*The sets are defined in $\mathbb{R}\_+^n$ $(n\geq 1)$. The relative interior of a convex set $C$ is denoted $\mathring C$.*


Let $S$ be the $n$-simplex:


$$S=\left\{x\in\mathbb{R}\_+^n,\,\sum\_{i=1}^n x\_i=1\right\}$$


and $E$ be a linear subspace such that $E\cap\mathring S\neq\varnothing$ (hence, $E$ contains a vector with positive coordinates) of codimension $2$ in $\mathbb{R}^n$. For $F$ a face of $E\cap S$, one can see that there exists a unique face $G$ of $S$ such that $\mathring F\subset\mathring G$. Hence, $E\cap G=F$.


**Question:** We assume that, in $\text{Aff}(G)$, $\text{Aff}(F)$ has codimension $2$. If a face $G'$ of $S$ satisfies $G'\cap E=F$, do $G'=G$?



Intersection of the simplex with a linear subspace of codimension $2$
The answer is *Yes*.


By your assumption there exists $x\in\mathring F\subset \mathring G$.
We also have $x\in F\subset G'$, and the only way a face $G'$ can contain an interior point of a face $G$ is if $G\subseteq G'$.


Now, consider a small neighborhood of $x$. In this neighborhood, $F$, $G$ and $E$ look like linear subspaces, while $G'$ looks like a cone that contains the subspace $G$. We will keep this local perspective for the rest of this proof (we can assume that $x$ is the origin of our local linear space).


Since all subspaces contain $F$, we can factor by $F$. We find


$$G/F\cap E/F = (G\cap E)/F=F/F = \{0\}.$$


Moreover, since $\DeclareMathOperator{\codim}{codim}\codim\_G F=2$, we have $\dim G/F=2$; and since $\codim\_{\Bbb R^n} E=2$, we have $\codim\_{\Bbb R^n/F} E/F=2$. We thus find the direct decomposition


$$G/F \oplus E/F = \Bbb R^n/F.$$


This suffices to conclude that also $G'\subseteq G$, hence $G=G'$: fix $y\in G'/F$.
Then there is a unique decomposition $y=y'+y''$ with $y'\in G/F$ and $y''\in E/F$. Since $y'\in G/F$ and $G/F$ is a linear space, we have $-y'\in G/F\subset G'/F$.
Since $G'/F$ is a cone, we have $y+(-y')=y''\in G'/F$. Thus


$$y''\in E/F\cap G'/F = (E\cap G')/F=F/F=\{0\}$$


and we conclude $y=y'\in G/F$.


