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 $p$.


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).


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 .

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 !

  • $\begingroup$ yikes - thanks! $\endgroup$
    – 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.)


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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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