Base-orderable matroid

Base-orderability imposes two requirements on the function ${\displaystyle f}$:

1. It should be a bijection;
2. For every ${\displaystyle a\in A\setminus B}$, both ${\displaystyle (A\setminus \{a\})\cup \{f(a)\}}$ and ${\displaystyle (B\setminus \{f(a)\})\cup \{a\}}$ should be bases.

Each of these properties alone is easy to satisfy:

1. All bases of a given matroid have the same cardinality, so there are n! bijections between them (where n is the common size of the bases). But it is not guaranteed that one of these bijections satisfies property 2.
2. All bases ${\displaystyle A}$ and ${\displaystyle B}$ of a matroid satisfy the symmetric basis exchange property, which is that for every ${\displaystyle a\in A\setminus B}$, there exists some ${\displaystyle f(a)\in B\setminus A}$, such that both ${\displaystyle (A\setminus \{a\})\cup \{f(a)\}}$ and ${\displaystyle (B\setminus \{f(a)\})\cup \{a\}}$ are bases. However, it is not guaranteed that the resulting function f be a bijection - it is possible that several ${\displaystyle a}$ are matched to the same ${\displaystyle f(a)}$.

Some matroids are not base-orderable. A notable example is the graphic matroid on the graph K₄, i.e., the matroid whose bases are the spanning trees of the clique on 4 vertices.[1] Denote the vertices of K₄ by 1,2,3,4, and its edges by 12,13,14,23,24,34. Note that the bases are:

• {12,13,14}, {12,13,24}, {12,13,34}; {12,14,23}, {12,14,34}; {12,23,24}, {12,23,34}; {12,24,34};
• {13,14,23}, {13,14,24}; {13,23,24}, {13,23,34}; {13,24,34};
• {14,23,24}, {14,23,34}; {14,24,34}.

Consider the two bases A = {12,23,34} and B = {13,14,24}, and suppose that there is a function f satisfying the exchange property (property 2 above). Then:

• f(12) must equal 14: it cannot be 24, since A \ {12} + {24} = {23,24,34} which is not a basis; it cannot be 13, since B \ {13} + {12} = {12,14,24} which is not a basis.
• f(34) must equal 14: it cannot be 24, since B \ {24} + {34} = {13,14,34} which is not a basis; it cannot be 13, since A \ {34} + {13} = {12,13,23} which is not a basis.

Then f is not a bijection - it maps two elements of A to the same element of B.

There are matroids that are base-orderable but not strongly-base-orderable.[4][1]

Every partition matroid is base-orderable. Every transversal matroid is strongly base-orderable.[2]

In base-orderable matroids, a feasible exchange bijection exists not only between bases but also between any two independent sets of the same cardinality, i.e., any two independent sets ${\displaystyle A}$ and ${\displaystyle B}$ such that ${\displaystyle |A|=|B|}$.

This can be proved by induction on the difference between the size of the sets and the size of a basis (recall that all bases of a matroid have the same size). If the difference is 0 then the sets are actually bases, and the property follows from the definition of base-orderable matroids. Otherwise by the augmentation property of a matroid, we can augment ${\displaystyle A}$ to an independent set ${\displaystyle A\cup \{x\}}$ and augment ${\displaystyle B}$ to an independent set ${\displaystyle B\cup \{y\}}$. Then, by the induction assumption there exists a feasible exchange bijection ${\displaystyle f}$ between ${\displaystyle A\cup \{x\}}$ and ${\displaystyle B\cup \{y\}}$. If ${\displaystyle f(x)=y}$, then the restriction of ${\displaystyle f}$ to ${\displaystyle A}$ and ${\displaystyle B}$ is a feasible exchange bijection. Otherwise, ${\displaystyle f^{-1}(y)\in A}$ and ${\displaystyle f(x)\in B}$, so ${\displaystyle f}$ can be modified by setting: ${\displaystyle f(f^{-1}(y)):=f(x)}$. Then, the restriction of the modified function to ${\displaystyle A}$ and ${\displaystyle B}$ is a feasible exchange bijection.

The class of base-orderable matroids is complete. This means that it is closed under the operations of minors, duals, direct sums, truncations, and induction by directed graphs.[1]^:2 It is also closed under restriction, union and truncation.[5]^:410

The same is true for the class of strongly-base-orderable matroids.