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In Grunbaums book on convex polytopes there is a theorem that states that you can reconstruct a polytope from it's $(d-2)$-skeleton. Here follows the proof (In bold, the two points I don't understand):

This proof is based on the fact that every subset of the $n$-sphere homeomorphic to the $(n-1)$-sphere divides the $n$-sphere into two parts (Jordan's separation theorem).

Let's take $\cal{C}$ a $(d-2)$-complex which is ${\cal{P}}^d$-realizable (${\cal{P}}^d$ is the set of all $d$-polytopes). We can assume ${\cal{C}} = \text{skel}_{d-2}P$ for some $P\in{\cal{P}}^d$. We wish to show that if $P'$ is another $d$-polytope such that $\cal{C}$ is combinatorially equivalent to $\text{skel}_{d-2}P'$, $P$ and $P'$ are combinatorially equivalent. We will do this by showing that the $(d-2)$-equivalence of $P$ and $P'$ can be extended to a $(d-1)$-equivalence (a combinatorial equivalence).

Let ${\cal{F}}_{d-1}$ be the set of facets of the polytope $P$ and index the facets from 1 to $f_{d-1}$. Then for each $i$, ${\cal{B}}(F_i)$ is a complex which, as a subset of $\mathbb{R}^d$, is homeomorphic to the $(d-2)$-sphere. The complexes ${\cal{B}}(F_i)$ have the following property:

(*)If $F$ is a face of $P$ such that all vertices of $F$ are in ${\cal{B}}(F_i)$, then either $F = F_i$ or $F\in{\cal{B}}(F_i)$.

Define ${\cal{B}}_i$ as the sub-complex of $\text{skel}_{d-2}P'$ which corresponds to ${\cal{B}}(F_i)$. By taking the image of the boundary of $P'$ by the homeorphism to the $(d-1)$-sphere, ${\cal{B}}_i$ is homeomorphic to a $(d-2)$-sphere embedded in the image $P'$. Therefore the image of $P'$ is divided into two connected parts. One of those components must contain no vertices of $P'$. But because of property (*), this component may not meet the relative interior of any $k$-face of $P'$ ($k\leqslant d-2$) ?WHY?. It follows that this component meets only one ($d-1)$-face $F'_i$ of $P'$ and ${\cal{B}}_i = {\cal{B}}(F'_i)$ ?WHY?. So from our equivalence on the 2-skeleton, we have found an equivalence on the facets and so on the whole polytope.

Any help would be welcome.

PS: I think I found another way of reasonning. I posted it in answers and would be gratefull if someone checks that it makes sense

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I resume at the point where we state that one of the two components must have no vertices in it.

Firstly, there must exist a facet $F$ which intersects this component. This implies that the facet must include the component and not the other way or we would find vertices in the component which we know is not true.

Now, the facet is of dimension $d-1$ and by applying the same reasonning as in the proof, it's border must split the sphere into two components with one of them having no vertices in it. So if the vertices of $B_i$ are not the vertices of $F$ this is a contradiction because it would mean that the polytope would have only one facet. Therefore, the facet $F$ has the same vertices as $B_i$ and so actually $B_i$ whas the border complex of $F$

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