Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

There is such statement:

Presheaf $\mathfrak F$ is a sheaf if it commutes with limits: $$\mathfrak F(\lim_{\to} U_\alpha)=\lim_{\gets}\mathfrak F(U_\alpha).$$

Can you rewrite this formula more exactly using following definitions.

Let $\Phi : \mathfrak J \to \mathfrak C$ is a functor between two categories.

Definition 1. Direct limit of $\Phi$ is an object $X\in\mathfrak C$ which represent functor $${\mathfrak C}^0\longrightarrow Sets$$ $$Y\longmapsto\hom_{Funct(\mathfrak J,\mathfrak C)}(\Delta Y, \Phi),$$ where $\Delta : \mathfrak C\to Funct(\mathfrak J,\mathfrak C)$ be a diagonal functor.

Definition 2. Analogically inverse limit corepresent functor $$\mathfrak C\longrightarrow Sets$$ $$Y\longmapsto\hom_{Funct(\mathfrak J,\mathfrak C)}(\Phi,\Delta Y).$$

Definition 3. At last presheaf on space $M$ is a functor from category of open subsets of $M$.

share|cite|improve this question

The statement is false. But it is correct that a sheaf is a presheaf which preserves a certain class of colimits. See also my answer here.

share|cite|improve this answer
It's quite irritating that this myth keeps showing up. I wonder where it comes from? – Zhen Lin Jan 9 '14 at 17:35

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.