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I whould like to know why $ \ \mathcal{C} : U \to \mathcal{C} ( U , \mathbb{R} ) $ is a sheaf ? $ U $ is an open set of $ E $ a $ \mathbb{R} $ - vector space which has a finite dimension. $ \mathcal{C} ( U , \mathbb{R} ) $ contains continous maps over $ U $, with values in $ \mathbb{R} $. Thanks a lot. P.S : sorry about my english language, i'm a foreign men from other country. :-)

Bryan
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    Have you tried checking the sheaf axioms? Where did you get stuck? – Alex Kruckman Jun 11 '13 at 21:30
  • I feel myself enable to establish the latest axiom which define what's a sheaf. More precisely, why is, if for avery open set $ V \subset E $ such that $ V = \displaystyle \bigcup_{i \in I} U_i $ : union of open sets, and $ (s_i ){i \in I} $ sections over $ U_i $ checking : $ {s{i}}{|U{i} \bigcap U_{j}} = {s_{j}}{|U{i} \bigcap U_{j}} $, then $ \exists ! s $ over $ V $ such that $ s_{|U_{i}} = s_i $ ?. So, my problem is to check the existence and the uniqueness. Thanks a lot. – Bryan Jun 11 '13 at 21:42
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    Actually this is one of the motivating examples for the definition of a sheaf. A good(!) introduction to sheaves makes you know that this is a sheaf before giving the precise definition of a sheaf ... – Martin Brandenburg Jun 11 '13 at 21:58
  • Ok ! Thanks to yourself, i'll apply your advices. Thank you. :-) – Bryan Jun 11 '13 at 22:03

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The proof goes mostly straightforward and this (nice functions with restriction of domain) is the "mother" example of all sheaves: Given $V=\bigcup U_i$ and $s_i\colon U_i\to\mathbb R$ with $s_i|_{U_i\cap U_j}=s_j|_{U_i\cap U_j}$ for $i.j\in I$, define $s\colon V\to \mathbb R$ by letting $s(x)=s_i(x)$ where $i\in I$ is arbitrary with $x\in U_i$. Such $i$ exists because $x\in V=\bigcup U_i$. The function is welldefined because if we can pick another $j\in I$ with $x\in U_j$, then $s_i(x)=s_j(x)$. Obviously $s|_{U_i}=s_i$ for all $i\in I$. This makes $s$ continuous for all $x\in U_i$, hence for all $x\in V$, i.e. $s$ is a section over $V$.

This $s$ is unique for if $s'\ne s$ then $s'(x)\ne s(x)$ for some $x$, then for $U_i$ with $x\in U_i$, we have $s'|_{U_i}(x)\ne s|_{U_i}(x)$, hence $s'|_{U_i}\ne s|_{U_i}=s_i$.

Hagen von Eitzen
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  • Great, thanks a lot ! ^-^ It's so easy. but, infortunatly, i'm not enough strong at mathematics. Thank you again for having helped me. Thanks. :-) – Bryan Jun 11 '13 at 21:56
  • @Hagen:+1 for your accuracy. I have a question: is it true that a presheaf whose sections are actual functions is necessarily a sheaf? – Brenin Jun 11 '13 at 22:38
  • @atricolf If, given a presheaf $P$ on a topological space $(X,\ \tau)$, by a section of $P$ you simply mean an element $s\in P(U)$ (where $U$ is some open subset of $X$) the answer is actually negative (in general). For example, take the presheaf $B_{\mathbb{R}^{n}}(-, \mathbb{R})$ which associates to every open subset $U$ of $\mathbb{R}^{n}$ (endowed with the standard euclidean topology) the set of all bounded functions $U\longrightarrow X$. Convince yourself this is not a sheaf :) – Marco Vergura Jun 12 '13 at 07:03