Co-trees of a simple graph 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).
 A: Fix a spanning tree $T\subset E$ in a connected graph $G=(V,E)$. It defines fundamental cycles $C_1,\ldots,C_k$, $k=|E|-|T|$: for any $e\in E\setminus T$ take the unique cycle in $T\cup \{e\}$.
Let first $k$ columns of C correspond to the edges in $E\setminus T$. Then the first $k$ columns form a matrix with $\pm 1$ on the diagonal and zeros outside the diagonal. Consider an $r\times r$ minor of C, we should prove that it either 0 or $\pm 1$. By the structure of the first $k$ columns, this reduces to the case when all $r$ columns of the minor correspond to $r$ edges $e_1,\ldots,e_r$ of a tree, let $r$ its rows correspond to $f_1,\ldots,f_r$ of $E\setminus T$. We contract all edges of $T$ except $e_1,\ldots,e_r$. I claim that if the set $f_1,\ldots,f_r$ does not contain a cycle, the corresponding rows are linearly independent over any field (thus the minor is $\pm 1$), and if they do contain a cycle, the rows are dependent with the coefficients $\pm 1$ (thus the minor is 0). For seeing this, we embed the span of the edges $e_1,\ldots,e_r$ to the hyperplane in the vertex space of the (new) tree $\tilde{T}$: each edge $e_i=uv$ corresponds to the vector $v-u$. With this embedding, the row which correspond to edge $f_j=uv$ corresponds to $\pm(v-u)$. The above claim is now clear.
Below is a couple of claims concerning the previous version of the question.
We have the following
Theorem 1. If set $S\subset E$ is a cotree (a complement of a spanning tree), then there exists a bijection $f\colon S\to \{1,\ldots,k\}$ such that $e\in C_{f(e)}$ for all $e\in S$.
Proof.
To prove that such a bijection exists, by Hall lemma (for the natural bipartite graph with parts $S$ and $\{1,\ldots,k\}$ corresponding to a relation $(e,i):e\in C_i$) it suffices to prove that the union $C_0$ of any $m=1,2,\ldots,k$ fundamental cycles contain at least $m$ elements of $S$. Denote $T_0=T\cap C_0$. Then $T_0$ is a forest, and it is an inclusion-maximal subforest of $C_0$ (since after adding to $T_0 $any edge in $C_0\setminus T$ you get a fundamental cycle). Assume that on the contrary $|S\cap C_0|\leqslant m-1$, this is equivalent to $|(E\setminus S)\cap C_0|\geqslant |C_0|-m+1=r+1$. But $E\setminus S$ is a tree, then $(E\setminus S)\cap C_0$ is a subforest of $C_0$, and it has more edges than an inclusion-maximal subforest $T_0$ of $C_0$. A contradiction.
The converse does not hold. For example, consider the graph (the black edges constitute a spanning tree, take the edge $e_1$ in the fundamental cycle $C_1$ and $e_2$ in $C_2$. The set $\{e_1,e_2\}$ is not a cotree, however.)
Also, the bijection is not always unique.
(the spanning tree is black) three edges $e_1,e_2,e_3$ may be paired with fundamental cycles by different ways.
Finally, I prove
Theorem 2. Assume that set $S\subset E$ is such that there exists unique bijection $f\colon S\to \{1,\ldots,k\}$ such that $e\in C_{f(e)}$ for all $e\in S$. Then $S$ is a cotree.
Proof. Consider the cut matroid of $G$: the bases are cotrees, the independent sets are subsets of cotrees. Let $e_1,\ldots,e_k$ be all elements of $E\setminus T$ enumerated so that $e_i\in C_i$. Note that $f\in C_i$ if and only if $(T\cup e_i)\setminus f$ is a tree, that is, $((E\setminus T)\setminus e_i)\cup f=f\cup \{e_j:j\ne i\}$ is a cotree. That is, if $f$ does not belong to a flat generated by $\{e_j:j\ne i\}$. Our matroid is a vector matroid, so this may be said as "the coefficient of $e_i$ in the expansion of $f$ in the basis $\{e_1,\ldots,e_k\}$ is non-zero". That is, we are given that the matrix of vectors of $S$ in the basis $\{e_1,\ldots,e_k\}$ has unique generalized diagonal with non-zero elements. This yields that the determinant of this matrix is non-zero, so, it is non-singular and $S$ constitutes a basis.
A: Consider 2 different spanning trees of the graph, and the corresponding cotrees and fundamental cycles. Then $$C_2=C_2|S_1*C_1$$ spells out the algebraic details of the following well-known fact: every fundamental cycle of Type 2 is a linear combination of fundamental cycles of Type 1 with coefficients from the ${0,1,-1}$ set; $C_2|S_1$ implies selecting only those columns/edges of $C_2$ which are in the first cotree (these are the coefficients of the linear combinations).
Now, selecting only columns which are in $S_2$ from each side of this equation yields $$C_2|S_2=C_2|S_1*C_1|S_2$$ Since the determinant of the LHS matrix is equal to either $1$ or $-1$ (there is exactly one way of matching each edge of a cotree with the corresponding fundamental cycle, only one of the $k!$ term of the determinant is non-zero). This implies that (integer-valued) determinants of each of the RHS matrices have the same property (being equal to $1$ or $-1$).
When the $k$ edges contain a bond, there is a linear combination of the corresponding columns of C (with coefficients from the $1,-1$ set) which yields a zero vector (a bond has to be traversed both ways by each cycle - the two contributions always cancel). This implies that these columns are linearly dependent and the corresponding determinant is thus equal to $0$.
The totally unimodular issue: It's known (and I take it for granted) that C is unimodular if and only if it's possible to find a linear combination of any selection of its rows such that (i) the coefficients are either $1$ or $-1$, (ii) the resulting elements are all from the $0,1,-1$ set.
To find such coefficients, one must first simply add the selected rows and apply 'mod 2' to each component of the answer (getting a vector of ones and zeros); this defines an undirected closed trail (revisiting vertices is allowed). Give this trail a consistent orientation (there may be several ways of doing it) by changing the sign of some of the $1$ components; the corresponding sign of the unique co-tree component of each row of C tells us whether the row should be added or subtracted - the answer will match the already constructed directed trail.
