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If $k$ is a field and $\Delta$ a finite simplicial complex with vertex set $x_1, \ldots, x_n$, the Stanley-Reiser ideal of $\Delta$ is

$$I_\Delta := \left\langle \prod_{i \in S}x_i : S \not \in \Delta \right\rangle \subset k[x_1, \ldots, x_n].$$

There is a bijective correspondence between simplicial complexes on a finite set $x_1, \ldots, x_n$ and monomial ideals of $k[x_1, \ldots, x_n]$ given by $\Delta\leftrightarrow I_\Delta$.

The Stanley-Reiser ring of $K$ is $k[\Delta] := k[x_1, \ldots, x_n]/I_\Delta$.

Are these constructions functorial? Concretely, if $f : \Delta_1 \to \Delta_2$ is a simplicial map between finite simplicial complexes, does this induce a $k$-algebra morphism between $k[\Delta_1]$ and $k[\Delta_2]$?

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To expand on my comment, yes this correspondence can be made functorial, but is a contravariant functor.

Proposition (See e.g. Proposition 3.1.5 cf. this answer) Let $f:\Gamma \to \Delta$ be a simplicial map where $\{1,2,\dots,m\}$ and $\{1,2,\dots,n\}$ are the vertex sets of $\Gamma$ and $\Delta$ respectively. Define the map $f^*:k[x_1,\dots,x_n] \to k[y_1,\dots,y_m]$ on generators by $f^*(x_i)=\displaystyle \sum_{j \in f^{-1}(i)} y_j$. Then $f^*$ induces a homomorphism $k[\Delta] \to k[\Gamma]$.

For example, let $\Gamma=\{\{1,\},\{2\}\}$ and $\Delta=\{\{1\}\}$, with simplicial map $f:\Gamma \to \Delta$ given by $f(1)=f(2)=1$. Note this is the example mentioned by Angina Seng in their answer. Then $k[\Delta]=k[y]$ and $k[\Gamma]=k[x_1,x_2]/(x_1x_2)$. The map $f^*$ induces the map $k[\Delta] \to k[\Gamma]$ given by $y \mapsto x_1+x_2$.

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    $\begingroup$ Awesome, thanks! I was trying to do something similar but hand't figured out how to "combine" all the preimages, i.e. I had shown this for morphisms that are injective on vertices. $\endgroup$ – guidoar Jul 9 '20 at 11:04
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I don't think so. If $\Delta_1$ is two isolated vertices, the Stanley-Reisner ring is $k[x_1,x_2]/(x_1x_2)$, and therein $x_1$ and $x_2$ are zero divisors.

This complex has a simplicial map to the one-point space with Stanley-Reisner ring $k[y]$ which is an integral domain. Certainly there's no ring homomorphism mapping each $x_i$ to $y$.

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  • $\begingroup$ I had convinced myself of the same, that if existing the functorial assigment would not come from sending $x_i$ to $x_{f(i)}$ and then factoring the map between polynomial rings to both quotients. I'll leave the answer open for a while just in case. $\endgroup$ – guidoar Jul 9 '20 at 8:41

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