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In Spanier AT (p.325), there is a notion $\Gamma(\mathcal{U})$ called the module of compatible $\mathcal{U}$ families of $\Gamma$ where $\mathcal{U} = \{U\}$ is a collection of open sets and $\Gamma$ is a presheaf of modules on $X$.

Definition. A compatible $\mathcal{U}$ family of $\Gamma$ is an indexed family $\{\gamma_U\in\Gamma(U)\}_{U\in\mathcal{U}}$ such that $\gamma_U\mid U\cap U' = \gamma_{U'}\mid U\cap U'$ for $U,U'\in\mathcal{U}$.

What exactly $\Gamma(\mathcal{U})$ is? I don't think it's a graded module but something else. Because of the lack of understanding, I can't even understand simple example:

The constant presheaf $G$ defined by a module $G$ is not generally a sheaf [if $U$ is a disconnected open set, $G(U)\not\simeq\hat{G}(U)$].

Could you help?

Given an $R$-module $G$, the constant presheaf $G$ on $X$ assigns to every nonempty open $U\subset X$, the module $G$.

one potato two potato
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A compatible $\mathcal{U}$ family of $\Gamma$ is a pack of local sections that glue together. The set of those families $\Gamma(\mathcal{U})$ inherits a module structure from $\Gamma$ (just add and multiply the sections at each open set). Moreover, applying restrictions on each open set makes it a direct system of modules.

Now the direct limit of this system is obtained as the quotient module of: $$\bigsqcup_{\mathcal{U}}\Gamma(\mathcal{U}) $$ where two families $\{\gamma_U\}\in\Gamma(\mathcal{U})$ and $\{\delta_U\}\in\Gamma(\mathcal{V})$ are equivalent if there is a common refinement of $\mathcal{U}$ and $\mathcal{V}$ on which they restrict to the same sections.

The example you mentioned is important to see what's happening. You start with the presheaf of constant functions with values in $G$, which is not a sheaf because constant functions won't glue to another constant function, if your base space has more than one connected component. This is what taking the direct limit above achieves: it adds those sections that should have glued together the constant functions, i.e. the locally constant functions. Indeed, a locally constant function such as $$f(x)=\begin{cases} 1, x<0\\ 0, x>0 \end{cases} $$ can be seen as the class represented by the family $\{\gamma _{(-\infty,0)}=1$, $\gamma_{(0,\infty)}=0\}$.

For an actual proof that the sheafification/completion of the constant presheaf is the sheaf of locally constant functions, see this or this. There, it is equivalently described as the sheaf of continuous functions to $G$, equipped with the discrete topology. You will, however, have to know at least the notion of stalk, which Spanier doesn't seem to introduce, from what I can tell.

andres1
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  • Thank you for your answer. I can see why $G$ is not a sheaf. But I rather want to see this by showing $G(U)\not\simeq\hat{G}(U)$. I think the LHS is just $G$ and the RHS is $G\oplus G$. Am I correct? – one potato two potato Aug 01 '21 at 17:18
  • The RHS is the module of locally constant functions from $U$ to $G$. Thus, you can see it as a direct sum of $n$ copies of $G$, where $n$ is number of connected components of $U$. So yes, they are not equal for a disconnected $U$. – andres1 Aug 01 '21 at 17:36
  • I'm suddenly confusing. Could you explain why $\hat{G}$ is the direct sum of $n$ copies of $G$? $G(\mathcal{U}) = G$ for all $\mathcal{U}$ (I think this is the part which I'm confusing) so the direct limit is $G$. – one potato two potato Aug 02 '21 at 12:38
  • $G(U) =G$ for an open set $U$ (by definition), while $G(\mathcal{U})$ is the module of $\mathcal{U} $ compatible families of $G$ for the collection of open sets $\mathcal{U}$. – andres1 Aug 02 '21 at 13:47
  • I think I still don't understand the definition of $\Gamma(\mathcal{U})$. $\Gamma(\mathcal{U})$ is just a direct product of $\Gamma(U)$ for $U\in\mathcal{U}$? – one potato two potato Aug 02 '21 at 14:48
  • No, it is simply the module of $\mathcal{U}$ compatible families, as we said. Think about a collection $\mathcal{U}$ formed of two disjoint open sets $U$, $V$. A $\mathcal{U}$ family would be here some constant functions $f:U\rightarrow G$, $g:V\rightarrow G$ that are equal on $U\cap V=\emptyset$. Then you understand how the module of these families is identified with $G\oplus G$? – andres1 Aug 02 '21 at 18:02
  • I think you and I understand the 'constant presheaf' differently. I wrote the definition of constant presheaf on the post. And $\gamma_U$ is an element of $\Gamma(U) = G$ not a function. BTW, for each $U\in\mathcal{U}$, $\gamma_U\in \Gamma(U)$ is defined uniquely? I mean if $\mathcal{U} = U$ the the compatible $\mathcal{U}$ family is just single element ${\gamma_U} ? – one potato two potato Aug 02 '21 at 19:36
  • Ok, so this is what you didn't get. You can think of an element $g\in G$ as the constant function from $U$ to $G$ with value $g$. This is the proper way to think about the constant presheaf, the presheaf of constant functions. Try to reread my answer with this in mind. – andres1 Aug 02 '21 at 20:00
  • Oh, now I can see what you're saying. So $\gamma_U\in\Gamma(U)$ is not really a unique element. There could be $\gamma^1_U,\gamma^2_U,...\in\Gamma(U)$. Will it be more intuitively easier to understand if I interpret $\gamma_U\in\Gamma(U)$ as a constant function $U\to \Gamma(U)$? – one potato two potato Aug 02 '21 at 23:56