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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 $\codim_{\Bbb R^n/F} E/F=2$. Therefore,

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

We now show that also $G'\subseteq G$: 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 $y=y'\in G/F$. Thus $G=G$'.

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 $y=y'\in G/F$.

The answer is Yes.

We are gonna use $\dim G = \dim F +2$ (the codimension of $F$ in $G$ is two) and $G\cap E=G'\cap E=F$. You further assume that there exists $x\in\mathring F\subset \mathring G$. But 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 $G\subseteq G'$.

The rest follows from dimension considerations:

\begin{align} \dim G - 2 &= \dim F \\&= \dim(G'\cap E) \\&\ge n - ((n - \dim G') + \smash{\overbrace{(n - \dim E)}^{=2}}) \\&= \dim G' - 2, \\[1em] &\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\implies \dim G'\le \dim G \end{align}

where the inequality is the usual estimation for the dimension of the intersection of linear subspaces (technically, $G'$ is not a linear subspace; but locally at $x$, $G'$ is a "closed half-subspace" and thus intersects $E$ as a "full" subspace would). Together with $G\subseteq G'$ we obtain $G=G'$.

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 $\codim_{\Bbb R^n/F} E/F=2$. Therefore,

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

We now show that also $G'\subseteq G$: 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 $y=y'\in G/F$. Thus $G=G$'.

The answer is Yes.

We are gonna use $\dim G = \dim F +2$ (the codimension of $F$ in $G$ is two) and $G\cap E=G'\cap E=F$. You further assume that there exists $x\in\mathring F\subset \mathring G$. But 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 $G\subseteq G'$.

The rest follows from dimension considerations:

\begin{align} \dim G - 2 &= \dim F \\&= \dim(G'\cap E) \\&\ge n - ((n - \dim G') + \smash{\overbrace{(n - \dim E)}^{=2}}) \\&= \dim G' - 2, \\[1em] &\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\implies \dim G'\le \dim G \end{align}

where the inequality is the usual estimation for the dimension of the intersection of linear subspaces (technically, only $E$ is a subspace, but locally at $x$, $G'$ is a "closed half-subspace", and thus intersects $E$ as a "full" subspace would). Together with $G\subseteq G'$ we obtain $G=G'$.

The answer is Yes.

We are gonna use $\dim G = \dim F +2$ (the codimension of $F$ in $G$ is two) and $G\cap E=G'\cap E=F$. You further assume that there exists $x\in\mathring F\subset \mathring G$. But 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 $G\subseteq G'$.

The rest follows from dimension considerations:

\begin{align} \dim G - 2 &= \dim F \\&= \dim(G'\cap E) \\&\ge n - ((n - \dim G') + \smash{\overbrace{(n - \dim E)}^{=2}}) \\&= \dim G' - 2, \\[1em] &\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\implies \dim G'\le \dim G \end{align}

where the inequality is the usual estimation for the dimension of the intersection of linear subspaces (technically, $G'$ is not a linear subspace; but locally at $x$, $G'$ is a "closed half-subspace" and thus intersects $E$ as a "full" subspace would). Together with $G\subseteq G'$ we obtain $G=G'$.