Mappings for an integral scheme with generic point Let $X$ be an integral scheme with generic point $\eta$. 
Show that for any $x \in X$, there is a canonical ring homomorphism $\mathcal{O}_{X,x}\rightarrow \mathcal{O}_{X,\eta}$. 
For any open subset $U \subset X$ and any point $x \in U$, show that the canonical homomorphisms $\mathcal{O}_{X}(U) \rightarrow \mathcal{O}_{X,x}$ and $\mathcal{O}_{X,x} \rightarrow \mathcal{O}_{X,\eta}$ are injective. 
Identifying $\mathcal{O}_{X}(U)$ and $\mathcal{O}_{X,x}$ with subrings of $\mathcal{O}_{X,\eta}$ show that $\mathcal{O}_{X}(U)=\bigcap_{x \in U} \mathcal{O}_{X,x}$ 
My idea for the third question was to reduce this to the case where $U$ is affine by means of an inverse limit argument. Then I can just work with primes in some domain $R$. 
The injectivity argument for $\mathcal{O}_{X,x} \rightarrow \mathcal{O}_{X,\eta}$ seems clear from taking $U=\textrm{Spec} R$ for $R$ a domain to be some open affine neighborhood of $x$. 
I'm a little unsure of how to approach the ring homomorphism argument and the other injectivity argument...
 A: This problem is probably easiest of you think of the local ring $\mathcal{O}_{X,x}$ as being the collection of pairs $(U,t)$, where $x\in U$ and $t\in\mathcal{O}_X (U)$, with the equivalence relation $(U,t) \sim (U',t')$ if $t|_{U\cap U'} = t'_{U\cap U'}$.
1) If an element of $\mathcal{O}_{X,x}$ has a representative $t\in \mathcal{O}_X (U)$ on some $x\in U\subset X$ with $U$ nonempty, then $\eta\in U$, giving an obvious map $\mathcal{O}_{X,x} \to \mathcal{O}_{X,\eta}$.  The only question is whether this map is well-defined.  But it is not so hard to show that the equivalence relation on $\mathcal{O}_{X,\eta}$ is stronger than the one on $\mathcal{O}_{X,x}$.
2) For the first one, non-injectivity would mean that there is some smaller open on which $t$ vanishes.  Since $t$ must fail to vanish on some affine open, and all nonempty open subsets intersect, we can work locally as you say.
The second is similar; we can work locally.
Both could also be done this way: The set of points on which some $t\in\mathcal{O}_X(U)$ vanishes is the complement of the closed subscheme defined by the annihilator.  But all annihilators are trivial on an integral scheme.
3) One direction is clear.  To show that $\bigcap_{x\in U} \mathcal{O}_{X,x} \subset \mathcal{O}_X(U)$,consider both as subsets of $\mathcal{O}_{X,\eta}$.  The left side is exactly the rational functions $f$ such that, for each $x\in U$, there is some open set $U_x$ containing $x$ on which $f_x$ is defined and restricts to $f\in\mathcal{O}_{X,\eta}$.  We can check that the collection $\{f_x\}$ satisfies the sheaf condition on the cover $\{U_x\}$, and glue.
