[proofplan] We prove both directions by translating between graph duality and matroid duality. If $G$ has a planar embedding, its geometric dual $G^\dagger$ has cycle matroid naturally isomorphic to $M(G)^*$, so the dual matroid is graphic. Conversely, if $M(G)^*$ is the cycle matroid of a graph $H$, we replace $H$ by a connected graph with the same cycle matroid and compare cycles and bonds in $G$ and $H$. Whitney's abstract duality theorem then gives a sphere embedding of $G$ whose geometric dual is $H$, hence $G$ is planar. [/proofplan] [step:Use a planar embedding to identify the dual matroid with a cycle matroid] Assume first that $G$ is planar. Choose a cellular embedding of $G$ in the sphere $S^2$, and let $G^\dagger$ be the geometric dual graph of this embedding. The edge set of $G^\dagger$ is naturally identified with $E(G)$ by sending each dual edge to the unique primal edge that it crosses. By the planar duality theorem for cycle matroids, applied to this sphere embedding, the cycle matroid of the geometric dual satisfies \begin{align*} M(G^\dagger) \cong M(G)^* \end{align*} under the above edge identification. Here we are citing a result not yet in the wiki: planar duality theorem for cycle matroids. Since $G^\dagger$ is a finite graph, $M(G)^*$ is graphic. [/step] [step:Replace the representing graph by a connected graph without changing its cycle matroid] Conversely, assume that $M(G)^*$ is graphic. Then there exists a finite graph $H_0$ and a matroid isomorphism \begin{align*} \varphi: M(G)^* \longrightarrow M(H_0). \end{align*} Using $\varphi$, identify the ground set $E(H_0)$ with $E(G)$, so that we may regard \begin{align*} M(H_0) = M(G)^* \end{align*} as matroids on the same edge set. If $H_0$ is disconnected, construct a graph $H$ by choosing one vertex in each connected component of $H_0$ and identifying all chosen vertices to a single vertex. This operation changes only the vertex set and incidence between formerly separate components at the identified vertex; it does not create or destroy any closed edge-walk whose edge set is a cycle. Therefore the circuits of $M(H)$ are exactly the circuits of $M(H_0)$, and hence \begin{align*} M(H) = M(H_0) = M(G)^*. \end{align*} Thus we may assume from now on that $H$ is connected. [guided] We start with the hypothesis that $M(G)^*$ is graphic. This means precisely that there is a finite graph $H_0$ whose cycle matroid is isomorphic to $M(G)^*$. A matroid isomorphism is a bijection of ground sets preserving independent sets, equivalently preserving circuits. We use that bijection to relabel the edges of $H_0$, so both matroids have the same ground set $E(G)$ and we may write \begin{align*} M(H_0) = M(G)^* \end{align*} after this relabelling. Whitney's abstract duality theorem is stated for connected graphs, so we next arrange connectedness without changing the cycle matroid. Suppose $H_0$ has connected components $C_1,\dots,C_m$. Choose a vertex $v_i \in V(C_i)$ for each $i \in \{1,\dots,m\}$, and define $H$ by identifying the vertices $v_1,\dots,v_m$ to a single vertex $v$ while leaving every edge of $H_0$ unchanged. Why does this preserve the cycle matroid? A circuit of a graph's cycle matroid is exactly the edge set of a graph-theoretic cycle. Identifying one vertex from each connected component cannot create a new cycle using edges from two different original components: to pass from one old component to another, a closed walk would have to go through the new vertex $v$, but inside each old component the walk portions remain separated by the same incidence data as before. Any simple cycle in $H$ therefore lies entirely inside one old component, and within that component it was already a simple cycle of $H_0$. Conversely, every cycle of $H_0$ is still a cycle of $H$, because no edge has been removed and no incidence inside a component has been changed except at the chosen vertex. Thus $H$ and $H_0$ have exactly the same cycle edge sets, hence the same cycle matroid: \begin{align*} M(H) = M(H_0). \end{align*} Combining this equality with the relabelled representation of $M(G)^*$ gives \begin{align*} M(H) = M(G)^*. \end{align*} The graph $H$ is connected by construction, so it is a connected graphic representative of the dual matroid. [/guided] [/step] [step:Translate cycles of $H$ into bonds of $G$] Let $\mathcal{C}(H)$ denote the set of circuit edge sets of the graph $H$, and let $\mathcal{B}(G)$ denote the set of bonds of $G$, meaning the minimal nonempty edge cuts of $G$. Since \begin{align*} M(H) = M(G)^*, \end{align*} the circuits of $H$ are exactly the circuits of the dual matroid $M(G)^*$. By the definition of dual matroid circuits, these are exactly the cocircuits of $M(G)$. For a finite connected graph, the cocircuits of its cycle matroid are exactly its bonds; here we are citing a result not yet in the wiki: bond-cocircuit theorem for cycle matroids. Therefore \begin{align*} \mathcal{C}(H) = \mathcal{B}(G). \end{align*} [/step] [step:Translate bonds of $H$ into cycles of $G$] Let $\mathcal{B}(H)$ denote the set of bonds of $H$, and let $\mathcal{C}(G)$ denote the set of circuit edge sets of $G$. Because $M(H) = M(G)^*$, dualizing gives \begin{align*} M(H)^* = M(G). \end{align*} The bonds of $H$ are the cocircuits of $M(H)$ by the bond-cocircuit theorem for cycle matroids. These cocircuits are precisely the circuits of $M(H)^*$, and since $M(H)^* = M(G)$, they are precisely the circuits of $M(G)$. Hence \begin{align*} \mathcal{B}(H) = \mathcal{C}(G). \end{align*} [/step] [step:Apply Whitney's abstract duality theorem to obtain a sphere embedding] The graphs $G$ and $H$ are finite and connected, they have the same edge set, and the preceding two steps show that cycles of one graph are exactly bonds of the other: \begin{align*} \mathcal{C}(H) = \mathcal{B}(G) \end{align*} and \begin{align*} \mathcal{B}(H) = \mathcal{C}(G). \end{align*} Thus $G$ and $H$ satisfy the hypotheses of Whitney's abstract duality theorem; here we are citing a result not yet in the wiki: Whitney's abstract duality theorem for graphs. Whitney's theorem gives a cellular embedding of $G$ in the sphere $S^2$ for which $H$ is a geometric dual graph. In particular, $G$ admits an embedding in $S^2$, and therefore $G$ is planar. This proves the converse implication and completes the proof. [/step]