3 deleted 39 characters in body

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. The paracompactness of $X$ is trivial.

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.

2 added 30 characters in body

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 right 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. The paracompactness of $X$ is trivial.

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.

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 the right definition. 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. The paracompactness of $X$ is trivial.

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.