# On the functoriality of assigning a simplicial complex to is Stanley-Reisner ring

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 $$k[\Delta_2]$$?

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 $$y \mapsto x_1+x_2$$.

• 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. – guidoar Jul 9 '20 at 11:04

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 $$y$$.

• 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. – guidoar Jul 9 '20 at 8:41