 S\eta=1\), the particular stalk maps \(h^0\) cannot arise from a sheaf retraction. Consequently there is no way to apply sections to that family contraction as though it were a contraction of sheaf maps.

10.4 Flabby terms over an integer stalk

Exercise 4 (advanced). On a one-point space with ring \(\mathbb Z\), take \(F=\mathbb Z/2\). Write every Godement differential and compute its cohomology. Prove that these flabby terms are not injective, and give a genuine injective resolution for comparison.

Solution. Both \(S\) and \(R\) are the identity on abelian groups. Thus \(T=1\), \(\eta=1\), and every coface is the identity. The sum in (4.1) is zero for even \(n\) and the identity for odd \(n\). The augmented complex is therefore \(0\to F\xrightarrow{1}F\xrightarrow{0}F\xrightarrow{1}F\xrightarrow{0}F\to\cdots\). It is exact, its unaugmented cohomology is \(F\) in degree zero and zero in positive degrees, and (5.1) is the identity in every degree. Sections on a point are the identity functor, so these are precisely its derived groups.

Every sheaf on a point is flabby: the only restrictions are identities and maps to the zero module on the empty open. But \(F\) is not injective. Consider the inclusion of abelian groups given by \(\mathbb Z\xrightarrow{\times2}\mathbb Z\) and the map from its domain sending one to the nonzero class of \(F\). An extension to the codomain would send some integer generator to \(a\) with \(2a=1\) in \(F\), which is impossible. Hence every Godement term in this example is noninjective.

A genuine injective resolution is \(0\to\mathbb Z/2\to\mathbb Q/\mathbb Z\xrightarrow{\times2}\mathbb Q/\mathbb Z\to0\), where the first map sends the nonzero class to \(1/2+\mathbb Z\). Multiplication by two is onto by dividing rational representatives, and its kernel consists exactly of zero and \(1/2+\mathbb Z\). The two larger groups are divisible and hence injective by the complete extension proof in Section 6 of the second prerequisite. This injective resolution and the flabby Godement resolution compute the same section cohomology, as guaranteed by the acyclic comparison, without requiring the flabby terms to be injective.