# Reconstructing an oriented matroid from its deletion and contraction

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, 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.

Found the answer in a paper 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.
• Also found the relevant place in the book: basically, my question can be rephrased as "when can a strong map be factored as an extension followed by a contraction", and the answer is "when $r(\mathcal{M}_1)\leq r(\mathcal{M}_2)+1$", see Proposition 7.7.4 and Exercise 7.30 here Sep 18, 2016 at 9:26