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Consider fundamental cycles (say $k$ of them) of a specific spanning tree of a simple graph (with $m$ edges) which is also connected and has no one-edge bonds.

Make the graph directed (in an arbitrary way) and each cycle directed (to follow consistently one of the two possible orientations); then create a matrix C with $k$ rows and $m$ columns whose elements indicate whether an edge is a part of the cycle (by $1$ or $-1$ when the two orientations agree or not, respectively) or not (by 0).

How would one prove the following:

Selecting $k$ columns of the matrix, the corresponding sub-determinant equals to $1$ or $-1$ when the corresponding edges contain no bond, equals to 0 otherwise.

A bit more challenging would be to prove that C is totally unimodular.

(My original contention that there was a unique bijection between edges of a cotree and fundamental cycles a spanning tree has been disproved below).

Consider fundamental cycles (say $k$ of them) of a specific spanning tree of a simple graph (with $m$ edges) which is also connected and has no one-edge bonds.

Make the graph directed (in an arbitrary way) and each cycle directed (to follow consistently one of the two possible orientations); then create a matrix C with $k$ rows and $m$ columns whose elements indicate whether an edge is a part of the cycle (by $1$ or $-1$ when the two orientations agree or not, respectively) or not (by 0).

How would one prove the following:

Selecting $k$ columns of the matrix, the corresponding sub-determinant equals to $1$ or $-1$ when the corresponding edges contain no bond, equals to 0 otherwise.

A bit more challenging would be to prove that C is totally unimodular.

(My original contention that there was a unique bijection between edges of a cotree and fundamental cycles a potentially different spanning tree has been disproved below).

Consider fundamental cycles (say $k$ of them) of a specific spanning tree of a simple graph (with $m$ edges) which is also connected and has no one-edge bonds.

Make the graph directed (in an arbitrary way) and each cycle directed (to follow consistently one of the two possible orientations); then create a matrix C with $k$ rows and $m$ columns whose elements indicate whether an edge is a part of the cycle (by $1$ or $-1$ when the two orientations agree or not, respectively) or not (by 0).

How would one prove the following:

Selecting $k$ columns of the matrix, the corresponding sub-determinant equals to $1$ or $-1$ when the corresponding edges contain no bond, equals to 0 otherwise.

A bit more challenging would be to prove that C is totally unimodular.

Consider fundamental cycles (say $k$ of them) of a specific spanning tree of a simple graph (with $m$ edges) which is also connected and has no one-edge bonds.

Make the graph directed (in an arbitrary way) and each cycle directed (to follow consistently one of the two possible orientations); then create a matrix C with $k$ rows and $m$ columns whose elements indicate whether an edge is a part of the cycle (by $1$ or $-1$ when the two orientations agree or not, respectively) or not (by 0).

How would one prove the following:

Selecting $k$ columns of the matrix, the corresponding sub-determinant equals to $1$ or $-1$ when the corresponding edges contain no bond, equals to 0 otherwise.

A bit more challenging would be to prove that C is totally unimodular.

(My original contention that there was a unique bijection between edges of a cotree and fundamental cycles a spanning tree has been disproved below).

Consider fundamental cycles (say $k$ of them) of a specific spanning tree of a simple graph (with $m$ edges) which is also connected and has no one-edge bonds.

Make the graph directed (in an arbitrary way) and each cycle directed (to follow consistently one of the two possible orientations); then create a matrix C with $k$ rows and $m$ columns whose elements indicate whether an edge is a part of the cycle (by $1$ or $-1$ when the two orientations agree or not, respectively) or not (by 0).

How would one prove the following:

Selecting $k$ columns of the matrix, the corresponding sub-determinant equals to $1$ or $-1$ when the corresponding edges contain no bond, equals to 0 otherwise.

A bit more challenging would be to prove that C is unimodular.

Consider fundamental cycles (say $k$ of them) of a specific spanning tree of a simple graph (with $m$ edges) which is also connected and has no one-edge bonds.

Make the graph directed (in an arbitrary way) and each cycle directed (to follow consistently one of the two possible orientations); then create a matrix C with $k$ rows and $m$ columns whose elements indicate whether an edge is a part of the cycle (by $1$ or $-1$ when the two orientations agree or not, respectively) or not (by 0).

How would one prove the following:

Selecting $k$ columns of the matrix, the corresponding sub-determinant equals to $1$ or $-1$ when the corresponding edges contain no bond, equals to 0 otherwise.

A bit more challenging would be to prove that C is totally unimodular.