The local ring at a point as an inverse limit of regular functions Let $\mathbb{F}$ be an algebraically closed field and $X$ an affine algebraic set. Fix $x \in X$. For an open set $U \subset X$ let $\mathcal{O}_X(U)$ denote the collection of regular functions on $U$. Given two open neighborhoods $U \subset V$ of $x$ then restriction defines a map of algebras $\mathcal{O}_X(V) \to \mathcal{O}_X(U)$. The family of algebras $\mathcal{O}_X(U)$ as $U$ runs over all open neighborhoods of $x$ is an inverse system with respect to these restriction maps. I want to show that $\mathcal{O}_x:= \mathbb{F}[X]_{M_x}$ is the inverse limit of this inverse system in the category of $\mathbb{F}$ algebras. Here $M_x \subset F[X]$ is the maximal ideal of functions which vanish at $x$. 
In order to show $\mathcal{O}_x$ is this inverse limit I want to find $\pi$, a collection of maps $\{\pi_U : U \subset X \text{ an open neighborhood of }x\}$ such that for each open set $U$ we have $\pi_U:\mathcal{O}_x \to \mathcal{O}_X(U)$. Further I need $(\mathcal{O}_x,\pi)$ to be a cone over my inverse system. However I'm struggling even to define a map $\pi_U:\mathcal{O}_x \to \mathcal{O}_X(U)$. It seems that it really suffices to define $\pi_U$ for $U = X$ where $\mathcal{O}_X(X) = \mathbb{F}[X]$ as, for $(\mathcal{O}_x,\pi)$ to be a cone we must have $\pi_U\left(\frac{f}{g}\right) = \pi_X\left(\frac{f}{g}\right)\bigg{|}_{U}$ for any open set $U$. 
After thinking about this for a bit, I think I might be misinterpreting something but I'm really not sure where. Any help would be appreciated. 
 A: We first define a collection of maps $\pi_U:\mathcal{O}_X(U) \to \mathcal{O}_x$ so that $(\mathcal{O}_x, \pi)$ is a cocone from $\{\mathcal{O}_X(U)\}$. Given $f \in \mathcal{O}_X(U)$, by definition of $\mathcal{O}_X(U)$ we can find $g,h \in \mathbb{F}[X]$ so that $f = \frac{g}{h}$ on a neighborhood of $x$ with $h(x) \neq 0$. If this is the case then let $\pi_U(f) = \frac{g}{h}$. I claim $\pi_U$ is well-defined. First note that since $h(x) \neq 0$, $\frac{g}{h} \in \mathcal{O}_x$. Next suppose that $f = \frac{g_1}{h_1}$ on $x \in U_1 \subset U$ and $f = \frac{g_2}{h_2}$ on $x \in U_2 \subset U$. Then we hae $\frac{g_1}{h_1} = \frac{g_2}{h_2}$ on $U_1 \cap U_2$. For a function $s \in \mathbb{F}[X]$ let $X_s := X \backslash \mathcal{V}(s)$. We know these principle open sets for a basis for the topology on $X$ so that we can assume that $\frac{g_1}{h_1} = \frac{g_2}{h_2}$ on $X_s$ with $x \in X_s$. We then have $s(g_1h_2 - g_2h_1) = 0$ on $X$ and thus this is equality in $\mathbb{F}[X]$. Since $x \in X_s$, $s(x) \neq 0$ and $\frac{g_1}{h_1} = \frac{g_2}{h_2}$ in $\mathcal{O}_x$. This shows $\pi_U$ is well-defined. 
Next we check that $(\mathcal{O}_x, \pi)$ is a cocone from $\{\mathcal{O}_X(U)\}$. Note that if $V \subset U$ then $\pi_{V}(f|_{V}) = \pi_U(f)$ as, if $f = \frac{g}{h}$ in a neighborhood $W \subset U$ of $x$ then $f|_{V} = \frac{g}{h}$ in $W \cap V$, a neighborhood of $x$. 
Finally we need to check $(\mathcal{O}_x,\pi)$ is universal with respect to cocones from $\{\mathcal{O}_X(U)\}$. Let $(Y,\sigma)$ be such a cocone. Given $\frac{f}{g} \in \mathcal{O}_x$ we have $\frac{f}{g} \in \mathcal{O}_X(X_g)$. Define $\alpha:\mathcal{O}_x \to Y$ by $\alpha\left(\frac{f}{g}\right) = \sigma_{X_g}\left(\frac{f}{g}\right)$. We need to check that $\alpha$ is well-defined. Suppose $\frac{f_1}{g_1} = \frac{f_2}{g_2}$ in $\mathcal{O}_x$. Then these two functions are equal on $X_{g_1g_2}$ so that, as $(Y, \sigma)$ is a cocone, 
$$ \sigma_{X_{g_1}}\left(\frac{f_1}{g_1}\right) = \sigma_{X_{g_1\cdot g_2}}\left(\frac{f_1}{g_1}\bigg{|}_{X_{g_1\cdot g_2}}\right) = \sigma_{X_{g_1\cdot g_2}}\left(\frac{f_2}{g_2}\bigg{|}_{X_{g_1\cdot g_2}}\right) = \sigma_{X_{g_2}}\left(\frac{f_2}{g_2}\right). $$
Further $\alpha \circ \pi_U = \sigma_U$. To see $\alpha$ is unique note that if $\alpha'\circ \pi_U = \sigma_U$ then $\frac{f}{g} \in \mathcal{O}_x$ then 
$$\alpha'\left(\frac{f}{g}\right) = \alpha' \left( \pi_{X_g}\left(\frac{f}{g}\right)\right) = \sigma_{X_g}\left(\frac{f}{g}\right) = \alpha\left(\frac{f}{g}\right).$$
