5
$\begingroup$

Let $\mathcal{F}$ be a presheaf on some topological space $X$. It is not hard to prove directly that the map $\mathcal{F}\rightarrow \mathcal{F}^{sh}$ induces an isomorphism of stalks (Here $\mathcal{F}^{sh}$ is the sheafification of $\mathcal{F}$).

I want to improve on using adjoints to simplify arguments, so I was hoping someone has a proof of this using adjoints. Perhaps we can use the fact that sheafification is left-adjoint to the forgetful functor, or that the inverse image functor is left adjoint to the push forward functor?

$\endgroup$

1 Answer 1

6
$\begingroup$

This is a consequence of two facts:

  • Adjoints are unique up to unique isomorphism.
  • The adjoint of a composite is the composite of the adjoints.

The key observation is that the functor that sends a presheaf to its stalk has the "same" right adjoint as the functor that sends a sheaf to its stalk, namely the skyscraper (pre)sheaf functor.

$\endgroup$
1
  • $\begingroup$ Where do we need the fact that "The adjoint of a composite is the composite of the adjoints."? $\endgroup$
    – No One
    Commented Dec 14, 2017 at 23:25

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .