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Let $\mathcal F$ be the presheaf image of some morphism of sheaves $\varphi:\mathcal G'\to\mathcal G$. Consider the sheafification morphism $\theta:\mathcal F\to\mathcal F^+=\text{im }\varphi$. Is $\theta(U)$ injective for any $U$? (using Hartshorne notation)

It seems so to me: $\ker\theta(U)$ consists of elements $s\in\mathcal F(U)$ that are zero at every stalk (if I understand the construction of the sheafification properly) and the image sheaf $\mathcal F^+=\text{im }\varphi$ can be identified with a subsheaf of the target sheaf $\mathcal G$ (Hartshorne Ex. II.1.4(b)). Then $s$, viewed as an element of $\mathcal G(U)$, is zero on stalks, so $s\equiv 0$ as a section of $\mathcal G$. Finally, $s=\varphi(U)(t)$ for some section $t\in\mathcal G'(U)$, so that should imply $t\in\ker\varphi(U)$, and hence $s=0$ as an element of $\mathcal F(U)=\text{im}\big(\varphi(U)\big)$.

Is this argument correct? If yes, is there a more direct argument?

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    $\begingroup$ Yes what you wrote is correct. More generally the sheafification morphism of any subpresheaf of a sheaf is injective. $\endgroup$ – Georges Elencwajg Dec 30 '14 at 17:09
  • $\begingroup$ You are welcome, aytio. $\endgroup$ – Georges Elencwajg Dec 30 '14 at 19:07

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