# Sheafification of presheaf with support is equal to sheafification with support?

Let $$X$$ be a topological space, $$Z \subseteq X$$ a closed subset and $$\mathcal F$$ a presheaf on $$X$$ with sheafification $$\mathcal F^+$$. We can define a subpresheaf $$\mathcal H^0_Z(\mathcal F^+)$$ of $$\mathcal F^+$$ by sending an open $$V \subseteq X$$ to the subgroup of $$\mathcal F^+(V)$$ consisting of all sections whose support is contained in $$Z$$. As known, this presheaf is a sheaf and is called the subsheaf of $$\mathcal F$$ with supports in $$Z$$.

Apparently we can do the same with the presheaf $$\mathcal F$$ (take sections with support in $$Z$$), and obtain a presheaf which I ambigously also denote by $$\mathcal H^0_Z(\mathcal F)$$. We can sheafify this presheaf, and I am wondering if the two sheaves are the same, i.e. if

$$\mathcal H^0_Z(\mathcal F)^+ = \mathcal H^0_Z(\mathcal F^+).$$

Is this true? I think it should be, because the canonical morphism $$\mathcal F \to \mathcal F^+$$ 'preserves supports in $$Z$$', and hence we get a morphism $$\mathcal H^0_Z(\mathcal F)^+ \to \mathcal H^0_Z(\mathcal F^+)$$ which should be an isomorphism, as the stalks should be equal.

Is this correct? Thank you very much in advance!

EDIT: Based on withoutfeather's comment, an idea would be to use the explicit representation of a section of the sheafification by a special family of germs and show that any element in $$\mathcal H^0_Z(\mathcal F^+)$$ can be represented in $$H^0_Z(\mathcal F)^+$$. I.e. Let $$U$$ be open and $$s \in \mathcal H^0_Z(\mathcal F^+)$$ where $$s = (s_x)_{x \in U} \in \prod_{x \in U}\mathcal F_x$$ (where the family has the special sheafification property) a section of $$\mathcal F^+$$ with support in $$Z$$. For any $$x \in U$$ there exists an open $$W \subseteq U$$ around $$x$$ and $$c \in \mathcal F(W)$$ such that $$c_y = s_y \ \forall y \in W$$. As $$s$$ has support in $$Z$$ and the restriction maps are given by projection, we have $$c_y = 0$$ for all $$y \notin Z$$. Hence $$c \in \mathcal H^0_Z(\mathcal F)(W)$$ which shows that we can represent each $$s_x$$ by a section of $$\mathcal H^0_Z(\mathcal F)$$, thus $$\mathcal H^0_Z(\mathcal F)^+(U)$$ agrees with $$\mathcal H^0_Z(\mathcal F^+)(U)$$.

• I think you are right, but the last verification on stalks may be troublesome. I find it easier to deal with sections instead. Sep 3, 2019 at 17:20
• Dear withoutfeather, thank you very much. I have edited my question with an idea based on your comment (as it is too long for the comment section), if you want to have a look. Sep 3, 2019 at 19:01