1
$\begingroup$

Let $X = \mathbb{R^n}$. For every open subset $U \subseteq X$ , consider a presheaf $ , \Gamma(U) := \{ f: U \rightarrow \mathbb{R} : f $ is differentiable$ \}$. I need to show that this presheaf is a sheaf.

Now since the concept of differentiability is local in nature, so for any open subset, say $V$ of $U$ where $U$ is an open subspace of $X$, the set $ \Gamma(V)$ should give rise to a sheaf. This is more of an intuitive explanation. I am unable to figure out how to mathematically prove the conditions that are required for a presheaf to become a sheaf. Any help is appreciated!

PS- I am not familiar with category theory so I wouldn't be able to get any answers/comments based on that.

$\endgroup$
1
$\begingroup$

If ya buy that it's a presheaf already, then ya just need to show it satisfies

1) Local determination 2) Gluing Lemma

So for 1, suppose we have a bunch of open sets $U_{\alpha}$ in $\mathbb{R}^n$ that cover an open set $U$ and we have two functions $f,g \in \Gamma(U)$ that, when restricted to each of these $U_{\alpha}$, agree with each other. We must show that f = g identically on $U$. But this is simple, because two functions are the same if they agree pointwise. So ya just needs to show that f(x) = g(x) for every x in $U$.

For 2, suppose we have an open cover $U_{\alpha}$ of a set $U$ and we have functions $f_{\alpha} \in \Gamma(U_\alpha)$ such that for any $\alpha$ and $\beta$ the restrictions of $f_\alpha$ and $f_\beta$ agree when mapped into $\Gamma(U_\alpha \cap U_\beta)$, (i.e., when restricted to their intersection). Under these conditions, we want to produce an f in $\Gamma(U)$ that agrees with the $f_\alpha$ when restricted. To do this, for $ x \in U$, define f(x) by $f(x) = f_{\alpha}(x) $, where $U_\alpha$ is any open set in our collection that contains $x$. This is possible because it forms an open covering. A little work needs to be done to show this is a well-defined, smooth function though.

$\endgroup$
2
$\begingroup$

You need to show the two missing properties: locality and gluing.

Locality is trivial for any presheaf where sections are functions and restriction is the usual restriction of functions: Assume $U=\bigcup_{i\in I} U_i$ and we have $s,t\in \Gamma(U)$ such that $\operatorname{res}^U_{U_i}(s)=\operatorname{res}^U_{U_i}(t)$. We want to show $s=t$. Since $s,t$ are functions with domain $U$, we need to show that $s(u)=t(u)$ holds for all $u\in U$. OK, so let $u\in U$. Then $u\in U_i$ for some $U_i$ (it's a covering after all). Then $s(u)=s|_{U_i}(u)=\operatorname{res}^U_{U_i}(s)(u)=\operatorname{res}^U_{U_i}(t)(u)=t|_{U_i}(u)=t(u)$, as desired.

It is also clear what happens with gluing: Mere functions can always be glued. Given $U=\bigcup_{i\in I} U_i$ and sections $s_i\in\Gamma(U_i)$ such that $\operatorname{res}^{U_i}_{U_i\cap U_j}(s_i)=\operatorname{res}^{U_j}_{U_i\cap U_j}(s_j)$ always holds, we can define a function $s\colon U\to \Bbb R$ by letting $s(u)=s_i(u)$ where $i\in I$ is arbitrary with $u\in U_i$. This is always possible because the $U_i$ form a cover, and this is well-defined precisely because of the given condition for restrictions. The problem is just that it is not sufficient to merely have a function $s\colon U\to \Bbb R$, but rather a differentiable function. But as you correctly notice, differentiability is a local property. The function $s$ is differentiable iff it is differentiable at every point $u\in U$, and in order to prove that (and even compute the derivative) it suffices to "know" $s$ only in an open neighbourhood of $u$ -- such as an $U_i$ with $u\in U_i$.

So indeed, for any spaces $X,Y$ and any local property $\Phi$ the presheaf given by $\Gamma(U)=\{\,f\colon U\to Y\mid f\text{ has property }\Phi\,\}$, together with $\operatorname{res}^U_V(s)=s|_V$ for $V\subseteq U$ and $s\in \Gamma(U)$, is always a sheaf in the above somewhat obvious way.

The "fun with sheaves" really begins when $\Gamma(U)$ is not actually a set of functions ...

$\endgroup$

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.