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Let be $\mathcal F$ a presheaf on a topological space $X$, the usual definition of the sheaf associated to $\mathcal F$ is the following

$\mathcal F^+(U)=\{\widetilde s: U\rightarrow Et(\mathcal F)\;|\; s\in \mathcal F(U),\;\textrm{and}\;\widetilde s(x)=s_x \}$

where $s_x$ is the germ at $x$ of $s$.

In the book "Algebraic geometry: an introduction - D.Perrin ", I have found another definition of the sheaf associated to a presheaf in the particular case of sheaves of functions.

Let be $\mathcal F$ a presheaf (on $X$) of functions with codomain the set $A$ then

$\mathcal F^+(U)=\{f:U\rightarrow A\;|\;\forall x\in U, \exists V\subseteq U\; \textrm{where $V$ is an open set containing $x$ and}\; \exists g\in\mathcal F(V)\;\textrm{such that}\; f|_V=g\}$

My question is the foillowing: why, in the case of sheaves of function, these definitions are the same? The nature of the elements of $\mathcal F^+(U)$ is different, infact in the first case we have function from $U$ to $Et(\mathcal F)$ instead in the second case the codomain is $A$.

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up vote 3 down vote accepted

If $\mathcal F$ is a presheaf of sets on a topological space $X$, the etale space construction shows that you can interpret $\mathcal F$ as a presheaf of continuous functions into the topological splace $Et(\mathcal F)$ :

If $s \in \mathcal F(U)$, you associate it to the function $\widetilde{s} : x \in U\mapsto s_x \in Et(\mathcal F)$. Write $\widetilde {\mathcal F}(U) = \{\widetilde{s}, s \in \mathcal F(U)\}$. Together with the obvious restriction maps, it is a presheaf on $X$ isomorphic to $\mathcal F$

Now that we have as a presheaf of functions into $Et(\mathcal F)$, we can use Perrin's definition : $\widetilde {\mathcal F}^+(U)$ is the set of functions $U \to Et(\mathcal F)$ that are locally elements of $\widetilde {\mathcal F}(U)$ :

$\widetilde {\mathcal F}^+(U) = \{f : U \to Et(\mathcal F) / \forall x \in U, \exists g \in \widetilde {\mathcal F} (V), g = f|_V \}$ , where $V$ is an open containing $x$. $ = \{ f : U \to Et(\mathcal F) / \forall x \in U, \exists g \in \widetilde {\mathcal F} (V), f_x = g_x \} \\ = \{ f : U \to Et(\mathcal F) / \forall x \in U, \exists s \in \mathcal F (V), f_x = \widetilde{s}_x \} \\ = \{ f : U \to Et(\mathcal F) / \forall x \in U, \exists s \in \mathcal F (V), f(x) = s_x \}$
because the germ of a function into the etale space is determined by its image at that point, and $\widetilde{s}(x) = s_x$.

Thus the two definitions end up with the same sheaf in the end.

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