Some questions on matroid I have an unknown questions as follows.
Thank you in advance.
Let $M=(E,I)$ be a matroid and let $B$ and $B′$ be two disjoint bases of $M$. Let $B$ be
partitioned into sets $Y_1$ and $Y_2$. Show that there exists a partition of $B′$ into sets $Z_1$ and $Z_2$ so that both $Y_1 \cup Z_2$ and $Z_1 \cup Y_2$ are bases of $M$.
 A: Here is a somewhat lengthy proof by induction.
Let $B = \{b_{1}, b_{2}, \dots b_{n}\}$ and $B' = \{c_{1}, c_{2}, \dots c_{n}\}$.
First, we show the base case of $Y_{1} = \{b_{i}\}$.
Consider $B - \{b_{i}\}$. By the basis-exchange axiom, we know there exists a $c_{j} \in B'$ with $(B - \{b_{i}\}) \cup \{c_{j}\}$ independent. I claim that $(B' - \{c_{j}\}) \cup \{b_{i}\}$ is also independent, giving the required partition ($Y_{1} = \{b_{i}\}$, $Y_{2} = (B - Y_{1})$, and similarly for $Z_{1}, Z_{2}$).
You can check this by looking at the rank function. If there is a $c_{k} \in B'$ with $r(\{b_{i}, c_{k}\}) = 1$,we can also see that $r(\{c_{j}, b_{i}\}) = 1$ so $r(\{b_{i}, c_{k}, c_{j}\}) = 1$. This then implies that $r((B' - \{c_{j}\}) \cup \{b_{i}\} \cup \{c_{j}\}) = r(B' \cup \{b_{i}\}) \leq |B'| - 1$ which is a contradiction as $r(B') = |B'|$ and rank is monotone increasing. Equivalently, no three elements can be parallel as at least two are in the same basis. 
For the induction step, consider the matroid $M'$ with bases $B - Y_{1}$ and $B' - Z_{1}$, where $Y_{1}, Z_{1}$ are parts in partitions of $B, B'$. By the base case, we can partition the bases of $M'$ with parts $Y_{1}', Z_{1}'$ such that $|(Y_{1} \cup Y_{1}')| = |Y_{1}| + 1$ and $|(Z_{1} \cup Z_{1}')| = |Z_{1}| + 1$. The sets $(Y_{1} \cup Y_{1}'), (Z_{1} \cup Z_{1}')$ now are parts for a partition of $B, B'$ satisfying the condition (we can let $Y_{2} = B - Y_{1}$).
