[proofplan] We prove the forward implication directly and then invoke the same implication for the dual matroid $M^*$. Write $G := E \setminus F$. If $F$ is a cyclic flat of $M$, then flatness of $F$ shows that deleting any element of $G$ from $G$ does not lower the dual rank, so no element of $G$ is a coloop of the restriction $M^*|G$. The cyclicity of $F$ shows that adjoining any element of $F$ to $G$ raises dual rank, so $G$ is closed in $M^*$. Thus $G$ is both cyclic and flat in $M^*$. [/proofplan] [step:Translate the hypotheses on $F$ into rank conditions in $M$] Let $r_M: 2^E \to \mathbb{Z}_{\ge 0}$ denote the rank function of the matroid $M$, and let $r_{M^*}: 2^E \to \mathbb{Z}_{\ge 0}$ denote the rank function of the dual matroid $M^*$. Define $G := E \setminus F$. Assume first that $F$ is a cyclic flat of $M$. Since $F$ is a flat of $M$, every element $e \in E \setminus F = G$ satisfies $r_M(F \cup \{e\}) = r_M(F) + 1$. Since $F$ is cyclic in $M$, the restriction $M|F$ has no coloops. Equivalently, for every element $e \in F$, $r_M(F \setminus \{e\}) = r_M(F)$. Indeed, an element $e \in F$ is a coloop of $M|F$ exactly when deleting $e$ lowers the rank of the restricted matroid, which is the condition $r_M(F \setminus \{e\}) = r_M(F) - 1$. [/step] [step:Show that no element of $G$ is a coloop in the restriction $M^*|G$] We use the dual rank formula for the dual matroid: for every subset $A \subset E$, \begin{align*} r_{M^*}(A) = |A| - r_M(E) + r_M(E \setminus A). \end{align*} Fix $e \in G$. Applying the formula to $A = G$ and to $A = G \setminus \{e\}$ gives \begin{align*} r_{M^*}(G) - r_{M^*}(G \setminus \{e\}) = 1 + r_M(E \setminus G) - r_M(E \setminus (G \setminus \{e\})). \end{align*} Because $E \setminus G = F$ and $E \setminus (G \setminus \{e\}) = F \cup \{e\}$, this becomes \begin{align*} r_{M^*}(G) - r_{M^*}(G \setminus \{e\}) = 1 + r_M(F) - r_M(F \cup \{e\}). \end{align*} By flatness of $F$, we have $r_M(F \cup \{e\}) = r_M(F) + 1$, hence \begin{align*} r_{M^*}(G) - r_{M^*}(G \setminus \{e\}) = 0. \end{align*} Thus deleting $e$ from $G$ does not lower the rank of the restriction $M^*|G$. Since $e \in G$ was arbitrary, $M^*|G$ has no coloops, so $G$ is cyclic in $M^*$. [guided] The goal of this step is to prove cyclicity of $G$ inside the dual matroid. For a subset $G$ of the ground set of a matroid, saying that $G$ is cyclic means that the restriction to $G$ has no coloops. Thus we must show that no element $e \in G$ is rank-essential in the restricted matroid $M^*|G$. Let $e \in G$ be arbitrary. In the restriction $M^*|G$, the rank of a subset of $G$ is computed using the same rank function $r_{M^*}$ as in the ambient matroid $M^*$. Therefore $e$ is a coloop of $M^*|G$ exactly when $r_{M^*}(G) - r_{M^*}(G \setminus \{e\}) = 1$. We will prove that this difference is instead equal to $0$. The dual rank formula for the dual matroid says that for every subset $A \subset E$, \begin{align*} r_{M^*}(A) = |A| - r_M(E) + r_M(E \setminus A). \end{align*} Applying this formula first to $A = G$ gives \begin{align*} r_{M^*}(G) = |G| - r_M(E) + r_M(E \setminus G). \end{align*} Applying it next to $A = G \setminus \{e\}$ gives \begin{align*} r_{M^*}(G \setminus \{e\}) = |G \setminus \{e\}| - r_M(E) + r_M(E \setminus (G \setminus \{e\})). \end{align*} Since $e \in G$, the finite cardinality satisfies $|G \setminus \{e\}| = |G| - 1$. Subtracting the second equality from the first gives \begin{align*} r_{M^*}(G) - r_{M^*}(G \setminus \{e\}) = 1 + r_M(E \setminus G) - r_M(E \setminus (G \setminus \{e\})). \end{align*} Now use the definition $G := E \setminus F$. This gives $E \setminus G = F$. Since $e \in G = E \setminus F$, removing $e$ from $G$ means that $e$ moves into the complement, so $E \setminus (G \setminus \{e\}) = F \cup \{e\}$. Substituting these two identities into the rank difference yields \begin{align*} r_{M^*}(G) - r_{M^*}(G \setminus \{e\}) = 1 + r_M(F) - r_M(F \cup \{e\}). \end{align*} Because $F$ is a flat of $M$ and $e \notin F$, adjoining $e$ to $F$ raises rank by one: \begin{align*} r_M(F \cup \{e\}) = r_M(F) + 1. \end{align*} Therefore \begin{align*} r_{M^*}(G) - r_{M^*}(G \setminus \{e\}) = 0. \end{align*} So $e$ is not a coloop of $M^*|G$. Since the argument works for every $e \in G$, the restriction $M^*|G$ has no coloops. Hence $G$ is cyclic in $M^*$. [/guided] [/step] [step:Show that $G$ is closed in the dual matroid] It remains to prove that $G$ is a flat of $M^*$. Let $e \in E \setminus G = F$. Applying the dual rank formula for the dual matroid to $A = G \cup \{e\}$ and to $A = G$ gives \begin{align*} r_{M^*}(G \cup \{e\}) - r_{M^*}(G) = 1 + r_M(E \setminus (G \cup \{e\})) - r_M(E \setminus G). \end{align*} Since $E \setminus G = F$ and $E \setminus (G \cup \{e\}) = F \setminus \{e\}$, this is \begin{align*} r_{M^*}(G \cup \{e\}) - r_{M^*}(G) = 1 + r_M(F \setminus \{e\}) - r_M(F). \end{align*} Because $F$ is cyclic in $M$, the element $e$ is not a coloop of $M|F$, so $r_M(F \setminus \{e\}) = r_M(F)$. Therefore \begin{align*} r_{M^*}(G \cup \{e\}) - r_{M^*}(G) = 1. \end{align*} Thus every element outside $G$ raises rank when adjoined to $G$ in $M^*$, so no element outside $G$ lies in the closure of $G$ in $M^*$. Hence $G$ is a flat of $M^*$. [/step] [step:Apply duality again to obtain the converse] The preceding two steps show that if $F$ is a cyclic flat of $M$, then $E \setminus F$ is a cyclic flat of $M^*$. Applying this implication to the dual matroid $M^*$ and the subset $G := E \setminus F$, and using the dual-dual identity $(M^*)^* = M$, gives: if $G$ is a cyclic flat of $M^*$, then $E \setminus G = F$ is a cyclic flat of $M$. Therefore $F$ is a cyclic flat of $M$ if and only if $E \setminus F$ is a cyclic flat of $M^*$. [/step]