What is the meaning of "open set" in the context of sheaves?
问题内容
I am reading up on some algebraic geometry and came across the above definition of sheaves. Some things confuse me.
- When the author says in 2) of Definition 4.1, "For each inclusion of open sets $V \subset U$..", does he mean that $V$ is an open set in $X$ or in the induced topology on $U$?
- In Defintion 4.2, is the open covering $\{ U_\alpha \}_{\alpha \in I}$ considered with $U_\alpha$ open in $X$ or $U_\alpha = U \cap X_\alpha$ for some open $X_\alpha$ in $X$?
In the beginning, I thought we always talked about the induced topology, both in the open covering in Definition 4.2 and the inclusion in definition 4.1. However, it struck me that $\mathscr{F}(U)$ is by definition an abelian group when $U$ is open in $X$, so in that case the function $\rho_{U,V}: \mathscr{F}(U) \to \mathscr{F}(V)$ wouldnt make sense (as $\mathscr{F}(V)$ is not defined).
Can someone help me clear up my confusion?
回答 (2)
When $U$ is open in $X$, we have that $V \subseteq U$ is open in the induced topology if and only if $V \subseteq X$ is open.
In particular, since for the foundational definitions of sheaves we're only ever looking at open subsets $U$, we see that the ambiguity you're worried about does not exist -- $V$ open in $X$ and $V$ open in the induced topology on $U$ agree! Of course, the same thing is true of your family $\{U_\alpha\}$ each of which is open in $X$ if and only if they're open in the induced topology on $U$.
It's a nice exercise in point set topology to prove this to yourself, but I'll include a proof under the break:
First, say that $V \subseteq U$ is open in the induced topology. That means $V = U \cap W$ for some $W$ open in $X$. But now $V$ is an intersection of two opens in $X$, thus is itself open in $X$. Next, say that $V$ is open in $X$. Then $V = U \cap V$ is the intersection of $U$ with an open of $X$, and thus is open in the induced topology on $U$.
That said, I think it's psychologically easier to only work with the opens of $X$. So when I read "$V \subseteq U$" I think of them both as being open subsets of $X$. Similarly, when I read "$\{U_\alpha\}$ covers $U$", I think of that as being a statement about open subsets of $X$. This is useful for a more categorical perspective on sheaves as "functors from the lattice of open subsets of $X$".
I hope this helps ^_^
- Given an open set $U$ of a topological space $X$, a subset $V\subseteq U$ is open relative to $U$ if and only if it is open relative to $X$. So there is no ambiguity, although as Chris Grossack argues, it might be more natural to think of "$V$ is open" as meaning that $V$ is open in $X$.
- Here, an "open cover" of $U$ is a family $\{U_{\alpha}\}_{\alpha\in\mathcal A}$ of subsets of $U$ such that (i) $U_\alpha$ is open in $X$ for all $\alpha\in\mathcal A$, and (ii) $\bigcup_{\alpha\in\mathcal A}U_\alpha=U$.
For (2), let me note that if $X$ is a topological space, and $E\subseteq X$, then sometimes an "open cover" of $E$ is defined more loosely as a family $\{U_{\alpha}\}_{\alpha\in\mathcal A}$ of open sets in $X$ such that $E\subseteq\bigcup_{\alpha\in \mathcal A}U_\alpha$. This is not what open cover means in this context.