[proofplan] We characterize flats in the cycle matroid by connectivity in the spanning subgraph determined by an edge set. This turns hyperplanes into maximal proper edge sets whose spanning subgraph has exactly one more connected component than $G$ itself. Taking complements converts these hyperplanes into precisely the minimal nonempty cuts, namely bonds. Finally, the matroid-theoretic relation between cocircuits and hyperplane complements identifies these bonds with the cocircuits of $M(G)$. [/proofplan] [step:Characterize closure in the cycle matroid by connected components] For a subset $A \subseteq E$, let $G[A]$ denote the spanning subgraph $(V,A)$. For vertices $u,v \in V$, write $u \sim_A v$ when $u$ and $v$ lie in the same connected component of $G[A]$. We claim that for every edge $e \in E \setminus A$ with endpoints $u,v \in V$, one has \begin{align*} e \in \operatorname{cl}_{M(G)}(A) \iff u \sim_A v. \end{align*} Indeed, in the cycle matroid $M(G)$, an edge belongs to $\operatorname{cl}_{M(G)}(A)$ exactly when adding it to $A$ does not increase the graphic rank. If $u \sim_A v$, then there is a path in $G[A]$ from $u$ to $v$, so adding $e$ creates a cycle and does not increase rank. Conversely, if $u$ and $v$ lie in distinct connected components of $G[A]$, then adding $e$ merges those two components, so the rank increases by $1$. Thus $A$ is a flat of $M(G)$ precisely when it contains every edge of $G$ whose endpoints lie in the same connected component of $G[A]$. [/step] [step:Identify the hyperplanes from the component partitions they induce] Let $H \subsetneq E$ be a flat of $M(G)$. Since $H$ is a flat, the preceding step implies that $H$ contains every edge whose endpoints lie in the same connected component of $G[H]$. Therefore the only edges missing from $H$ are edges joining distinct connected components of $G[H]$. Let $c_G$ denote the number of connected components of $G$, and let $c_H$ denote the number of connected components of $G[H]$. The rank of the cycle matroid satisfies \begin{align*} r_{M(G)}(A) = |V| - c_A \end{align*} for every $A \subseteq E$, where $c_A$ is the number of connected components of $G[A]$. Hence $H$ is a hyperplane exactly when it is a maximal proper flat, equivalently when \begin{align*} r_{M(G)}(H) = r_{M(G)}(E) - 1. \end{align*} Using the rank formula, this is equivalent to \begin{align*} |V| - c_H = |V| - c_G - 1, \end{align*} and therefore to \begin{align*} c_H = c_G + 1. \end{align*} So a hyperplane of $M(G)$ is exactly a flat whose spanning subgraph has one more connected component than $G$. [guided] The point of this step is to translate the matroid word “hyperplane” into graph language. A hyperplane is a maximal proper flat, and in any matroid this is the same as a flat of rank one less than the rank of the whole ground set. For the cycle matroid of a finite graph, the rank of an edge set is controlled by connected components. If $A \subseteq E$ and $c_A$ denotes the number of connected components of the spanning subgraph $G[A] = (V,A)$, then \begin{align*} r_{M(G)}(A) = |V| - c_A. \end{align*} In particular, if $c_G$ is the number of connected components of the original graph $G$, then \begin{align*} r_{M(G)}(E) = |V| - c_G. \end{align*} Now let $H \subsetneq E$ be a flat. The condition that $H$ is a hyperplane is \begin{align*} r_{M(G)}(H) = r_{M(G)}(E) - 1. \end{align*} Substituting the graphic rank formula gives \begin{align*} |V| - c_H = |V| - c_G - 1. \end{align*} Cancelling $|V|$ from both sides gives \begin{align*} c_H = c_G + 1. \end{align*} Thus the graph $G[H]$ has exactly one more connected component than $G$. The flat condition also matters: it says $H$ already contains every edge internal to each connected component of $G[H]$. Therefore the complement $E \setminus H$ consists exactly of the edges crossing between the two pieces created by splitting one connected component of $G$ into two. [/guided] [/step] [step:Show that complements of hyperplanes are bonds] Let $H$ be a hyperplane of $M(G)$, and define \begin{align*} B := E \setminus H. \end{align*} By the previous step, $G[H]$ has exactly one more connected component than $G$. Since $H$ is a flat, every edge of $G$ internal to a connected component of $G[H]$ belongs to $H$. Hence every edge in $B$ joins two distinct connected components of $G[H]$. Because $c_H = c_G + 1$, the set $B$ separates exactly one connected component of $G$ into two connected components of $G[H]$. Let $X \subseteq V$ be the vertex set of one of these two components. Then the edges crossing from $X$ to $V \setminus X$ are precisely the edges missing from $H$, so \begin{align*} B = \delta_G(X). \end{align*} It remains to prove minimality. Suppose $B_0 \subsetneq B$ is a nonempty edge cut. Then $E \setminus B_0$ is a proper superset of $H$. Since $H$ is a maximal proper flat, adding any edge of $B$ to $H$ raises the rank to $r_{M(G)}(E)$, which reconnects the two components separated by $B$. Thus $E \setminus B_0$ has the same connected components as $G$, so $B_0$ cannot be an edge cut. Therefore $B$ is an inclusion-minimal nonempty edge cut, hence a bond. [/step] [step:Show that every bond is the complement of a hyperplane] Let $B \subseteq E$ be a bond. By definition, there exists $X \subseteq V$ such that \begin{align*} B = \delta_G(X), \end{align*} and $B$ is nonempty and inclusion-minimal among edge cuts. Define \begin{align*} H := E \setminus B. \end{align*} Since $B$ is an edge cut, removing $B$ increases the number of connected components by at least $1$. Since $B$ is minimal, removing any proper subset of $B$ does not disconnect the graph further. Hence removing $B$ increases the number of connected components by exactly $1$, so \begin{align*} c_H = c_G + 1. \end{align*} The set $H$ is a flat. Indeed, if an edge $e \in E$ has endpoints in the same connected component of $G[H]$, then $e$ cannot cross the cut $\delta_G(X)=B$, so $e \in H$. Thus $H$ contains every edge whose endpoints are connected in $G[H]$, which is the flat criterion from the first step. Since $H$ is a flat and $c_H = c_G + 1$, the rank computation in the second step gives \begin{align*} r_{M(G)}(H) = r_{M(G)}(E) - 1. \end{align*} Therefore $H$ is a hyperplane of $M(G)$. [/step] [step:Translate hyperplane complements into cocircuits] We use the standard matroid fact that, in a finite matroid, the cocircuits are exactly the complements of hyperplanes. Applying this to the finite matroid $M(G)$ on ground set $E$, a subset $C \subseteq E$ is a cocircuit of $M(G)$ if and only if $E \setminus C$ is a hyperplane of $M(G)$. The preceding two steps prove that the complements of hyperplanes of $M(G)$ are exactly the bonds of $G$. Hence $C$ is a cocircuit of $M(G)$ if and only if $C$ is a bond of $G$. This proves the theorem. [/step]