Let $M$ be a matroid and $X$ is a subset of the ground set of $M$.
Someone claims that $$M.X:=\{\tau\subseteq X: \tau \cup \sigma \in M \text{ for all $\sigma\in M$ with $\sigma\cap X=\emptyset$}\}$$ is a matroid, which is called contraction of $M$ onto $X$.
It is easy to check that $\emptyset\in M.X$ and it is downward-closed. I am wondering how to verify the augmentation property?
Let $A,B\in M.X$ with $|A|>|B|$. Then we know there exists $a\in A\setminus B$ such that $B\cup\{a\}\in M$. But how do we verify that $B\cup\{a\}\cup\sigma \in M$ for all $\sigma\in M$ with $\sigma\cap X=\emptyset$?
(One also claims that $M.X=(M^*|X)^*$, where $M|X$ means restricting the independent set of $M$ to $X$ and $M^*$ stands for the dual of $M$. Does this offer any hint?)
Let $E$ be the groud set of $M$.
$\{\sigma\in M \text{ with } \sigma\cap X=\emptyset\}$ is a matroid (noted $M\setminus X$), this is fairly easy to show.
Let $b$ be a base of this matroid. Let $\tau \in X$ such that $\tau\cup b\in M$. For all $b'$ base of $M\setminus X$, using the augmentation property, we get $\tau \cup b' \in M$ (because we can only augment $b'$ with elements of $\tau$). By inclusion, we get $\tau \cup \sigma \in M\;\forall \sigma \in M\setminus X$. Finally, we can rewrite $M.X=\{\tau\subseteq X: \tau \cup \sigma \in M \text{ for all $\sigma\in M$ with $\sigma\cap X=\emptyset$}\} = \{\tau\subseteq X: \tau \cup b\in M\}$
Now, to show the augmentation property, we have $A\cup b\in M$ and $B\cup b\in M$. Using augmentation property, there exists $a \in A$ such that $B\cup b\cup \{a\}\in M$.
Another way to prove that contraction gives a matroid is by exploiting the fact that $M.X = (M^*\setminus E - X)^*$ where $^*$ denotes matroid duality.