Suppose that $\mathcal{M}\_1$ and $\mathcal M\_2$ are two oriented matroids on the same ground set $E$. Under what conditions on $\mathcal{M}\_1$ and $\mathcal{M}\_2$ is there an oriented matroid $\mathcal{M}$ on the ground set $E\cup \{e\}$ such that $\mathcal{M}-e=\mathcal{M}\_1$ and $\mathcal{M}/e=\mathcal{M}\_2$?


I feel like I've seen the answer to this in [the book](http://epubs.siam.org/doi/abs/10.1137/1036173), and that the necessary and sufficient condition was 
$\mathcal{T}(\mathcal{M}\_2)\subset \mathcal{T}(\mathcal{M}\_1),$ where $\mathcal{T}(\mathcal{M})$ is the set of topes of $\mathcal{M}$. However, I've been trying to find it in the book again without any success for a very long time.



Reconstructing an oriented matroid from its deletion and contraction
Found the answer in a [paper](https://www-m10.ma.tum.de/foswiki/pub/Lehrstuhl/PublikationenJRG/13_BohneDress.pdf) by Ziegler & Richter-Gebert, see Theorem 4.1. Basically, they show that if two oriented matroids $\mathcal{M}\_1$, $\mathcal{M}\_2$ of ranks $r$ and $r-1$ respectively satisfy $\mathcal{L}(\mathcal{M}\_2)\subset \mathcal{L}(\mathcal{M}\_1)$ then there exists a (essentially unique) oriented matroid $\mathcal{M}$ with $\mathcal{M}\setminus e=\mathcal{M}\_1$ and $\mathcal{M}/e=\mathcal{M}\_2$. So the necessary and sufficient condition is the inclusion of covectors. 


