This is a sillier question but I'm working through Vakil's notes, exercise 2.4L and I'm having trouble seeing why sheafification is left adjoint to the forgetful functor.

I have done the exercises up till now and see why sheafification as constructed satisfies the universal property. But I don't see how the universal property gives us the adjointness a-priori.

As far as I understand, for the two functors to be an adjoint pair, we need to have a bijection

$\phi : Mor(\mathcal{F}^{sh}, \mathcal{G}) \rightarrow Mor(\mathcal{F}, \mathcal{G}^{pre})$.

Where $\mathcal{G}$ is a sheaf and $\mathcal{F}$ is a presheaf. Now I know that the universal property allows us to construct a map of sheaves $\mathcal{F}^{sh} \rightarrow \mathcal{G}$ from a map of presheaves $\mathcal{F}^{pre} \rightarrow \mathcal{G}$. So this is the backwards map in the bijection. But what is the forward map? Is every map from the sheafification of $\mathcal{F}$ to $\mathcal{G}$ an induced map of the presheaves obtained from the universal property? Surely we don't just apply the forgetful functor to $\mathcal{F}^{sh}$ and $\mathcal{G}$ and take the same map, since sheafification and forgetful functor aren't inverses.

What's the exact map from the left side to the right side?


1 Answer 1


$\DeclareMathOperator{\sh}{sh}\DeclareMathOperator{\pre}{pre}$Fix a presheaf $\cal F$, and a sheaf $\cal G$. The mantra about sheafification is: "a morphism of sheaves $\cal F^{\sh} \to \cal G$ is the same as a morphism of presheaves $\cal F \to \cal G$".
(In your notation, the latter is written as $\cal F \to \cal G^{\pre}$.)

From what you've written, you seem to understand how a presheaf morphism $\cal F \to \cal G$ gives a sheaf morphism $\cal F^{\sh} \to \cal G$.

The other direction is what you are confused about. But that is actually even simpler if you keep track of the details.
Indeed, recall that the sheafification is not just a sheaf $\cal F^{\sh}$ but also a presheaf morphism $\cal F \xrightarrow{\sh} \cal F^{\sh}$. Thus, given a morphism of sheaves $\cal F^{\sh} \to \cal G$, it is also a morphism of presheaves and composing as $$\cal F \xrightarrow{\sh} \cal F^{\sh} \to \cal G$$ gives us the presheaf morphism that we wanted.

Can you now verify that this satisfies the naturality? (The fact that this correspondence is one-to-one should be more or less clear.)

  • 2
    $\begingroup$ Ah I see! I forgot that we're assuming the sheafification comes with a morphism in the definition. Thanks for pointing that out! $\endgroup$
    – User20354
    Jul 13, 2021 at 19:54

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