Proposition 6

Let B be a topological space, $\mathscr{F}$ a sheaf on B and $(A_i)_{i\in I}$ a locally finite closed covering of B. In order that the sheaf $\mathscr{F}$ be soft, it is necessary and sufficient that, for every $i\in I$, the induced sheaf $\mathscr{F}_{A_i}$ be soft.

The condition is obviously necessary. Let us prove that it is sufficient. Let us denote by $p: E\rightarrow B$ the étale B-space $E_{\mathscr{F}}$ associated with the sheaf $\mathscr{F}$. Let A be a closed subspace of B and $s: A\rightarrow E$ a continuous section of $p$ over A; it is a question of proving that $s$ possesses a continuous extension to B. For every subset J of I, put $A_J=\bigcup_{i\in J}A_i$; the set $A_J$ is closed in B (TG, I, p. 6, prop. 4).

Let $\mathscr{S}$ be the set of pairs $(J, t)$ where J is a subset of I and $t$ a continuous section of E over $A_J$ which coincides with $s$ in $A\cap A_J$. Let us make $\mathscr{S}$ into an ordered set by the ordering relation denoted by $\leqslant$ for which $(J, t)\leqslant (J', t')$ if $J\subset J'$ and $t'|A_J=t$. For $\sigma = (J, t)\in \mathscr{S}$, we write $J_{\sigma}= J$ and $t_{\sigma}=t$. Let us show that the ordered set $\mathscr{S}$ is inductive. Let S be a totally ordered subset of $\mathscr{S}$. Put $J =\bigcup_{\sigma\in S}J_{\sigma}$; this is a subset of I. One then defines a section $t$ of E over $A_J$ by setting $t(x) =t_{\sigma}(x)$, if $x\in A_{J_{\sigma}}$; thus $t|A\cap A_J=s$. Let $j\in J$ and let $\sigma \in S$ be such that $j\in J_{\sigma}$; since $t|A_j=t_{\sigma}|A_j$, the restriction of $t$ to $A_j$ is continuous. It then follows from TG, I, p. 19, Prop. 4 that $t$ is continuous. Thus, $(J, t)$ is an element of $\mathscr{S}$; by construction, it is an upper bound of S. This proves that the set $\mathscr{S}$ is inductive. It therefore possesses a maximal element $(J, t)$ (E, III, p. 20, th. 2).

Let us reason by contradiction, supposing that J $= I\not$ . Let $i$ be an element of I-J. Put $A'= (A_i\cap A)\cup (A_i\cap A_J)$ and define a section $s'$ of E over $A'$ by:

$'s(a)$ for $a\in A_i\cap A$,

$t(a)$ for $a\in A_i\cap A_J$,

which is possible since $s$ and $t$ coincide on $A\cap A_J$. Moreover, since $A_i\cap A$ and $A_i\cap A_J$ are closed, the section $s'$ is continuous (TG, I, p. 19, prop. 4). By hypothesis, there exists a continuous section $s_i: A_i\rightarrow E$ extending $s'$. Since the restrictions of $s_i$ and $t$ to $A_J\cap A_i$ are equal, the continuous section $t': A_{J\cup \{i\}}\rightarrow E$ which coincides with $t$ on $A_J$ and with $s_i$ on $A_i$ is a continuous section of $p$ over $A_{J\cup \{i\}}$, extending $s|A\cap A_{J\cup \{i\}}$. We then have $(J, t)<(J\cup \{i\}, t')$, which contradicts the hypothesis that $(J, t)$ is maximal.

Thus, J = I, hence $A_J= B$ and $t$ is a continuous section of E over B extending $s$.