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Let $\mathcal{F}$ be a pre-sheaf on $X$. It seems that if we let $\mathcal{F}^+(U)$ to be the set of all maps $U \rightarrow \cup_{p \in U} \mathcal{F}_p$, where $\mathcal{F}_p$ is the stalk of $\mathcal{F}$ at p with the single requirement that $s(p) \in \mathcal{F}_p$, then we have a sheaf on $X$. Where exactly does the second requirement in the sheafification comes into play?

Edited: The second requirement is that any map $s:U\rightarrow \cup_{p \in U} \mathcal{F}_p$ satisfies the following property: for any point $p \in U$ there exists a neighborhood $V \subseteq U$ of $p$ and $t \in \mathcal{F}(V)$ such that for any $q \in V$ we have $t_q=s(q)$, where $t_q$ is the germ of $t$ at $p$.

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4 Answers 4

up vote 5 down vote accepted

Indeed $\mathcal{F}^+$ is a sheaf on $X$.
However it has absurdly large sets of sections: since there is no constraint on the choice of $s(p)$ when $p$ varies, essentially all links with $\mathcal F$ are dissolved.

It has however one redeeming feature: it serves as a huge container for the correct sheafification $\mathcal F^{sh}$ of $\mathcal F$.
Indeed by definition $\mathcal F^{sh}(U)\subset \mathcal F^+(U)$ consists of those $s\in \mathcal{F}^+(U)$ such that locally near each $p_0\in U$ there exists $t\in \mathcal F(V)$ ($p_0\in V\subset U$) with $s(p)=t_p$ for all $p\in V$.
The second axiom for sheaves is then automatically satisfied for $\mathcal F^{sh}$.
Moreover the construction immediately yields a morphism of presheaves $\mathcal F\to \mathcal F^{sh}$ which is injective iff the first axiom for sheaves is verified for the presheaf $\mathcal F$.

To sum up visually, we have canonical morphisms of presheaves : $$\mathcal{F}\to \mathcal F^{sh} \hookrightarrow \mathcal{F}^+$$ (and the last two presheaves are sheaves).

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A very optional complement
For the sake of completeness and for the record, I'd like to analyze the nature of the functor (from presheaves to sheaves) $\mathcal F\to \mathcal F^+$ on an example and show that it is subtler than one might think.

Let $X$ be a topological space, $\mathcal C$ be the sheaf of real-valued continuous functions on $X$ and $\mathcal Disc $ the sheaf of all real-valued functions, maybe discontinuous: $\mathcal Disc (U)=\mathbb R^U$.
We have a morphism of sheaves $\mathcal C^+\to \mathcal Disc$ given by
$$\mathcal C^+(U)\to \mathcal Disc ( U): s\mapsto \tilde s \text { where} \;\tilde s(x)=(s(x))(x)$$

The strange but logical notation $(s(x))(x)$ means that $s(x)=f_x\in \mathcal C_x$ is the germ at $x$ of some continuous function $f$ defined near $x$ and that you then take the value $f(x)=f_x(x)$ of that function at $x$ and obtain the real number $\tilde s(x)$.

It is then true that for every $U\subset X$ the map $\mathcal C^+(U)\to \mathcal Disc ( U)$ is surjective [take germs of constant functions in the domain], so that a fortiori the morphism of sheaves $\mathcal C^+\to \mathcal Disc $ is surjective.
But the morphism of sheaves $\mathcal C^+\to \mathcal Disc $ is not injective, because the morphism $\mathcal C(X)\to \mathcal Disc ( X)$ (for example) is not injective.
Indeed choose $x_0\in X$ and take for $s\in \mathcal C^+(X)$ the collection of germs $s(x)=0_x$ for $x\neq x_0$ and $s(x_0)=g_{x_0}$ where $g$ is a non-zero continuous function (defined near $x_0$) satisfying $g(x_0)=0$.
Then $\tilde s=0\in \mathcal Disc ( X)$ although $s\neq 0\in \mathcal C^+(X)$

If you start with the sheaf $\mathcal C_b$ of locally bounded continuous functions on $X$, you will find $\mathcal C_b^+=\mathcal C^+$ and the same analysis applies .

Conclusion
You may forget all the details above and just remember that $\mathcal C^+$ is not the sheaf $\mathcal Disc$ of arbitrary [and possibly very discontinuous] functions on $X$, but an even much more horrible sheaf !

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yikes - thanks! –  user29743 Sep 26 '12 at 15:04

The requirement ensures that the sections of $\mathcal F^{sh}$ are locally induced by sections of the presheaf $\mathcal F.$ Otherwise, we would get arbitrary functions $U\to \cup_{p\in U}\mathcal F_p$, which could bear no relation to the original presheaf, which we want to sheafify. (We want the "smallest" sheaf that glues sections of the presheaf to form a sheaf.)

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It often helps for intuition with sheaves and presheaves to think about the case where the sheaf or presheaf in question is a sub-presheaf of some sheaf of functions on $X$.

For example, my favorite presheaf that's not a sheaf is the presheaf of "bounded continuous real-valued functions." This isn't a sheaf because being locally bounded does not imply that a function is globally bounded. Its correct sheafification is clearly the sheaf of continuous real-valued function (this is the smallest subsheaf of the sheaf of all not-necessarily-continuous functions which contains our presheaf); your construction would instead give the sheaf of all (not necessarily continuous) functions.

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Dear countinghaus: your last sentence is (in a very subtle way) not quite correct and thinking about it motivated my second answer. –  Georges Elencwajg Sep 25 '12 at 21:29

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