Why are there exactly two cocircuits for a given basis $B$ and $e\in B$? I am very unfamiliar with the topic of oriented matroids and am just learning about it. I want to prove the following result which is needed to define fundamental cocircuits. Alas, I did not succeed in doing so, so far. Therefore, I would appreciate any inputs on how to obtain the result.

Claim: Let $\mathcal{M} = (E, \mathcal{F})$ be a oriented matroid and $\mathcal{D}$ the set of its cocircuits. For every basis $B$ (of the underlying matroid) and every $e\in B$ there exist exactly two cocircuits $X, -X \in \mathcal{D}$ such that $B\backslash e \subseteq X^0$. In this case $X_e\neq 0$.

For clarity I will describe the setting I am working in and include the respective definitions which I want to use to obtain the above result. In particular, I am looking a way to obtain the claim that does not need any duality results. I am working through/using the following PhD thesis: http://e-collection.library.ethz.ch/eserv/eth:24433/eth-24433-02.pdf from which I took both the above claim and the subsequent definitions.
The text there just says that for the proof of the claim one should use the definition of basis and cocircuit axiom (C2), cf. below, so I feel that this whole thing should not take more than a few lines.
Thank you for any advice!
Here are now the relevant definitions:


where for two sign  vectors $X, Y \in \{-,+,0\}^E$
$$
(X\circ Y)_e := \begin{cases} X_e &\mbox{if } X_e \neq 0 \\
Y_e & \mbox{otherwise.} \end{cases}
$$
and
$$
D(X,Y) := \{e \in E;\; X_e = -Y_e \neq 0\}
$$




Here, $X^0$ denotes the zero support of $X$.






 A: Let $B$ be a basis of the underlying matroid $\underline{\mathcal{M}} = (E, \mathcal{A})$. and let $e \in B$. Then $\mathcal{A}$ contains the flat $\overline{B\backslash e}$. This means that there exists $Y \in \mathcal{F}$ such that $\overline{B \backslash e} = Y^0$ and one has $B\backslash e \subseteq Y^0$. Now it follows $Y_e \neq 0$, since otherwise $B \subseteq Y^0$ and thus $\overline{Y^0} = \overline{B\backslash e} = \overline{B}$, a contradiction. By Lemma 0.6.2 in the linked PDF there exists a cocircuit $X \in \mathcal{D}$ such that $X \preceq Y$ and $X_e = Y_e$. We also have $B\backslash e \subseteq X^0$ which gives the existence. It remains to show that for any other cocircuit $Z$ satisfying this property we have $X = Z$ or $X = -Z$.
Assume $Z$ is another cocircuit satisfying $B\backslash e \subseteq Z^0$ such that $X \neq Z$. The above consideration implies $Z_e \neq 0$ as well. Applying cocircuit elimination (C3) to $X$, $-Z$ and $e$ yields a cocircuit $\tilde{X}$ that satisfies $B\backslash e \subseteq \tilde{X}^0$ and $\tilde{X}_e = 0$, a contradiction. Hence, either $Z = X$ or $Z = -X$.
