I want to understand why the following statement is true:
Let $C$ be a circuit of a matroid $M.$ For e in $E(M)$:
If $e \notin C,$ then $C$ is the union of circuits of $M/e.$ I understood why the following statement is true:
$e \in C,$ then $e$ is a loop of $M,$ or $C - e$ is a circuit of $M/e.$
I know that circuits of $M/e$ (by proposition 3.1.10) consist of minimal nonempty members of $\{ C - T: C \in \mathcal C(M)\}.$ Which I understand as the circuits in M such that when I remove from them e they remain minimal dependent in M/e. (I hope my understanding is correct, please correct me if I am wrong).
Still I do not know how I will get to the idea of unions of circiuts. Can someone help me please?
Equivalently, we need to prove that every $f \in C$ is contained in some circuit that is a subset of $C$ in $M/e$. Assume the contrary, this implies that for every independent set $J$ contained within $C$ in $M/e$ such that $f \not \in J$, we have $J \cup f$ also being independent. Alternatively, for every independent set $J$ contained within $C \cup e$ in $M$ such that $e \in J$ and $f \not \in J$, we have $J \cup f$ also being independent.
Why is this a problem? Well, first note that if $(C - f) \cup e$ is independent, then we would have $C \cup e$ also being independent, which is impossible since $C$ is a circuit. Thus, $C-f$ is independent but $(C-f) \cup e$ is dependent. Let $D$ be the fundamental circuit of $(C-f)$ with respect to $e$. Note that $D \neq \{e\}$ (if $e$ were a loop, contraction would not be well-defined). Let $g \in D$ be such that $g \neq e$. Now, $C$ and $D$ are two circuits with $g \in C \cap D$. It follows that there exists some circuit contained in $C \cup D - g$.
Since $C \cup D - g$ is dependent and contains both $e$ and $f$, we know by our assumption that $C \cup D - g - f$ is also dependent. However, this is contained in $(C - f) \cup e$ and must thus contain $D$, and this is certainly not the case since it does not contain $g$. Thus, we have the required contradiction.