Let $P \subseteq \mathbb{R}^d$ and $Q \subseteq \mathbb{R}^e$ be (convex) polytopes. Define them to be affinely isomorphic if there is an affine map $f \colon \mathbb{R}^d \to \mathbb{R}^e$ whose restriction to $P$ gives a bijection $P \to Q$. I want to show that if $P$ and $Q$ are affinely isomorphic then their face lattices are isomorphic.
Any help? I was able to show that $f$ induces a bijection between the vertices of the polytopes but then I got stuck. I also tried to show this directly from the definitions but that seems really difficult. A face of $P$ is of the form $F = P \cap H$ where $H = \{x \in \mathbb{R}^d \mid cx = c_0\}$ where $c$ is some row vector and $c_0 \in \mathbb{R}$, and where $cx \le c_0$ for all $x \in P$. If I take the image of $F$ under $f$ then it is not clear to me that the image is a face of $Q$ since I can't seem to find a suitable linear equation on the codomain side.
Edit. I finally found a way to prove this that I feel satisfied with. A lemma that needs to be proved before this proof is that if $f \colon P \to Q$ is a bijection that extends to an affine map $\mathbb{R}^d \to \mathbb{R}^e$, then $f^{-1} \colon Q \to P$ also extends to an affine map $\mathbb{R}^e \to \mathbb{R}^d$. Proving this lemma is no trivial task though, but I'll skip it.
Suppose $P \subseteq \mathbb{R}^d$ and $Q \subseteq \mathbb{R}^e$ be polytopes, an suppose $f \colon P \to Q$ is a bijection that extends to an affine map $\mathbb{R}^e \to \mathbb{R}^d$. Let $f(x) = L(x) + b$ for some linear $L \colon \mathbb{R}^d \to \mathbb{R}^e$ and vector $b \in \mathbb{R}^e$. Now $f^{-1} \colon Q \to P$ extends to an affine map $g \colon \mathbb{R}^e \to \mathbb{R}^d$. Let $g(x) = T(x) + a$ where $T \colon \mathbb{R}^e \to \mathbb{R}^d$ is linear and $a \in \mathbb{R}^d$. Now \begin{equation*} g(f(x)) = T(L(x) + b) + a = x \end{equation*} for all $x \in P$. Let $F$ be a face of $P$. Now \begin{equation*} F = P \cap \{ x \in \mathbb{R}^d \mid \alpha(x) = c \} \end{equation*} where $\alpha \in (\mathbb{R}^d)^*$ and $c \in \mathbb{R}$ and where $\alpha(x) \le c$ for all $x \in P$. We claim that $f(F)$ is a face of $Q$ given by \begin{equation*} f(F) = Q \cap \{ y \in \mathbb{R}^e \mid \alpha(T(y)) = c - \alpha(a) \}. \tag{1} \end{equation*} First we check that the inequality $\alpha(T(y)) \le c - \alpha(a)$ is valid for $Q$. Let $y \in Q$. Now $y = L(x) + b$ for some $x \in P$. Thus \begin{align*} \alpha(T(y)) = \alpha(T(L(x) + b)) = \alpha(x - a) = \alpha(x) - \alpha(a) \le c - \alpha(a). \end{align*} Next we show that we have equality in (1). Let $f(x) \in f(F)$ where $x \in F$. Now $f(x) \in Q$ and \begin{equation*} \alpha(T(f(x))) = \alpha(T(L(x) + b)) = \alpha(x-a) = \alpha(x) - \alpha(a) = c - \alpha(a). \end{equation*} Then let $y \in Q$ be such that $\alpha(T(y)) = c - \alpha(a)$. Now $y = f(x)$ for some $x \in P$. We claim that $x \in F$. For this we note that \begin{equation*} c - \alpha(a) = \alpha(T(y)) = \alpha(T(f(x))) = \alpha(x) - \alpha(a) \end{equation*} and thus $\alpha(x) = c$. So $x \in F$. We therefore have the equality in (1). We have now seen that $f$ maps faces of $P$ to faces of $Q$. By symmetry, $f^{-1}$ maps faces of $Q$ to faces of $P$. Thus the map $F \mapsto f(F)$ is a bijection $L(P) \to L(Q)$. Furthermore, it and its inverse are order preserving since \begin{equation*} F \subseteq G \iff f(F) \subseteq f(G) \end{equation*} for all faces $F,G \in L(P)$. Hence $L(P) \cong L(Q)$.