# The “Hartshornian” sheafification of a sheaf

Given a sheaf $\mathscr F$ on $X$, how does one show that its sheafification (in the sense of Hartshorne) is isomorphic to it? The most obvious thing to do is a universal property argument: since $\mathscr F$ is already a sheaf, to the identity $id_{\mathscr F}:\mathscr{F\to F}$ there corresponds a unique morphism of presheaves (and thus sheaves) $\psi:\mathscr{F^+\to F}$ such that $$\psi\circ\theta=id_{\mathscr F}\tag{1}$$ where $\theta:\mathscr{F\to F^+}$ is canonical. So $\psi$ has a left-inverse. In the category of sheaves do I need to do extra work to show that $\psi$ is right-invertible, or do I get this for free (that is, without referencing the explicit construction) from $(1)$?

$\theta \circ \psi\colon \mathscr{F}^+ \to \mathscr{F}^+$ is a morphism such that $(\theta \circ \psi) \circ \theta = \theta \circ (\psi \circ \theta) = \theta \circ \operatorname{id}_\mathscr{F} = \theta$. $\operatorname{id}_{\mathscr{F}'}$ is another such morphism and you can apply uniqueness.
• Thanks. So the conclusion is that $f\circ g=id$ implies $g\circ f=id$ in any category? – theage Aug 5 '15 at 19:06