Restricting a Soft Sheaf to an Open is again Soft?

Question

Hi everyone! Answered to my satisfaction in the comments - thanks nosr and Jacob Bell! :)

Let $X$ be Hausdorff, locally compact, paracompact. Consider $\mathcal{F}$ a soft sheaf on $X$: as there are a variety of definitions of soft sheaf, let me emphasize my definition (taken from Gelfand-Manin, Methods of Homological Algebra, p.38): $\mathcal{F}$ is soft if for any closed $Z \subset X$, the restriction map $$\mathcal{F}(X) \rightarrow \varinjlim_{U \supset Z} \mathcal{F}(U)$$is a surjection.

For an open embedding $j: U \rightarrow X$, is it true that $j^* \mathcal{F}$ is again soft?

Thanks so much!

What I understand/motivation: I can prove this when $U$ is again paracompact - while (Hausdorff) and (Hausdorff + locally compact) both are inherited by open subsets, my impression was that paracompactness does not always descend, hence my question. Also, to be fair Gelfand-Manin also ask that $X$ be separable - will this affect the answer?

By the way, the motivation for my question is in trying to understand their proof of Verdier duality; on p. 231 they state that if $L$ is soft, then so is $L \otimes j_!j^*\mathbb{Z}$. My approach would be $$L \otimes j_! j^* \mathbb{Z} \simeq j_! j^* L$$ and Gelfand-Manin state in an exercise (which I'm still puzzling out) that $f_!$ between locally compact Hausdorff spaces sends soft sheaves to soft sheaves.

One last soft (hah!) question: if I use the definition of a soft sheaf found here: http://amathew.wordpress.com/2011/06/10/soft-sheaves/ literally everything I'm doing becomes much easier, and the way Gelfand-Manin use the word 'deduce' in Exercise 1(b) on p. 236 makes me think this was in fact the definition they want to have (otherwise the 'if' statement as far as I can tell requires a reasonably involved paracompactness argument - and they don't define a soft sheaf on a not paracompact space, yet don't ask that the spaces in that exercise be paracompact, which adds to the confusion!). Has anyone run into problems with GM's notion of softness before when trying to read this section/does anyone have a reason not to simply supplant their definition with Akhil's?

Answer 1

This is more of a comment than an answer, but it won't fit in the comment section.

Gelfand-Manin's definition of sections over a closed set looks suspicious. In GM, for any subset $Z\subseteq X$, $$\mathcal{F}(Z):=\varinjlim_{V\supseteq Z}\mathcal{F}(V).$$ In general, this is different from $\Gamma(Z, \mathcal{F}\mid_Z)$, where $\mathcal{F}\mid_Z$ is the pullback of $\mathcal{F}$ by along the canonical inclusion $Z\hookrightarrow X$.

The following example brings up the question whether this is what one want to use in the definition of softness. Let $X$ be the four element set $\{x,y,z,w\}$. We give $X$ a structure of topological space by specifying $$X, \{x, z, w\}, \{y, z, w\}, \{z, w\}, \{z\}, \{w\}, \emptyset$$ as open sets.

Given a sheaf $\mathcal{F}$ on $X$, it is uniquely determined by its stalks at the four points. Obviously, $\Gamma(\{z,w\}, \mathcal{F})=\mathcal{F}_z\times \mathcal{F}_w$. Since $\{x,z,w\}$ is the smallest open set containing $x$, $\Gamma(\{x,z,w\}, \mathcal{F})=\mathcal{F}_x$ . We obtain a map $\mathcal{F}_x\to \mathcal{F}_z\times \mathcal{F}_w$ from the restriction $\Gamma(\{x,z,w\}, \mathcal{F})\to \Gamma(\{z,w\}, \mathcal{F})$. Similar statements holds for $y$ as well. We have $$\Gamma(X, \mathcal{F})=\mathcal{F}_x\times_{(\mathcal{F}_z\times \mathcal{F}_w)} \mathcal{F}_y \;.$$ On the other hand, given abelian groups $\mathcal{F}_x, \cdots, \mathcal{F}_w$ with above maps, we have a unique sheaf whose stalks are as specified.

Let $\mathcal{F}$ be the constant sheaf $\mathbb{Z}_X$ on $X$. Then all the stalks are $\mathbb{Z}$, and $\mathcal{F}_x\to \mathcal{F}_z\times \mathcal{F}_w$ is given by the diagonal map $\Delta: n\mapsto (n,n)$. Therefore $\Gamma(X, \mathcal{F})=\mathbb{Z}$.

We note that $\mathcal{F}$ is soft under MG's definition. The nontrivial closed sets are $\{x\}, \{y\}, \{x,y\}, \{x,y, z\}, \{x,y,w\}$. The only open set containing $\{x,y\}$ is $X$. One easily sees that the group of sections over every nontrivial closed set is $\mathbb{Z}$.

Now let $\mathcal{G}$ be the sheaf specified by the following conditions: $$\mathcal{G}_x=\mathcal{G}_y=\mathbb{Z}^2, \quad \mathcal{G}_z=\mathcal{G}_w=\mathbb{Z},$$ and $\mathcal{G}_x\to \mathcal{G}_z\times \mathcal{G}_w$ is the identity map (same for $y$). We have $\Gamma(X, \mathcal{G})=\mathbb{Z}^2$. The restriction maps from $X$ to the 3-elements open sets are identity again.

The map $\Delta: \Gamma(X, \mathcal{F})\to \Gamma(X, \mathcal{G})$ embeds $\mathcal{F}$ into $G$ as a subsheaf. Let $\mathcal{H}$ be the quotient sheaf. We have $$ \mathcal{H}_x=\mathcal{H}_y=\mathbb{Z}, \quad \mathcal{H}_z=\mathcal{H}_w=0.$$ So $\Gamma(X, \mathcal{H})=\mathbb{Z}^2$, this means that the sequence $$0\to \Gamma(X, \mathcal{F}) \to \Gamma(X, \mathcal{G}) \to \Gamma(X, \mathcal{H}) $$ is not right exact, which contradicts the desired property of soft sheaves (EX I.5.2c of Gelfand-Mannin).

If we take $\mathcal{F}(Z):=\Gamma(Z, \mathcal{F}\mid_Z)$ in the definition of soft sheaves, then the constant sheaf above is not soft. Indeed, the closed set $Z=\{x,y\}$ in the induced topology is discrete, so $\Gamma(Z, \mathcal{F})=\mathbb{Z}^2$ but we have $\Gamma(X, \mathcal{F})=\mathbb{Z}$.

Of course, $X$ is not Hausdorff in this example.