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1) Consideremos $\mathbb{C}$ con la topología usual. Definimos el prehaz de las $\textbf{funciones acotadas}$ $\mathcal{F}:\textbf{Top}(X)\to \textbf{Ab}$ de la siguiente manera:

$$\mathcal{F}(U)=\{f:U\to \mathbb{C} \mid f\text{ acotada}\}.$$

Este prehaz no es un haz.

2) Consideremos $\mathbb{C}$ con la topología usual. Definimos el prehaz de las $\textbf{funciones acotadas}$ $\mathcal{F}:\textbf{Top}(X)\to \textbf{Ab}$ de la siguiente manera:

$$\mathcal{F}(U)=\{f:U\to \mathbb{C} \mid f\text{ holomorfa y acotada}\}.$$

Este prehaz no es un haz.

Más ejemplos de prehaces que no sean haces?

Muchas gracias

Translation:

1): Consider $\mathbb{C}$ with the usual topology. Define the presheaf of bounded functions $\mathcal{F}:\textbf{Top}(X)\to \textbf{Ab}$ as follows:

$\mathcal{F}(U)= \lbrace f:U\to \mathbb{C} \mid f \text{ is bounded} \rbrace$

This presheaf is not a sheaf.

2): Consider $\mathbb{C}$ with the usual topology. Define the presheaf of bounded functions $\mathcal{F}:\textbf{Top}(X)\to \textbf{Ab}$ as follows:

$\mathcal{F}(U)= \lbrace f:U\to \mathbb{C} \mid f \text{ is bounded and holomorphic} \rbrace$

This presheaf is not a sheaf.

More examples of presheaves which are not sheaves?

Thanks a lot.

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    $\begingroup$ Your other questions are in English. You're more likely to get help if you post this one that way. Also: when an answer satisfies you, please accept it (the check mark). You can also upvote that answer and any others you like (the up arrow). $\endgroup$ – Ethan Bolker Sep 20 '19 at 0:03
  • $\begingroup$ Why is $\mathcal F$ in (2) not a sheaf? Also, are these functions $U\to\Bbb C$ also continuous in (1)? (I guess, it's needed since the topology of $\Bbb C$ was mentioned.) $\endgroup$ – Berci Sep 20 '19 at 0:26
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    $\begingroup$ @Tomais Translated your question to english such that the majority here is able to read and understand it properly. As your other questions are in english as well, it does not seem like that would be a problem for you. $\endgroup$ – ThorWittich Sep 20 '19 at 1:35
  • $\begingroup$ @ThorWittich why didn't you leave the original text and provide the translation below? $\endgroup$ – user347489 Sep 20 '19 at 1:49
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    $\begingroup$ I don't think that was yours to decide, @ThorWittich : math.meta.stackexchange.com/questions/19292/… I do think it's commendable from you to translate the question, but not at the expense of completely deleting the intended version in Spanish. $\endgroup$ – user347489 Sep 20 '19 at 1:54
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Let me add one way to create some examples:

If $\mathcal{F}$ is a sheaf (of for example abelian groups, rings, etc.), then $\mathcal{F}(\emptyset)$ will be the terminal object in the category where the sheaf takes its values. That means you can find examples of presheaves which are not sheaves by preventing $\mathcal{F}(\emptyset)$ to be the terminal object. A very prominent example for that is given by constant presheaves:

Take for example some non-trivial abelian group $G$, and define a presheaf of abelian groups by $\mathcal{F}(U) = G$ (the restriction maps being the identity $\text{id}_G$)

In general this example not only fails to be a sheaf because of the above reason, but also because gluability tends to fail.

More examples arise in the following setting:

If you have a morphism $\varphi \colon \mathcal{F} \rightarrow \mathcal{G}$ of for example sheaves of abelian groups, then the presheaves $\text{im}(\varphi)$ and $\text{coker}(\varphi)$ defined by $\text{im}(\varphi)(U) = \text{im}(\varphi_U)$ and $\text{coker}(\varphi)(U) = \text{coker}(\varphi_U)$ will not be sheaves in general, where once again gluability tends to be the problem.

You can for example take a look at the topological space $X = \mathbb{C}$ you were considering yourself and the morphism $\text{exp} \colon \mathcal{O}_X \rightarrow \mathcal{O}^{\times}_X,$ where $\mathcal{O}_X$ is the sheaf of holomorphic functions and $\text{exp}$ is given by locally taking the exponential.

More examples arise from other constructions. When you try to mimic a construction known from for example abelian groups, then it will not necessarily respect the sheaf axiom and just be a presheaf at first.

That also shows why we will need to work with sheafification in general.

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  • $\begingroup$ Thank you very much to all. $\endgroup$ – Tomais Sep 20 '19 at 4:28
  • $\begingroup$ ThorWittich: In this link it says that the constant presheaf are sheaf en.wikipedia.org/wiki/Constant_sheaf $\endgroup$ – Tomais Sep 20 '19 at 4:33
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    $\begingroup$ @Tomais No. They are saying that the constant sheaf is the sheafification of the constant presheaf. Do not get confused by the name. Sheafifying does not not change the stalks, but the values on the opens sets will be different if it was not a sheaf before. $\endgroup$ – ThorWittich Sep 20 '19 at 10:24
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    $\begingroup$ The link you posted even gave explicit counterexamples (section: A detailed example). $\endgroup$ – ThorWittich Sep 20 '19 at 10:31

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