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

  • $\begingroup$ I think you are right, but the last verification on stalks may be troublesome. I find it easier to deal with sections instead. $\endgroup$ Commented Sep 3, 2019 at 17:20
  • 1
    $\begingroup$ 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. $\endgroup$
    – johnnycrab
    Commented Sep 3, 2019 at 19:01


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