Stanley-Reisner ring of a simplicial complex is a functor?

Let $K$ bea field and $[n]=\{1,\ldots,n\}$ and $K[x]=K[x_1,\ldots,x_n]$. For $\sigma=\{i_1,\ldots,i_k\}\subseteq [n]$, denote $x_\sigma=x_{i_1}\cdots x_{i_k}=\prod_{i\in\sigma}x_i\in K[x]$. Let $\Delta$ and $\Delta'$ be (abstract finite) simplicial complexes on $[n]$ and $[n']$ respectively. Let $f: \Delta\rightarrow \Delta'$ be a simplicial map, i.e. a map $f:[n]\rightarrow[n']$ for which $\forall \sigma\in\Delta: f(\sigma)\in\Delta'$ holds.

The Stanley-Reisner ideal of $\Delta$ is $I_\Delta:=\langle\langle x_\sigma;\; [n]\supseteq\sigma\notin\Delta\rangle\rangle$, and Stanley-Reisner ring of $\Delta$ is $$K[\Delta]:=K[x]/I_\Delta= K[x_1,\ldots,x_n \:|\: x_\sigma;\; [n]\supseteq\sigma\notin\Delta].$$

The coordinate ring of an affine variety $A\subseteq\mathbb{A}^n_K$ is $K[A]=K[x]/I_A$, where $I_A$ is the vanishing ideal $\{f\in K[x]; f|_A=0\}$. Thus the notation is the same as that of the Stanley-Reisner ring. Since the coordinate ring $K[-]$ is a functor from the category of affine varieties over an algebraically closed field to the category of finitely generated commutative unital reduced $K$-algebras, defined for $f: A\rightarrow A'$ by $K[f]: K[A']\rightarrow K[A]$ that sends $\alpha:A'\rightarrow K$ to $\alpha\circ f: A\rightarrow K$. This is in fact an antiequivalence of categories.

The immediate impulse is to try to make the Stanley-Reisner ring an antiequivalence of categories. But I'm having trouble making it even a functor. I didn't find anything in the literature (Bruns & Herzog, Stanley, Herzog & Hibi, Miller & Sturmfels) regarding functors.

1st try: We let $K[f]: K[\Delta]\rightarrow K[\Delta']$ send $x_i\mapsto x_{f(i)}$. But then for $\sigma\notin\Delta$, this map sends $x_\sigma\mapsto x_{f(\sigma)}$ (here $f(\sigma)$ is considered as a multiset, i.e. if $f(i)=f(j)$, then $x_{f(\sigma)}$ contains both $x_{f(i)}$ and $x_{f(j)}$), and we do not necessarily have $f(\sigma)\notin\Delta'$, meaning that $K[f]$ is not well defined on the quotient of $K[x_1,\ldots,x_n]$.

2nd try: We let $K[f]: K[\Delta']\rightarrow K[\Delta]$ send $x_{i'}\mapsto x_{f^{-1}(i')}$, where $f^{-1}$ denotes the preimage. Then for $\sigma'\notin\Delta'$, this map sends $x_{\sigma'}\mapsto x_{f^{-1}(\sigma')}$. We wish to have $f^{-1}(\sigma')\notin\Delta$. If this does not hold, $f^{-1}(\sigma')\in\Delta$, then $\sigma'\supseteq f(f^{-1}(\sigma'))\in\Delta'$, so I don't get any contradiction, and we don't have well-definedness.

Question: How can the Stanle-Reisner ring be made into a functor (preferrably in a way that it becomes an antiequivalence between the category of simplicial complexes and category of finitely generated commutative unital monomially related reduced $K$-algebras)?

Question: Can the Stanley-Reisner ideal be seen as an ideal of polynomial functions $\alpha: \Delta\rightarrow K$?

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I think I'd want to deal with partially defined functions $f: [n] \to [n']$, with the property that if $F$ is a face of $\Delta$, then $f(F)$ is a face of $\Delta'$. The linear extension of such an $f$ is a linear map from $K^n \to K^{n'}$, taking the $i$th basis vector to the $f(i)$th, or to zero if $f(i)$ is not defined.

The Stanley-Reisner ideal is the functions vanishing on the union $X_\Delta \subseteq K^n$ of coordinate spaces $\bigcup_{F\in\Delta} K^F$, where $K^F$ denotes the linear span of {the $i$th basis vector : $i\in F$}. (Does that answer your second question?) Then the linear extension of $f$ takes $X_\Delta$ into $X_{\Delta'}$.

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What is $X_\Delta$? What is $N$ and $F^N$? –  Leon Lampret Apr 14 '13 at 2:22
Oops, $N$ should be $n$ and $F^N$ should be $K^n$. $X_\Delta$ is, as stated, that union of coordinate spaces. –  Allen Knutson Apr 14 '13 at 13:43
The functor $K[-]$ from the category of simplicial complexes to the category of finitely generated commutative unital monomially related reduced $K$-algebras is not an equivalence of categories: if $\Delta=\{0\}$ is the one-point simplicial complex, then there is only one simplicial map $\Delta\rightarrow\Delta$, but there are many homomorphisms $K[\Delta]=K[x]\rightarrow K[\Delta]=K[x]$, namely for any polynomial $f(x)$ there is a homomorphism of K-algebras $x\mapsto f(x)$. I suspect this functor is an anti-equivalence from the category of simplicial complexes to ... graded algebras.