Flasque kernel lifts quotient sections

Let $S$ be the set of pairs $(W,t)$ with $W\subseteq U$ open and $t\in \mathcal{G}(W)$ such that ${\phi }_W(t)=s{\mathrm{\mid }}_W$, ordered by $(W,t)\le (W^{\mathrm{\prime }},t^{\mathrm{\prime }})$ when $W\subseteq W^{\mathrm{\prime }}$ and $t^{\mathrm{\prime }}{\mathrm{\mid }}_W=t$. This is a partial order (reflexivity, antisymmetry by locality of restrictions, transitivity by the restriction identities). It is nonempty: the pair $(\mathrm{\varnothing },0)$ lies in $S$ by [F7], since $\mathcal{G}(\mathrm{\varnothing })$ and $\mathcal{H}(\mathrm{\varnothing })$ are one-element groups and $s{\mathrm{\mid }}_{\mathrm{\varnothing }}=0$.

Every chain in $S$ has an upper bound. Let $\{(W_i,t_i){\}}_{i\in I}$ be a chain, a set-indexed family of elements of $S$, and put $W:={\bigcup }_i{W}_i$. For $i,j\in I$ the chain contains an element $(W_k,t_k)$ with $W_i\cup W_j\subseteq W_k$, so $t_i{\mathrm{\mid }}_{W_i\cap W_j}=t_k{\mathrm{\mid }}_{W_i\cap W_j}=t_j{\mathrm{\mid }}_{W_i\cap W_j}$; the family is therefore compatible and [F4] glues it to a single $t\in \mathcal{G}(W)$ with $t{\mathrm{\mid }}_{W_i}=t_i$ for all $i$. Then ${\phi }_W(t){\mathrm{\mid }}_{W_i}={\phi }_{W_i}(t_i)=s{\mathrm{\mid }}_{W_i}$ for every $i$, so ${\phi }_W(t)=s{\mathrm{\mid }}_W$ by the locality half of the sheaf condition [F4], and $(W,t)\in S$ is an upper bound of the chain. [F3, F4, step 1.1]

By [F5] the Axiom of Choice gives Zorn's lemma, so the nonempty poset $S$ of step 1.1, in which every chain has an upper bound by step 2.1, has a maximal element $(W,t)$. Suppose $W\mathrm{\ne }U$ and choose $x\in U\setminus W$. Since the given sequence is exact at $\mathcal{H}$, the stalk map ${\phi }_x:{\mathcal{G}}_x\to {\mathcal{H}}_x$ is surjective by [F1, F6]; the germ $s_x$ therefore has a preimage, which is represented by some $t_1\in \mathcal{G}(W_1)$ on an open neighbourhood $W_1\subseteq U$ of $x$ with ${\phi }_{W_1}(t_1)=s{\mathrm{\mid }}_{W_1}$ after shrinking $W_1$ if necessary (equality of germs is equality on a smaller neighbourhood, [F6]). Let $O:=W\cap W_1$; then ${\phi }_O(t_1{\mathrm{\mid }}_O-t{\mathrm{\mid }}_O)=s{\mathrm{\mid }}_O-s{\mathrm{\mid }}_O=0$, so $u:=t_1{\mathrm{\mid }}_O-t{\mathrm{\mid }}_O$ lies in $\ker ({\phi }_O)=\mathcal{F}(O)$ by [F2]. Since $\mathcal{F}$ is flasque, $u$ extends to some $\tilde{u}\in \mathcal{F}(W_1)\subseteq \mathcal{G}(W_1)$ [F3]. Then $t_1-\tilde{u}\in \mathcal{G}(W_1)$ still satisfies ${\phi }_{W_1}(t_1-\tilde{u})=s{\mathrm{\mid }}_{W_1}$, and on $O$ one has $(t_1-\tilde{u}){\mathrm{\mid }}_O=t_1{\mathrm{\mid }}_O-u=t{\mathrm{\mid }}_O$. Hence $t$ on $W$ and $t_1-\tilde{u}$ on $W_1$ are compatible and glue by [F4] to a section $t^{\mathrm{\prime }}\in \mathcal{G}(W\cup W_1)$ with $t^{\mathrm{\prime }}{\mathrm{\mid }}_W=t$ and $t^{\mathrm{\prime }}{\mathrm{\mid }}_{W_1}=t_1-\tilde{u}$; since $W\cup W_1⊋W$ and ${\phi }_{W\cup W_1}(t^{\mathrm{\prime }})$ agrees with $s$ on the cover $\{W,W_1\}$, we get $(W\cup W_1,t^{\mathrm{\prime }})>(W,t)$ in $S$, contradicting maximality. [F1, F2, F3, F4, F5, F6, step 2.1]

Therefore $W=U$, and the maximal element provides $t\in \mathcal{G}(U)$ with ${\phi }_U(t)=s{\mathrm{\mid }}_U=s$. As $s\in \mathcal{H}(U)$ was arbitrary, ${\phi }_U$ is surjective; since $U$ was an arbitrary open subset, part 1 holds. [step 3.1, construct]

For part 2 assume in addition that $\mathcal{G}$ is flasque. Let $U\subseteq V$ be open and let $s\in \mathcal{H}(U)$. By step 4.1 there is $t\in \mathcal{G}(U)$ with ${\phi }_U(t)=s$; since $\mathcal{G}$ is flasque, $t$ extends to some $t^{\mathrm{\prime }}\in \mathcal{G}(V)$ with $t^{\mathrm{\prime }}{\mathrm{\mid }}_U=t$ [F3]. Then ${\phi }_V(t^{\mathrm{\prime }})\in \mathcal{H}(V)$ restricts to ${\phi }_U(t^{\mathrm{\prime }}{\mathrm{\mid }}_U)={\phi }_U(t)=s$, because $\phi$ is a morphism of sheaves and hence commutes with restrictions. So every section over $U$ extends to $V$ and $\mathcal{H}$ is flasque [F3]. [F3, step 4.1, given] ∎