Help understanding injectivity of function.
问题内容
I fail to understand the highlighted statement in my screenshot below. If $U \subset Y$ is a non empty open subset, then the natural map $g: \mathscr{O}_Y(U) \to k(Y)$ given by $(U,f) \mapsto [U,f]$ is naturally injective. Indeed, if $g((U,f_1)) = g((U,f_2))$, i.e $[U, f_1] = [U,f_2]$, then $f_1$ and $f_2$ restrict to the same function on $U \cap U = U$ by definition, and so we must have $(U, f_1) = (U, f_2)$. How is $(ii)$ from lemma 2.2 actually used here? (To be found at the very top of the screenshot, I couldnt fit the whole Lemma).
回答 (1)
It's due to how the text is written.
Notice that if you stick to the definition given, then the transitivity of the relation follows from the lemma (if $[f, U]$ is equivalent to $[g,V]$, and $[g,V]$, is equivalent to $[h, W]$ you need the lemma to prove that $[f, U]$ is equivalent to $[h, W]$ as they will only be equal on $V\cap U\cap W$ a priori instead of the full $U\cap W$, so to define the relation the way you want, you need to assume that $Y$ is irreducible. In that case your argument works for proving the injectivity, but the lemma is needed for the well-defined-ness of the relation.
But there's another possible way, that is to define the relation by $[f, U]$ and $[g,V]$ equivalent if there exists $W \subseteq V\cap U$ st $f=g$ when restricted to $W$, in that case, this is an equivalence relation "on the nose", but you then need the lemma to prove the injectivity.
Indeed in that case if $[f, U]=[g,U]$ in $k(Y)$, that means that you may find $W\subseteq U$ such that $[f, W]=[g,W]$, and then the lemma is useful, to imply that $f$ was already equal to $g$ on $U$.
Noticing that $[f, U]=[g,U]$ if $f$ and $g$ restrict to the same function a possibly smaller open subset than $U$, might not be a bad thing to keep in mind (it will make things like the defnition of stalks pretty obvious, and the relationship with directed colimits will also be clear)