A question on sheaves Let $k$ be a field and let $ψ$ be the inclusion $ψ:k[x] \to k[x, \frac{1}{x}]$. 
Because of $ψ$ we can have a continuous map in the Zariski topology:
$ψ^α:Spec(k[x, \frac{1}{x}]) \to X_x$, where $X_x=D(x)=\{P \in Spec(k[x]):x \notin P\}$
And we also can go from the restriction sheaf $O_{Spec(k[x])|X_x}(X_g)$ for some $X_g \subset X_x$ to $O_{Spec(k[x, \frac{1}{x}])}(Y_g)$, where $Y_g$ is the respective Zariski-open for $k[x, \frac{1}{x}]$.
I'm reading in some notes that $O_{Spec(k[x])|X_x}$ and $O_{Spec(k[x, \frac{1}{x}])}$ are equal, so my question is, is this last mapping 1-1 and onto?
 A: Let $Y$ be the underlying space of $\operatorname{Spec}(k[x,x^{-1}])$.  Let $g \in k[x]$ such that $X_g \subset X_x$.  Let $\bar g$ be the image of $g$ in $k[x,x^{-1}]$.
The ring homomorphism $k[x] \rightarrow k[x,x^{-1}]$ induces a morphism of schemes $(Y,\mathcal O_Y) \rightarrow (X,\mathcal O_X)$ whose image of underlying spaces is equal to $X_x$.  There is a unique isomorphism of schemes $(Y,\mathcal O_Y) \rightarrow (X_x,\mathcal O_{X|X_x})$ such that the morphisms $Y \rightarrow X, Y \rightarrow X_x$, and the inclusion morphism $X_x \rightarrow X$ commute.
Similarly, there are isomorphisms of schemes
$$(X_g, \mathcal O_{X|X_g}) = (X_g, \mathcal O_{X|X_x|X_g}) \rightarrow \operatorname{Spec}(k[x]_g)$$
$$(Y_{\bar g}, \mathcal O_{Y|Y_{\bar g}}) \rightarrow \operatorname{Spec}(k[x,x^{-1}]_{\bar g})$$
The transitivity of localization gives you a ring isomorphism $k[g]_g \rightarrow k[x,x^{-1}]_{\bar g}$, hence an isomorphism of the corresponding schemes.  Putting these last three isomorphisms of schemes together gives you an isomorphism of schemes
$$(X_g, \mathcal O_{X|X_g}) \rightarrow (Y_{\bar g}, \mathcal O_{Y|Y_{\bar g}})$$
which you can painfully check is the morphism you were asking about.  It is the unique morphism such that 
$$\begin{matrix}   (X_x,\mathcal O_{X|X_x}) & \rightarrow & (Y,\mathcal O_Y)\\ \uparrow & & \uparrow \\ (X_g, \mathcal O_{X|X_g}) & \rightarrow & (Y_{\bar g}, \mathcal O_{Y|Y_{\bar g}}) \end{matrix}$$ commutes.
