Personal remark. I have not checked the isomorphism with the singular cohomology yet, but I should skip it for now and read these notes to fill this gap later. This is really great, because singular cohomology can be often flexible since we can use continuous deformations without changing the cohomology itself!
Our goal is to show that given any real manifold $M,$ any $\mathscr{C}^{\infty}_{M}$-module $\mathscr{F}$ is acyclic. In particular, this will show that the sheaves $\mathscr{A}^{p}$ and $\mathscr{A}^{p,q}$ (for complex manifolds) are acyclic. Our strategy is to show that
- any $\mathscr{C}^{\infty}_{M}$-module is "soft", and
- any soft sheaf is acyclic.
This posting will be more or less a regurgitation of these notes.
Restriction of sheaves to a subset. Let $\mathscr{F}$ be a sheaf valued in $\textbf{Ab}$ on a topological space $X.$ For any subset $S \subset X,$ its inclusion map $i_{S} : S \hookrightarrow X$ is continuous with the subspace topology. We denote by $\mathscr{F}|_{S} := i_{S}^{-1}\mathscr{F},$ which is a sheaf on $S,$ and write $\mathscr{F}(S) := \Gamma(S, \mathscr{F}|_{S}).$ From the last part of this posting, we have a concrete description of the set $\Gamma(S, \mathscr{F}|_{S})$: it consists of maps $$f : S \rightarrow \bigsqcup_{x \in S}\mathscr{F}_{x}$$ such that for any $x \in S,$ we may find an open $U \ni x$ in $X$ and $g \in \Gamma(U, \mathscr{F})$ such that $f(x) = g_{x}$ for all $x \in U \cap S.$ For convenience, we shall often write $$\Gamma(S, \mathscr{F}|_{S}) = \mathscr{F}(S),$$ which corresponds with our previous convention when $S$ is an open subset of $X.$ The global section of the sheaf map $$\mathscr{F} \rightarrow (i_{S})_{*}(\mathscr{F}|_{S})$$ (of sheaves on $X$) is given by $\Gamma(X, \mathscr{F}) \rightarrow \Gamma(S, \mathscr{F}|_{S}),$ which sends $h \mapsto h|_{S}.$ (This corresponds to the identity of $\mathscr{F}|_{S}$ in the adjunction formula in 2.7.B of Vakil.) We call this map the restriction map and note that this is the usual restriction map, if $S$ happens to be an open subset of $X.$
Personal remark. I think we should use pullbacks of modules (instead of inverse images of sheaves) to do similar activities as above when we are given ringed spaces, but I won't dig into these details, as we are not going to need this (as far as I reckon).
Remark. For our discussion, we only need to know restriction maps and technically do not need to know the inverse image sheaves. However, defining inverse image sheaves without the foundational story of sheaf theory seemed too ad-hoc and did not make musch sense to me, so I have summarized it in this posting. I will use the language in this summary because it seems impossible to avoid it for the discussions in this post.
Soft sheaves. A sheaf $\mathscr{F}$ valued in $\textbf{Ab}$ on a topological space $X$ is called soft if the restriction $\mathscr{F}(X) \rightarrow \mathscr{F}(Z)$ is surjective for every closed subset $Z \subset X.$
Paracompact spaces. A topological space $X$ is paracompact if
Corollary. For any smooth real manifold $M,$ we have $$H^{i}(M, \underline{\mathbb{R}}) \simeq H^{i}(\Gamma(M, \mathscr{A}^{\bullet}_{M})) =: H^{i}_{\mathrm{dR}}(M)$$ for all $i \geq 0.$
Corollary. For any complex manifold $M,$ we have $$H^{q}(M, \Omega^{p}_{M}) \simeq H^{q}(\Gamma(M, \mathscr{A}^{p,\bullet}_{M}))$$ for all $p, q \geq 0.$
Proof of Theorem 1. We will use one more word: a sheaf $\mathscr{A}$ on $X$ valued in $\textbf{Ab}$ is said to be flasque (or flabby) if for every pair of open subsets $U \hookrightarrow V$ in $X,$ the induced restriction map $\Gamma(V, \mathscr{A}) \rightarrow \Gamma(U, \mathscr{A})$ is surjective.
We proceed by induction on $i \geq 1.$ We use the following facts.
Fact 1. On a paracompact topological space, flasque sheaves are soft.
Fact 2. On any topological space, flasque sheaves are acyclic.
Fact 3. Given any soft sheaf $\mathscr{E}$ on a paracompact topological space $X,$ there is an exact sequence $0 \rightarrow \mathscr{E} \rightarrow \mathscr{A}$ in the category of sheaves on $X$ valued in $\textbf{Ab}$, where $\mathscr{A}$ is flasque.
Fact 4. On a paracompact topological space, given an exact sequence $$0 \rightarrow \mathscr{E} \rightarrow \mathscr{A} \rightarrow \mathscr{B} \rightarrow 0,$$ if $\mathscr{E}$ and $\mathscr{A}$ are soft, then so is $\mathscr{B}.$
Fact 5. On a paracompact topological space, given any exact sequence $$0 \rightarrow \mathscr{A} \rightarrow \mathscr{B} \rightarrow \mathscr{C} \rightarrow 0,$$ if $\mathscr{A}$ is soft, then the induced sequence on the global sections $$0 \rightarrow \Gamma(X, \mathscr{A}) \rightarrow \Gamma(X, \mathscr{B}) \rightarrow \Gamma(X, \mathscr{C}) \rightarrow 0$$ is exact.
Proof of Theorem 1. Now, using Fact 1, 2 and 4, we have an exact sequence $$0 \rightarrow \mathscr{E} \rightarrow \mathscr{A} \rightarrow \mathscr{B} \rightarrow 0,$$ where $\mathscr{A}$ is flasque and $\mathscr{B}$ is soft.
Consider (coming from the sketchy argument from this posting) the long exact sequence $$0 \rightarrow \Gamma(X, \mathscr{E}) \rightarrow \Gamma(X, \mathscr{A}) \rightarrow \Gamma(X, \mathscr{B}) \xrightarrow{\alpha} H^{1}(X, \mathscr{E}) \rightarrow H^{1}(X, \mathscr{A}) \rightarrow \cdots.$$ By Fact 5, we note that $\alpha$ is surjective and by Fact 2, we see $H^{i}(X, \mathscr{A}) = 0$ for $i \geq 1.$ Using this and the long exact sequence, we see that $H^{1}(X, \mathscr{E}) = 0$ and $$H^{i}(X, \mathscr{B}) \simeq H^{i+1}(X, \mathscr{E})$$ for all $i \geq 1.$ Since $\mathscr{B}$ is soft, applying induction on $i \geq 1$ to the statement we want to prove, we are done. $\Box$
Proof of Theorem 2. We need the smooth version of Uryshon's lemma. We follow Tu's book (Problem 13.3 (a)).
Lemma (Smooth Uryshon). Let $M$ be a smooth real manifold. Fix any disjoint closed sets $A, B \subset M.$ There is $f \in \Gamma(M, \mathscr{C}^{\infty}_{M})$ such that $f(A) = \{0\}$ and $f(B) = \{1\}.$
Proof. Let $U := M \setminus A$ and $V := M \setminus B,$ which form an open cover of $M.$ Choose a $\mathscr{C}^{\infty}$ partition of unity $\varphi_{U}, \varphi_{V}$ subordinate to this open cover (e.g., Theorem 13.7 (b) in Tu's book). This means that $\varphi_{U}, \varphi_{V} \in \Gamma(M, \mathscr{C}^{\infty}_{M})$ such that
Since $\mathrm{supp}(\varphi_{V}) \subset V = M \setminus B,$ we note that $\varphi_{V} = 0$ on $B.$ Similarly, we see that $\varphi_{U} = 0$ on $A,$ so $\varphi_{V} = 1$ on $A.$ Thus, taking $f := \varphi_{V}$ will do the job. $\Box$
We also need another technical fact.
Fact 6. Let $X$ be any paracompact topological space and $\mathscr{F}$ be a sheaf on $X$ valued in $\textbf{Ab}.$ Then for every closed subset $Z \subset X$ and $s \in \mathscr{F}(Z),$ there is an open subset $U \supset Z$ in $X$ and $t \in \mathscr{F}(U)$ such that $t|_{U} = s.$
We now prove Theorem 2.
Proof of Theorem 2. Let $\mathscr{F}$ be any $\mathscr{C}_{M}^{\infty}$-module. Fix any closed subset $Z \subset M$ and let $s \in \mathscr{F}(Z).$ Our goal is to extend this to a section in $\mathscr{F}(M).$ Using Fact 6, we may first extend a little bit. Namely, there is an open subset $U \supset Z$ in $M$ and $t \in \mathscr{F}(U)$ such that $t|_{Z} = s.$ Since $M$ is paracompact, it is normal (e.g., see the last part of this posting) so that we can find open subsets $U_{1}, U_{2} \subset M$ such that $$Z \subset U_{1} \subset \bar{U_{1}} \subset U_{2} \subset \bar{U_{2}} \subset U.$$ Apply Smooth Uryshon to find $f \in \Gamma(M, \mathscr{C}^{\infty}_{M})$ such that $f(\bar{U_{1}}) = \{1\}$ and $f(M \setminus U_{2}) = \{0\}.$
We now crucially use the $\mathscr{C}_{M}^{\infty}$-module structure of on $\mathscr{F}.$ Namely, writing $f = f|_{U}$ as well, we have $ft \in \mathscr{F}(U).$ What's so special about this section? We have $(ft)|_{U \setminus \bar{U_{2}}} = 0$ because $$U \setminus \bar{U_{2}} \subset M \setminus \bar{U_{2}} \subset M \setminus U_{2},$$ where $f$ vanishes. This implies that $0 \in \Gamma(M \setminus \bar{U_{2}}, \mathscr{F})$ and $ft \in \Gamma(U, \mathscr{F})$ restrict to the same section on the intersection $(M \setminus \bar{U_{2}}) \cap U = U \setminus \bar{U_{2}}.$ Since $\mathscr{F}$ is a sheaf and $(M \setminus \bar{U_{2}}) \cup U = M,$ this means that there is a unique $g \in \Gamma(M, \mathscr{C}^{\infty}_{M})$ such that $g|_{M \setminus \bar{U_{2}}} = 0$ and $g|_{U} = ft.$ Since $f = 1$ on $U_{1} \supset Z,$ we must have $g|_{Z} = (ft)|_{Z} = t|_{Z} = s,$ which finishes the proof. $\Box$
Proof of technical facts. First, we note that Fact 6 implies Fact 1.
Fact 6 implies Fact 1. Let $\mathscr{F}$ be flasque. To show it is soft, let $Z \subset X$ be a closed subset. Given any $s \in \mathscr{F}(Z),$ we want to show that $s$ extends to an element in $\mathscr{F}(X).$ By Fact 1, we may find an open subset $U \supset Z$ in $X$ with $t \in \mathscr{F}(U)$ such that $t|_{Z} = s.$ Since $\mathscr{F}$ is flasque, we may find some $t' \in \mathscr{F}(X)$ such that $t'|_{U} = t,$ so $t'|_{Z} = s.$ Thus, we have shown that $\mathscr{F}$ is soft. $\Box$
Proof of Fact 6. Using the definition of $\mathscr{F}(Z),$ we may find a family $\{U_{i}\}_{i \in I}$ of open subsets of $X$ such that $Z \subset \bigcup_{i \in I}U_{i}$ with $s_{i} \in \mathscr{F}(Z)$ satisfying $s_{i}|_{U_{i} \cap Z} = s|_{U_{i} \cap Z}$ for each $i \in I.$ Then $\{U_{i}\}_{i \in I} \cup \{X \setminus Z\}$ forms an open cover of $X.$ Since $X$ is paracompact, this open cover must have a locally finite subcover. We may keep the open set $X \setminus Z$ in this subcover, and the union of other members cover $Z,$ so this procedure allows us to rechoose $U_{i}$ so that $\{U_{i}\}_{i \in I} \cup \{X \setminus Z\}$ is a locally finite open cover of $X.$
For every $x \in Z$, we have finitely many $U_{i_{1}}, \dots, U_{i_{r}}$ in this cover that contain $x$. Using paracompactness (or more precisely normality) of $X,$ we may find open $V_{i_{j}} \ni x$ in $U_{i_{j}}$ such that $\overline{V_{i_{j}}} \subset U_{i_{j}}.$ We note that the family $\{V_{i}\}_{i \in I} \cup \{X \setminus Z\}$ form an open cover of $X$ such that $$\overline{V_{i}} \subset U_{i}$$ and $Z \subset \bigcup_{i \in I}V_{i}.$
Personal remark. If we did not have local finiteness, we might still be okay in the situation above by some axiom of choice.
Fix $x \in X.$ Choose an open neighborhood $W_{x} \ni x$ in $X$ such that
Write $s^{(x)} := s_{i}|_{W_{x}}.$ In particular, we have $s^{(x)}_{x} = (s_{i'})_{x} = s(x) \in \mathscr{F}_{x}.$ Whenever $x \notin \overline{V_{i}},$ we may replace $W_{x}$ by $W_{x} \setminus \overline{V_{i}}$ and modify $s^{(x)}$ accordingly, which would keep all the properties above. Since there can only be finitely many $\overline{V_{i}}$ intersecting $W_{x},$ this process will end after finitely many times, so we may assume that
Whenever $\overline{V_{i}} \cap W_{x} \neq \emptyset,$ we may replace $W_{x}$ with $W_{x} \cap U_{i},$ keeping all the properties given above. Since this process will end in finitely many times, we can also assume that
Whenever $\overline{V_{i}} \cap W_{x} \neq \emptyset,$ we have $s^{(x)}_{x} = s(x) = (s_{i})_{x},$ so by replacing $W_{x}$ with a small open neighborhood of $x$ in $W_{x}$ where $s^{(x)}$ and $s_{i}$ coincide, we may assume that $s^{(x)} = s_{i}|_{W_{x}}.$
Consider $U = \bigcup_{x \in Z}W_{x} \supset Z.$ Since $\mathscr{F}$ is a sheaf, to finish the entire proof, it is enough to show that $s^{(x)} = s^{(y)}$ on $W_{x} \cap W_{y}$ for all $x, y \in Z.$ Given any $z \in W_{x} \cap W_{y},$ we have some $V_{i} \ni x$ due to the above setting. This implies that $\overline{V_{i}} \neq W_{x} \neq \emptyset$ and $\overline{V_{i}} \neq W_{y} \neq \emptyset,$ so $W_{x}, W_{y} \subset U_{i},$ while $s_{i}|_{W_{x}} = s^{(x)}$ and $s_{i}|_{W_{y}} = s^{(y)}$ from the setting. This implies that $s^{(x)} = s_{i} = s^{(y)}$ on $W_{x} \cap W_{y},$ as desired. $\Box$
Proof of Fact 2. We follow these notes. Fix a topological space $X.$ First, we have a few facts.
Fact 2.1. Given any exact sequence $$0 \rightarrow \mathscr{F} \rightarrow \mathscr{G} \rightarrow \mathscr{H} \rightarrow 0$$ of sheaves (valued in $\textbf{Ab}$) on $X,$ if $\mathscr{F}$ is flasque, then we get an exact sequence $$0 \rightarrow \Gamma(X, \mathscr{F}) \rightarrow \Gamma(X, \mathscr{G}) \rightarrow \Gamma(X, \mathscr{H}) \rightarrow 0$$ when we take the global section functor.
Fact 2.2. Any injective sheaf on $X$ is flasque.
Fact 2.3. Given any exact sequence $$0 \rightarrow \mathscr{F} \rightarrow \mathscr{G} \rightarrow \mathscr{H} \rightarrow 0$$ of sheaves, if $\mathscr{F}$ and $\mathscr{G}$ are flasque, then $\mathscr{H}$ is flasque.
Proof of Fact 2. Let $\mathscr{F}$ be a flasque sheaf. Since $\textbf{Sh}_{X}$ has enough injectives (e.g., see Stacks Project), we may form an exact sequence $$0 \rightarrow \mathscr{F} \rightarrow \mathscr{I} \rightarrow \mathscr{Q} \rightarrow 0.$$ Consider the induced long exact sequence $$0 \rightarrow \Gamma(\mathscr{F}) \rightarrow \Gamma(\mathscr{I}) \rightarrow \Gamma(\mathscr{Q}) \rightarrow H^{1}(\mathscr{F}) \rightarrow H^{1}(\mathscr{I}) \rightarrow \cdots$$ of abelian groups, where we suppressed $X$ in our notations. By Fact 2.1, the map $\Gamma(\mathscr{Q}) \rightarrow H^{1}(\mathscr{F})$ is the zero map and since $\mathscr{I}$ is injective, we have $H^{1}(\mathscr{I}) = 0.$ Thus, we have $H^{1}(\mathscr{F}) = 0.$ For $i \geq 2,$ we have the exact sequence $$0 = \mathscr{H}^{i-1}(\mathscr{I}) \rightarrow H^{i-1}(\mathscr{H}) \rightarrow H^{i}(\mathscr{F}) \rightarrow \mathscr{H}^{i}(\mathscr{I}) = 0,$$ so $$H^{i-1}(\mathscr{H}) \simeq H^{i}(\mathscr{F}).$$ By Fact 2.2 and Fact 2.3, we note that $\mathscr{H}$ is also flasque, so arguing by induction, we may finish the proof. $\Box$
Proof of Fact 2.1. Let $s \in \Gamma(X, \mathscr{H}).$ Since taking the stalk at any point gives rise to an exact sequence, we may find an open cover $X = \bigcup_{i \in I}U_{i}$ with $f_{i} \in \Gamma(X, \mathscr{G})$ such that $f_{i} \mapsto s|_{U_{i}}$ for each $i \in I.$ Now, the pairs $(U_{i}, f_{i})$ form a partial ordering given by the inclusion of open subsets. Every chain has an upper bound, because $\mathscr{G}$ is a sheaf. Using Zorn's lemma, we may choose a maximal element $(U, f)$ among these pairs. In particular, we have $f \mapsto s|_{U}.$
To finish the proof, it is enough to show that $U = X.$ Suppose not. Then since $X = \bigcup_{i \in I}U_{i},$ there must be a nonempty $U_{i} \not\subset U.$ We have $$(f - f_{i})|_{U_{i} \cap U} \mapsto 0 \in \Gamma(U_{i} \cap U, \mathscr{H}),$$ so there must be (unique) $h \in \Gamma(U_{i}, \cap U, \mathscr{F})$ such that $h \mapsto (f - f_{i})|_{U_{i} \cap U}.$ But then since $\mathscr{F}$ is flasque, there is $\tilde{h} \in \Gamma(U_{i}, \mathscr{F})$ such that $\tilde{h}|_{U_{i} \cap U} = h.$ Now, consider $\tilde{h} \mapsto g \in \Gamma(U_{i}, \mathscr{G}).$ Then $f_{i} + g \in \Gamma(U_{i}, \mathscr{G})$ and $f \in \Gamma(U, \mathscr{G})$ coincide on $U_{i} \cap U,$ so we can glue them as a unique section $t \in \Gamma(U_{i} \cup U, \mathscr{G}),$ using the fact that $\mathscr{G}$ is a sheaf. We have $t \mapsto s|_{U_{i} \cup U},$ but then since $U \subsetneq U_{i} \cup U,$ we get a contradiction to the maximality of $(U, f).$ This finishes the proof. $\Box$
Proof of Fact 2.2. The proof is available in this posting. $\Box$
Proof of Fact 3. This immediately follows from Fact 2.2, because we may just take $\mathscr{A}$ to be an injective sheaf.
Proof that Fact 5 implies Fact 4. This is surprisingly simple. See Lemma 2.5 in these notes. $\Box$
It now remains to prove Fact 5.
Proof of Fact 5. We first prove a general lemma we will use.
Lemma. Let $X$ be any topological space. Say we are given a finite union $X = Z_{1} \cup \cdots \cup Z_{r}$ where each $Z_{i} \subset X$ is a closed subset. For any sheaf $\mathscr{F}$ on $X$ valued in $\textbf{Ab},$ the following gives an equalizer diagram (in $\textbf{Ab}$) $$0 \rightarrow \mathscr{F}(X) \rightarrow \prod_{i=1}^{r}\mathscr{F}(Z_{i}) \rightrightarrows \prod_{1 \leq i, j \leq r}\mathscr{F}(Z_{i} \cap Z_{j}),$$ where the map from $\mathscr{F}(X)$ is given by $h \mapsto (h|_{Z_{i}})_{1 \leq i \leq r}$ and the following pair of maps are given by
Restriction of sheaves to a subset. Let $\mathscr{F}$ be a sheaf valued in $\textbf{Ab}$ on a topological space $X.$ For any subset $S \subset X,$ its inclusion map $i_{S} : S \hookrightarrow X$ is continuous with the subspace topology. We denote by $\mathscr{F}|_{S} := i_{S}^{-1}\mathscr{F},$ which is a sheaf on $S,$ and write $\mathscr{F}(S) := \Gamma(S, \mathscr{F}|_{S}).$ From the last part of this posting, we have a concrete description of the set $\Gamma(S, \mathscr{F}|_{S})$: it consists of maps $$f : S \rightarrow \bigsqcup_{x \in S}\mathscr{F}_{x}$$ such that for any $x \in S,$ we may find an open $U \ni x$ in $X$ and $g \in \Gamma(U, \mathscr{F})$ such that $f(x) = g_{x}$ for all $x \in U \cap S.$ For convenience, we shall often write $$\Gamma(S, \mathscr{F}|_{S}) = \mathscr{F}(S),$$ which corresponds with our previous convention when $S$ is an open subset of $X.$ The global section of the sheaf map $$\mathscr{F} \rightarrow (i_{S})_{*}(\mathscr{F}|_{S})$$ (of sheaves on $X$) is given by $\Gamma(X, \mathscr{F}) \rightarrow \Gamma(S, \mathscr{F}|_{S}),$ which sends $h \mapsto h|_{S}.$ (This corresponds to the identity of $\mathscr{F}|_{S}$ in the adjunction formula in 2.7.B of Vakil.) We call this map the restriction map and note that this is the usual restriction map, if $S$ happens to be an open subset of $X.$
Personal remark. I think we should use pullbacks of modules (instead of inverse images of sheaves) to do similar activities as above when we are given ringed spaces, but I won't dig into these details, as we are not going to need this (as far as I reckon).
Remark. For our discussion, we only need to know restriction maps and technically do not need to know the inverse image sheaves. However, defining inverse image sheaves without the foundational story of sheaf theory seemed too ad-hoc and did not make musch sense to me, so I have summarized it in this posting. I will use the language in this summary because it seems impossible to avoid it for the discussions in this post.
Soft sheaves. A sheaf $\mathscr{F}$ valued in $\textbf{Ab}$ on a topological space $X$ is called soft if the restriction $\mathscr{F}(X) \rightarrow \mathscr{F}(Z)$ is surjective for every closed subset $Z \subset X.$
Paracompact spaces. A topological space $X$ is paracompact if
- $X$ is Hausdorff and
- every open cover $X = \bigcup_{i \in I}U_{i},$ there is a locally finite open cover $X = \bigcup_{j \in I}V_{j},$ refining the given cover.
Personal remark. The second condition is mouthful. Saying that $\{V_{j}\}_{j \in J}$ is locally finite in $X$ means that for any $x \in X,$ there are at most finitely many $V_{j}$ that contain $x.$ Saying that $\{V_{j}\}_{j \in J}$ refines $\{U_{i}\}_{i \in I}$ means that for each $V_{j}$ there is $U_{i} \supset V_{j}.$ This seems to me counter-intuitive at first, but I suppose I can think like "every small guy should refine some big guy" because if some $V_{j}$ does not lie in any $U_{i},$ it does not look like we are "refining" when we draw a picture.
Example. If $X$ is compact, then $X$ is paracompact, because already we can find a finite open cover of $X.$
Example. If $X$ is a CW complex, then $X$ is paracompact. (I have not checked this yet, but I left the reference here.)
Example. Any smooth manifold is paracompact. More generally, any topological space that is locally compact, Hausdorff, and second-countable is paracompact, and the proof from Lee's book is written here.
Our goal is to prove the following two theorems:
Our goal is to prove the following two theorems:
Theorem 1. Let $X$ be a paracompact topological space. Given any sheaf $\mathscr{E}$ valued in $\textbf{Ab},$ we have $H^{i}(X, \mathscr{E}) = 0$ for all $i \geq 1.$ In particular, (say by the arguments in this posting) for any sheaf $\mathscr{F}$ valued in $\textbf{Ab},$ any resolution $$0 \rightarrow \mathscr{F} \rightarrow \mathscr{E}^{0} \rightarrow \mathscr{E}^{1} \rightarrow \mathscr{E}^{2} \rightarrow \cdots$$ with soft $\mathscr{E}^{i}$ (for all $i$) computes the cohomology of $\mathscr{F}$: $$H^{i}(X, \mathscr{F}) \simeq H^{i}(\Gamma(X, \mathscr{E}^{\bullet}))$$ for all $i \geq 0.$
Given a smooth real manifold $M$ (in particular, Hausdorff with a countable basis), consider the sheaf $\mathscr{C}_{M}^{\infty}$ of $\mathbb{R}$-valued smooth functions (on open subsets of $M$). Note that $(M, \mathscr{C}_{M}^{\infty})$ is a (locally) ringed space.
Theorem 2. Every $\mathscr{C}_{M}^{\infty}$-module is a soft sheaf.
Given a smooth real manifold $M$ (in particular, Hausdorff with a countable basis), consider the sheaf $\mathscr{C}_{M}^{\infty}$ of $\mathbb{R}$-valued smooth functions (on open subsets of $M$). Note that $(M, \mathscr{C}_{M}^{\infty})$ is a (locally) ringed space.
Theorem 2. Every $\mathscr{C}_{M}^{\infty}$-module is a soft sheaf.
Corollary. For any complex manifold $M,$ we have $$H^{q}(M, \Omega^{p}_{M}) \simeq H^{q}(\Gamma(M, \mathscr{A}^{p,\bullet}_{M}))$$ for all $p, q \geq 0.$
Proof of Theorem 1. We will use one more word: a sheaf $\mathscr{A}$ on $X$ valued in $\textbf{Ab}$ is said to be flasque (or flabby) if for every pair of open subsets $U \hookrightarrow V$ in $X,$ the induced restriction map $\Gamma(V, \mathscr{A}) \rightarrow \Gamma(U, \mathscr{A})$ is surjective.
We proceed by induction on $i \geq 1.$ We use the following facts.
Fact 1. On a paracompact topological space, flasque sheaves are soft.
Fact 2. On any topological space, flasque sheaves are acyclic.
Fact 3. Given any soft sheaf $\mathscr{E}$ on a paracompact topological space $X,$ there is an exact sequence $0 \rightarrow \mathscr{E} \rightarrow \mathscr{A}$ in the category of sheaves on $X$ valued in $\textbf{Ab}$, where $\mathscr{A}$ is flasque.
Fact 4. On a paracompact topological space, given an exact sequence $$0 \rightarrow \mathscr{E} \rightarrow \mathscr{A} \rightarrow \mathscr{B} \rightarrow 0,$$ if $\mathscr{E}$ and $\mathscr{A}$ are soft, then so is $\mathscr{B}.$
Fact 5. On a paracompact topological space, given any exact sequence $$0 \rightarrow \mathscr{A} \rightarrow \mathscr{B} \rightarrow \mathscr{C} \rightarrow 0,$$ if $\mathscr{A}$ is soft, then the induced sequence on the global sections $$0 \rightarrow \Gamma(X, \mathscr{A}) \rightarrow \Gamma(X, \mathscr{B}) \rightarrow \Gamma(X, \mathscr{C}) \rightarrow 0$$ is exact.
Proof of Theorem 1. Now, using Fact 1, 2 and 4, we have an exact sequence $$0 \rightarrow \mathscr{E} \rightarrow \mathscr{A} \rightarrow \mathscr{B} \rightarrow 0,$$ where $\mathscr{A}$ is flasque and $\mathscr{B}$ is soft.
Consider (coming from the sketchy argument from this posting) the long exact sequence $$0 \rightarrow \Gamma(X, \mathscr{E}) \rightarrow \Gamma(X, \mathscr{A}) \rightarrow \Gamma(X, \mathscr{B}) \xrightarrow{\alpha} H^{1}(X, \mathscr{E}) \rightarrow H^{1}(X, \mathscr{A}) \rightarrow \cdots.$$ By Fact 5, we note that $\alpha$ is surjective and by Fact 2, we see $H^{i}(X, \mathscr{A}) = 0$ for $i \geq 1.$ Using this and the long exact sequence, we see that $H^{1}(X, \mathscr{E}) = 0$ and $$H^{i}(X, \mathscr{B}) \simeq H^{i+1}(X, \mathscr{E})$$ for all $i \geq 1.$ Since $\mathscr{B}$ is soft, applying induction on $i \geq 1$ to the statement we want to prove, we are done. $\Box$
Proof of Theorem 2. We need the smooth version of Uryshon's lemma. We follow Tu's book (Problem 13.3 (a)).
Lemma (Smooth Uryshon). Let $M$ be a smooth real manifold. Fix any disjoint closed sets $A, B \subset M.$ There is $f \in \Gamma(M, \mathscr{C}^{\infty}_{M})$ such that $f(A) = \{0\}$ and $f(B) = \{1\}.$
Proof. Let $U := M \setminus A$ and $V := M \setminus B,$ which form an open cover of $M.$ Choose a $\mathscr{C}^{\infty}$ partition of unity $\varphi_{U}, \varphi_{V}$ subordinate to this open cover (e.g., Theorem 13.7 (b) in Tu's book). This means that $\varphi_{U}, \varphi_{V} \in \Gamma(M, \mathscr{C}^{\infty}_{M})$ such that
- $\mathrm{supp}(\varphi_{U}) \subset U$ and $\mathrm{supp}(\varphi_{V}) \subset V,$ and
- $\varphi_{U} + \varphi_{V} = 1$ everywhere on $M.$
Since $\mathrm{supp}(\varphi_{V}) \subset V = M \setminus B,$ we note that $\varphi_{V} = 0$ on $B.$ Similarly, we see that $\varphi_{U} = 0$ on $A,$ so $\varphi_{V} = 1$ on $A.$ Thus, taking $f := \varphi_{V}$ will do the job. $\Box$
We also need another technical fact.
Fact 6. Let $X$ be any paracompact topological space and $\mathscr{F}$ be a sheaf on $X$ valued in $\textbf{Ab}.$ Then for every closed subset $Z \subset X$ and $s \in \mathscr{F}(Z),$ there is an open subset $U \supset Z$ in $X$ and $t \in \mathscr{F}(U)$ such that $t|_{U} = s.$
We now prove Theorem 2.
Proof of Theorem 2. Let $\mathscr{F}$ be any $\mathscr{C}_{M}^{\infty}$-module. Fix any closed subset $Z \subset M$ and let $s \in \mathscr{F}(Z).$ Our goal is to extend this to a section in $\mathscr{F}(M).$ Using Fact 6, we may first extend a little bit. Namely, there is an open subset $U \supset Z$ in $M$ and $t \in \mathscr{F}(U)$ such that $t|_{Z} = s.$ Since $M$ is paracompact, it is normal (e.g., see the last part of this posting) so that we can find open subsets $U_{1}, U_{2} \subset M$ such that $$Z \subset U_{1} \subset \bar{U_{1}} \subset U_{2} \subset \bar{U_{2}} \subset U.$$ Apply Smooth Uryshon to find $f \in \Gamma(M, \mathscr{C}^{\infty}_{M})$ such that $f(\bar{U_{1}}) = \{1\}$ and $f(M \setminus U_{2}) = \{0\}.$
We now crucially use the $\mathscr{C}_{M}^{\infty}$-module structure of on $\mathscr{F}.$ Namely, writing $f = f|_{U}$ as well, we have $ft \in \mathscr{F}(U).$ What's so special about this section? We have $(ft)|_{U \setminus \bar{U_{2}}} = 0$ because $$U \setminus \bar{U_{2}} \subset M \setminus \bar{U_{2}} \subset M \setminus U_{2},$$ where $f$ vanishes. This implies that $0 \in \Gamma(M \setminus \bar{U_{2}}, \mathscr{F})$ and $ft \in \Gamma(U, \mathscr{F})$ restrict to the same section on the intersection $(M \setminus \bar{U_{2}}) \cap U = U \setminus \bar{U_{2}}.$ Since $\mathscr{F}$ is a sheaf and $(M \setminus \bar{U_{2}}) \cup U = M,$ this means that there is a unique $g \in \Gamma(M, \mathscr{C}^{\infty}_{M})$ such that $g|_{M \setminus \bar{U_{2}}} = 0$ and $g|_{U} = ft.$ Since $f = 1$ on $U_{1} \supset Z,$ we must have $g|_{Z} = (ft)|_{Z} = t|_{Z} = s,$ which finishes the proof. $\Box$
Proof of technical facts. First, we note that Fact 6 implies Fact 1.
Fact 6 implies Fact 1. Let $\mathscr{F}$ be flasque. To show it is soft, let $Z \subset X$ be a closed subset. Given any $s \in \mathscr{F}(Z),$ we want to show that $s$ extends to an element in $\mathscr{F}(X).$ By Fact 1, we may find an open subset $U \supset Z$ in $X$ with $t \in \mathscr{F}(U)$ such that $t|_{Z} = s.$ Since $\mathscr{F}$ is flasque, we may find some $t' \in \mathscr{F}(X)$ such that $t'|_{U} = t,$ so $t'|_{Z} = s.$ Thus, we have shown that $\mathscr{F}$ is soft. $\Box$
Proof of Fact 6. Using the definition of $\mathscr{F}(Z),$ we may find a family $\{U_{i}\}_{i \in I}$ of open subsets of $X$ such that $Z \subset \bigcup_{i \in I}U_{i}$ with $s_{i} \in \mathscr{F}(Z)$ satisfying $s_{i}|_{U_{i} \cap Z} = s|_{U_{i} \cap Z}$ for each $i \in I.$ Then $\{U_{i}\}_{i \in I} \cup \{X \setminus Z\}$ forms an open cover of $X.$ Since $X$ is paracompact, this open cover must have a locally finite subcover. We may keep the open set $X \setminus Z$ in this subcover, and the union of other members cover $Z,$ so this procedure allows us to rechoose $U_{i}$ so that $\{U_{i}\}_{i \in I} \cup \{X \setminus Z\}$ is a locally finite open cover of $X.$
For every $x \in Z$, we have finitely many $U_{i_{1}}, \dots, U_{i_{r}}$ in this cover that contain $x$. Using paracompactness (or more precisely normality) of $X,$ we may find open $V_{i_{j}} \ni x$ in $U_{i_{j}}$ such that $\overline{V_{i_{j}}} \subset U_{i_{j}}.$ We note that the family $\{V_{i}\}_{i \in I} \cup \{X \setminus Z\}$ form an open cover of $X$ such that $$\overline{V_{i}} \subset U_{i}$$ and $Z \subset \bigcup_{i \in I}V_{i}.$
Personal remark. If we did not have local finiteness, we might still be okay in the situation above by some axiom of choice.
Fix $x \in X.$ Choose an open neighborhood $W_{x} \ni x$ in $X$ such that
- $W_{x}$ intersects only finitely many $U_{i}$ and
- $W_{x}$ is contained in some $V_{i'}.$
- if $\overline{V_{i}} \cap W_{x} \neq \emptyset,$ then $x \in \overline{V_{i}}.$
Whenever $\overline{V_{i}} \cap W_{x} \neq \emptyset,$ we may replace $W_{x}$ with $W_{x} \cap U_{i},$ keeping all the properties given above. Since this process will end in finitely many times, we can also assume that
- if $\overline{V_{i}} \cap W_{x} \neq \emptyset,$ then $W_{x} \subset U_{i}.$
Consider $U = \bigcup_{x \in Z}W_{x} \supset Z.$ Since $\mathscr{F}$ is a sheaf, to finish the entire proof, it is enough to show that $s^{(x)} = s^{(y)}$ on $W_{x} \cap W_{y}$ for all $x, y \in Z.$ Given any $z \in W_{x} \cap W_{y},$ we have some $V_{i} \ni x$ due to the above setting. This implies that $\overline{V_{i}} \neq W_{x} \neq \emptyset$ and $\overline{V_{i}} \neq W_{y} \neq \emptyset,$ so $W_{x}, W_{y} \subset U_{i},$ while $s_{i}|_{W_{x}} = s^{(x)}$ and $s_{i}|_{W_{y}} = s^{(y)}$ from the setting. This implies that $s^{(x)} = s_{i} = s^{(y)}$ on $W_{x} \cap W_{y},$ as desired. $\Box$
Proof of Fact 2. We follow these notes. Fix a topological space $X.$ First, we have a few facts.
Fact 2.1. Given any exact sequence $$0 \rightarrow \mathscr{F} \rightarrow \mathscr{G} \rightarrow \mathscr{H} \rightarrow 0$$ of sheaves (valued in $\textbf{Ab}$) on $X,$ if $\mathscr{F}$ is flasque, then we get an exact sequence $$0 \rightarrow \Gamma(X, \mathscr{F}) \rightarrow \Gamma(X, \mathscr{G}) \rightarrow \Gamma(X, \mathscr{H}) \rightarrow 0$$ when we take the global section functor.
Fact 2.2. Any injective sheaf on $X$ is flasque.
Fact 2.3. Given any exact sequence $$0 \rightarrow \mathscr{F} \rightarrow \mathscr{G} \rightarrow \mathscr{H} \rightarrow 0$$ of sheaves, if $\mathscr{F}$ and $\mathscr{G}$ are flasque, then $\mathscr{H}$ is flasque.
Proof of Fact 2. Let $\mathscr{F}$ be a flasque sheaf. Since $\textbf{Sh}_{X}$ has enough injectives (e.g., see Stacks Project), we may form an exact sequence $$0 \rightarrow \mathscr{F} \rightarrow \mathscr{I} \rightarrow \mathscr{Q} \rightarrow 0.$$ Consider the induced long exact sequence $$0 \rightarrow \Gamma(\mathscr{F}) \rightarrow \Gamma(\mathscr{I}) \rightarrow \Gamma(\mathscr{Q}) \rightarrow H^{1}(\mathscr{F}) \rightarrow H^{1}(\mathscr{I}) \rightarrow \cdots$$ of abelian groups, where we suppressed $X$ in our notations. By Fact 2.1, the map $\Gamma(\mathscr{Q}) \rightarrow H^{1}(\mathscr{F})$ is the zero map and since $\mathscr{I}$ is injective, we have $H^{1}(\mathscr{I}) = 0.$ Thus, we have $H^{1}(\mathscr{F}) = 0.$ For $i \geq 2,$ we have the exact sequence $$0 = \mathscr{H}^{i-1}(\mathscr{I}) \rightarrow H^{i-1}(\mathscr{H}) \rightarrow H^{i}(\mathscr{F}) \rightarrow \mathscr{H}^{i}(\mathscr{I}) = 0,$$ so $$H^{i-1}(\mathscr{H}) \simeq H^{i}(\mathscr{F}).$$ By Fact 2.2 and Fact 2.3, we note that $\mathscr{H}$ is also flasque, so arguing by induction, we may finish the proof. $\Box$
Proof of Fact 2.1. Let $s \in \Gamma(X, \mathscr{H}).$ Since taking the stalk at any point gives rise to an exact sequence, we may find an open cover $X = \bigcup_{i \in I}U_{i}$ with $f_{i} \in \Gamma(X, \mathscr{G})$ such that $f_{i} \mapsto s|_{U_{i}}$ for each $i \in I.$ Now, the pairs $(U_{i}, f_{i})$ form a partial ordering given by the inclusion of open subsets. Every chain has an upper bound, because $\mathscr{G}$ is a sheaf. Using Zorn's lemma, we may choose a maximal element $(U, f)$ among these pairs. In particular, we have $f \mapsto s|_{U}.$
To finish the proof, it is enough to show that $U = X.$ Suppose not. Then since $X = \bigcup_{i \in I}U_{i},$ there must be a nonempty $U_{i} \not\subset U.$ We have $$(f - f_{i})|_{U_{i} \cap U} \mapsto 0 \in \Gamma(U_{i} \cap U, \mathscr{H}),$$ so there must be (unique) $h \in \Gamma(U_{i}, \cap U, \mathscr{F})$ such that $h \mapsto (f - f_{i})|_{U_{i} \cap U}.$ But then since $\mathscr{F}$ is flasque, there is $\tilde{h} \in \Gamma(U_{i}, \mathscr{F})$ such that $\tilde{h}|_{U_{i} \cap U} = h.$ Now, consider $\tilde{h} \mapsto g \in \Gamma(U_{i}, \mathscr{G}).$ Then $f_{i} + g \in \Gamma(U_{i}, \mathscr{G})$ and $f \in \Gamma(U, \mathscr{G})$ coincide on $U_{i} \cap U,$ so we can glue them as a unique section $t \in \Gamma(U_{i} \cup U, \mathscr{G}),$ using the fact that $\mathscr{G}$ is a sheaf. We have $t \mapsto s|_{U_{i} \cup U},$ but then since $U \subsetneq U_{i} \cup U,$ we get a contradiction to the maximality of $(U, f).$ This finishes the proof. $\Box$
Proof of Fact 2.2. The proof is available in this posting. $\Box$
Proof of Fact 3. This immediately follows from Fact 2.2, because we may just take $\mathscr{A}$ to be an injective sheaf.
Proof that Fact 5 implies Fact 4. This is surprisingly simple. See Lemma 2.5 in these notes. $\Box$
It now remains to prove Fact 5.
Proof of Fact 5. We first prove a general lemma we will use.
Lemma. Let $X$ be any topological space. Say we are given a finite union $X = Z_{1} \cup \cdots \cup Z_{r}$ where each $Z_{i} \subset X$ is a closed subset. For any sheaf $\mathscr{F}$ on $X$ valued in $\textbf{Ab},$ the following gives an equalizer diagram (in $\textbf{Ab}$) $$0 \rightarrow \mathscr{F}(X) \rightarrow \prod_{i=1}^{r}\mathscr{F}(Z_{i}) \rightrightarrows \prod_{1 \leq i, j \leq r}\mathscr{F}(Z_{i} \cap Z_{j}),$$ where the map from $\mathscr{F}(X)$ is given by $h \mapsto (h|_{Z_{i}})_{1 \leq i \leq r}$ and the following pair of maps are given by
- $(s_{i})_{i=1}^{r} \mapsto (s_{i}|_{Z_{i} \cap Z_{j}})_{1 \leq i, j \leq r}$ and
- $(s_{i})_{i=1}^{r} \mapsto (s_{j}|_{Z_{i} \cap Z_{j}})_{1 \leq i, j \leq r}.$
Proof. We write $Z_{ij} := Z_{i} \cap Z_{j}.$ Let $(s_{i}) \in \prod_{i=1}^{r}\mathscr{F}(Z_{i})$ with $s_{i}|_{Z_{ij}} = s_{j}|_{Z_{ij}}$ for all $1 \leq i, j \leq r.$ Define $$s : X \rightarrow \bigsqcup_{x \in X}\mathscr{F}_{x}$$ by $s|_{Z_{i}} = s_{i}.$ This is a well-defined set map, and to show $s \in \Gamma(X, \mathscr{F}),$ or equivalently, we need to show that $s$ is continuous. (Note that uniqueness is already determined.) Since $s|_{Z_{i}} = s_{i}$ are continuous and $Z_{1}, \dots, Z_{r}$ form a finite cover of closed subsets of $X,$ it follows that $s$ is continuous. $\Box$
Finally, we prove Fact 5.
Proof of Fact 5. Write $$0 \rightarrow \mathscr{A} \xrightarrow{\phi} \mathscr{B} \xrightarrow{\psi} \mathscr{C} \rightarrow 0$$ to mean the given exact sequence. Let $c \in \Gamma(X, \mathscr{C}).$ Our goal is to show that there is $b \in \Gamma(X, \mathscr{B})$ such that $\psi(b) = c$ on $X.$ Exactness at $\mathscr{C}$ lets us choose an open cover $X = \bigcup_{i \in I}U_{i}$ with sections $b_{i} \in \Gamma(U_{i}, \mathscr{B})$ such that $\psi_{U_{i}}(b_{i}) = c|_{U_{i}}$ for each $i \in I.$ Since $X$ is paracompact, we may assume that $\{U_{i}\}_{i \in I}$ is locally finite. Since paracompactness implies normality, we may find an open cover $X = \bigcup_{i \in I}V_{i}$ such that $\overline{V_{i}} \subset U_{i}$ for each $i \in I.$ (We need closed subsets to utilize the softness of $\mathscr{A}.$)
Given any subset $J \subset I,$ we write $Z_{J} := \bigcup_{j \in J} \overline{V_{j}}.$ Note that the local finiteness of $\{V_{j}\}_{j \in J}$ implies that $$Z_{J} = \bigcup_{j \in J} \overline{V_{j}} = \overline{\bigcup_{j \in J} V_{j}},$$ so $Z_{J} \subset X$ is a closed subset. (See Exercise 13.7 of Tu.)
Consider the set of pairs $(J, b)$ where $J \subset I$ and $b \in \mathscr{B}(Z_{J})$ such that $\psi(b) = c$ on $Z_{J}.$ Such set has a partial order by declaring $(J, b) \leq (J', b')$ to mean $J \subset J'$ and $b'|_{Z_{J}} = b.$ Given any chain $\{(J_{\alpha}, b_{\alpha}) : \alpha \in \mathcal{L}\},$ let $J := \bigcup_{\alpha} J_{\alpha},$ and define $b : Z_{J} \rightarrow \mathscr{B}$ by locally defining it to be $b_{\alpha}$ on $Z_{J_{\alpha}}.$ For any closed subset $E \subset \mathscr{B},$ note that $$b^{-1}(E) \cap Z_{J_{\alpha}} = b_{\alpha}^{-1}(E)$$ is closed because $b_{\alpha}$ is continuous. Note that $\{Z_{J_{\alpha}} : \alpha \in \mathcal{L}\}$ is locally finite, so $$b^{-1}(E) = \bigcup_{\alpha}b_{\alpha}^{-1}(E) = \bigcup_{\alpha}\overline{b_{\alpha}^{-1}(E)} = \overline{\bigcup_{\alpha}b_{\alpha}^{-1}(E)},$$ so $b^{-1}(E) \subset X$ is closed, so $b$ is continuous. This gives us $b \in \mathscr{B}(Z_{J}),$ so we get an upper bound.
Personal remark. Even though I am following Wells (p. 52), I cannot understand the author's argument that we can find an upperbound using the gluing axiom of sheaves for $\mathscr{B}.$ This seems to be an important part of this proof, as now we are about to use Zorn's lemma.
Now, by Zorn's lemma, we may choose a maximal element $(J, b)$ in the partially ordered set. (Note that we reset the notation $(J, b).$) To finish the proof, it is enough to show that $J = I,$ because we then have $Z_{I} = X$ so that $\psi(b) = c$ on $X.$
Suppose not: $J \subsetneq I.$ Then there is $i \in I \setminus J,$ so in particular, we have $$Z_{J} \subsetneq Z_{J} \cup \overline{V_{j}} = Z_{J \cup \{i\}}.$$ Our strategy is to reach a contradiction to the maximality of $(J, b)$ by finding some $b'' \in \mathscr{B}(Z_{J \cup \{i\}})$ such that $\psi(b'') = c$ on $Z_{J \cup \{i\}}.$
We have $\psi(b_{i}) = c = \psi(b)$ on $\overline{V_{i}} \cap Z_{J},$ so$$\psi(b_{i} - b) = 0$$ on $\overline{V_{i}} \cap Z_{J}.$ With some thought, this implies that there is $a' \in \mathscr{A}(\overline{V_{i}} \cap Z_{J})$ such that $$b_{i} - b = \phi(a')$$ on $\overline{V_{i}} \cap Z_{J}.$ Since $\overline{V_{i}} \cap Z_{J} \subset X$ is closed, softness of $\mathscr{A}$ implies that there is $a \in \mathscr{A}(X)$ such that $$a|_{\overline{V_{i}} \cap Z_{J}} = a' \in \mathscr{A}(\overline{V_{i}} \cap Z_{J}).$$ Now, consider $b'_{i} := b_{i} - \phi(a'|_{U_{i}}) \in \mathscr{A}(U_{i}).$ Then $$\psi(b'_{i}) = \psi(b_{i}) - \psi(\phi(a')) = \psi(b_{i})$$ on $U_{i}.$ Moreover, we have $$b'_{i}|_{\overline{V_{i}} \cap Z_{J}} = (b_{i} - \phi(a'))|_{\overline{V_{i}} \cap Z_{J}} = b|_{\overline{V_{i}} \cap Z_{J}}.$$ Hence by Lemma, we may find $b'' \in \mathscr{B}(\overline{V}_{i} \cup Z_{J}) = \mathscr{B}(Z_{J \cup \{i\}})$ such that $b''|_{\overline{V_{i}}} = b'_{i}$ and $b''|_{Z_{J}} = b.$ This implies that $\psi(b'') = c$ on $Z_{J \cup \{i\}} = \overline{V_{i}} \cup Z_{J},$ so we get $(J \cup \{i\}, b'') > (J, b),$ contradicting the maximality of $(J, b).$ This finishes the proof. $\Box$
Finally, we prove Fact 5.
Proof of Fact 5. Write $$0 \rightarrow \mathscr{A} \xrightarrow{\phi} \mathscr{B} \xrightarrow{\psi} \mathscr{C} \rightarrow 0$$ to mean the given exact sequence. Let $c \in \Gamma(X, \mathscr{C}).$ Our goal is to show that there is $b \in \Gamma(X, \mathscr{B})$ such that $\psi(b) = c$ on $X.$ Exactness at $\mathscr{C}$ lets us choose an open cover $X = \bigcup_{i \in I}U_{i}$ with sections $b_{i} \in \Gamma(U_{i}, \mathscr{B})$ such that $\psi_{U_{i}}(b_{i}) = c|_{U_{i}}$ for each $i \in I.$ Since $X$ is paracompact, we may assume that $\{U_{i}\}_{i \in I}$ is locally finite. Since paracompactness implies normality, we may find an open cover $X = \bigcup_{i \in I}V_{i}$ such that $\overline{V_{i}} \subset U_{i}$ for each $i \in I.$ (We need closed subsets to utilize the softness of $\mathscr{A}.$)
Given any subset $J \subset I,$ we write $Z_{J} := \bigcup_{j \in J} \overline{V_{j}}.$ Note that the local finiteness of $\{V_{j}\}_{j \in J}$ implies that $$Z_{J} = \bigcup_{j \in J} \overline{V_{j}} = \overline{\bigcup_{j \in J} V_{j}},$$ so $Z_{J} \subset X$ is a closed subset. (See Exercise 13.7 of Tu.)
Consider the set of pairs $(J, b)$ where $J \subset I$ and $b \in \mathscr{B}(Z_{J})$ such that $\psi(b) = c$ on $Z_{J}.$ Such set has a partial order by declaring $(J, b) \leq (J', b')$ to mean $J \subset J'$ and $b'|_{Z_{J}} = b.$ Given any chain $\{(J_{\alpha}, b_{\alpha}) : \alpha \in \mathcal{L}\},$ let $J := \bigcup_{\alpha} J_{\alpha},$ and define $b : Z_{J} \rightarrow \mathscr{B}$ by locally defining it to be $b_{\alpha}$ on $Z_{J_{\alpha}}.$ For any closed subset $E \subset \mathscr{B},$ note that $$b^{-1}(E) \cap Z_{J_{\alpha}} = b_{\alpha}^{-1}(E)$$ is closed because $b_{\alpha}$ is continuous. Note that $\{Z_{J_{\alpha}} : \alpha \in \mathcal{L}\}$ is locally finite, so $$b^{-1}(E) = \bigcup_{\alpha}b_{\alpha}^{-1}(E) = \bigcup_{\alpha}\overline{b_{\alpha}^{-1}(E)} = \overline{\bigcup_{\alpha}b_{\alpha}^{-1}(E)},$$ so $b^{-1}(E) \subset X$ is closed, so $b$ is continuous. This gives us $b \in \mathscr{B}(Z_{J}),$ so we get an upper bound.
Personal remark. Even though I am following Wells (p. 52), I cannot understand the author's argument that we can find an upperbound using the gluing axiom of sheaves for $\mathscr{B}.$ This seems to be an important part of this proof, as now we are about to use Zorn's lemma.
Now, by Zorn's lemma, we may choose a maximal element $(J, b)$ in the partially ordered set. (Note that we reset the notation $(J, b).$) To finish the proof, it is enough to show that $J = I,$ because we then have $Z_{I} = X$ so that $\psi(b) = c$ on $X.$
Suppose not: $J \subsetneq I.$ Then there is $i \in I \setminus J,$ so in particular, we have $$Z_{J} \subsetneq Z_{J} \cup \overline{V_{j}} = Z_{J \cup \{i\}}.$$ Our strategy is to reach a contradiction to the maximality of $(J, b)$ by finding some $b'' \in \mathscr{B}(Z_{J \cup \{i\}})$ such that $\psi(b'') = c$ on $Z_{J \cup \{i\}}.$
We have $\psi(b_{i}) = c = \psi(b)$ on $\overline{V_{i}} \cap Z_{J},$ so$$\psi(b_{i} - b) = 0$$ on $\overline{V_{i}} \cap Z_{J}.$ With some thought, this implies that there is $a' \in \mathscr{A}(\overline{V_{i}} \cap Z_{J})$ such that $$b_{i} - b = \phi(a')$$ on $\overline{V_{i}} \cap Z_{J}.$ Since $\overline{V_{i}} \cap Z_{J} \subset X$ is closed, softness of $\mathscr{A}$ implies that there is $a \in \mathscr{A}(X)$ such that $$a|_{\overline{V_{i}} \cap Z_{J}} = a' \in \mathscr{A}(\overline{V_{i}} \cap Z_{J}).$$ Now, consider $b'_{i} := b_{i} - \phi(a'|_{U_{i}}) \in \mathscr{A}(U_{i}).$ Then $$\psi(b'_{i}) = \psi(b_{i}) - \psi(\phi(a')) = \psi(b_{i})$$ on $U_{i}.$ Moreover, we have $$b'_{i}|_{\overline{V_{i}} \cap Z_{J}} = (b_{i} - \phi(a'))|_{\overline{V_{i}} \cap Z_{J}} = b|_{\overline{V_{i}} \cap Z_{J}}.$$ Hence by Lemma, we may find $b'' \in \mathscr{B}(\overline{V}_{i} \cup Z_{J}) = \mathscr{B}(Z_{J \cup \{i\}})$ such that $b''|_{\overline{V_{i}}} = b'_{i}$ and $b''|_{Z_{J}} = b.$ This implies that $\psi(b'') = c$ on $Z_{J \cup \{i\}} = \overline{V_{i}} \cup Z_{J},$ so we get $(J \cup \{i\}, b'') > (J, b),$ contradicting the maximality of $(J, b).$ This finishes the proof. $\Box$
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