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According to Hartshorne's book a sheafification is:

I saw a different definition elsewhere

Definition 2

Given a presheaf $\mathcal F$ in a topological space $X$ we can associate a $\mathcal F$ a sheaf called $\mathcal F^+$:

$\mathcal F^+(U):=\{t\in D_{\mathcal F}(U);t_x\in Im((\varphi_{\mathcal F})_x),\forall x \in U\}$.

What I'm trying to understand is why saying the sheaf he constructs with (1) and (2) in the Hartshorne's proof is equivalent to the sheaf of the definition 2.

My attempt to solve this question

$\varphi_{\mathcal F}$ is defined as $\varphi_{\mathcal F}:\mathcal F\to D_{\mathcal F}$ and when we apply an open subset $U$, we have $\varphi_{\mathcal F}(U):\mathcal F(U)\to D_{\mathcal F}(U)$, where $D_{\mathcal F}(U)=\prod_{x\in U} \mathcal F_x$ and we know that $t_x\in Im((\varphi_{\mathcal F})_x),\forall x \in U\Leftrightarrow t\in Im((\varphi_{\mathcal F}))$

Since the image of $\varphi_{\mathcal F}$ is an element of $\prod_{x\in U} \mathcal F_x$ the part (1) is ok, since these functions $s$ can be regarded as elements of $\prod_{x\in U} \mathcal F_x$ because of (1). (see definition of products as functions).

I'm having troubles to prove that the sheaf of the definition 2 has as a property the part 2 in the Hartshorne's proof and vice-versa.

Thanks

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3  
Again this just comes from the definition of stalks ... are you really sure you want and can learn algebraic geometry this way when you even have to ask so many question about the absolute basics which answer themselves by looking at the definitions? Perhaps wait a year or so? –  Martin Brandenburg Sep 20 '13 at 9:07
    
@MartinBrandenburg Thank you for your concerns, but I don't have choice. The worse part is that I have to study this in the same time with basic algebraic geometry without any previous course of commutative algebra. –  user42912 Sep 20 '13 at 9:32
    
@MartinBrandenburg However I think it will be ok, this has already happened with me before in some particular courses, the definitions and theorems are painful until I get used with the subject. –  user42912 Sep 20 '13 at 9:36
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@user42912 I really suggest that you not look at Hartshorne to study this material. If you don't have any background in commutative algebra, you shouldn't even be studying algebraic geometry. –  user38268 Sep 20 '13 at 15:23
    
@user38268 and Martin I will talk with my advisor about that, he said to me to study this without any commutative algebra background, that's why I said I didn't have choice. Thank you two for your concerns. Any way, what I will suggest to my advisor is study just the basic definitions and theorems of scheme theory and go to commutative algebra and basic algebraic geometry, what do you think? Is it a good strategy? –  user42912 Sep 21 '13 at 10:37

1 Answer 1

up vote 1 down vote accepted

Hartshorne's (2) corresponds to your $t_x \in \def\Im{\mathop{\rm Im}}\Im(\phi_{\mathcal F})_x$, all $x \in U$. To see this, let $x \in U$. Then $t_x \in \Im(\phi_{\mathcal F})_x$, that is, there is an open $V \ni x$ and a $s \in \mathcal F(V)$ such that $t_x = s_x \in \mathcal F_x$. This gives us - by definition of the stalk - an open $W \subseteq U \cap V$ containg $x%$ with $t|_W = s|_W$. So $s_y = t_y$ for all $y \in W$ and (2) is fulfilled.

On the other hand, suppose (2) holds, then given $x \in U$ and $t \in D_{\mathcal F}(U)$, choose $s \in \mathcal F(V)$ with $x\in V \subseteq U$ and $t|_V = s|_V$. Then $t_x = s_x \in \Im(\phi_{\mathcal F})_x$.

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Why $t_x \in \Im(\phi_{\mathcal F})_x$ implies that there is an open $V \ni x$ and a $s \in \mathcal F(V)$ such that $t_x = s_x \in \mathcal F_x$? –  user42912 Sep 20 '13 at 12:20
    
By definition, ${\rm Im}(\phi_{\mathcal F})_x$ is the direct limit of the $\phi_{\mathcal F}(V)\bigl(\mathcal F(V)\bigr)$ as $V \to x$, so each element is represented by an $\phi_{\mathcal F}(V)(s)$ with $s \in \mathcal F(V)$. –  martini Sep 20 '13 at 13:16

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