I am personally fond of matroid duality. Let $M$ be a matroid with ground set $E$. The dual of $M$ is the matroid with ground set $E$ and whose bases are the complements of bases of $M$. It is easy to verify that the dual of $M$ is indeed a matroid and we immediately have that $M^{dd}=M$.

Matroid duality illustrates that deletion and contraction are actually dual operations. That is, deletion corresponds to contraction in the dual and vice versa.

It also nicely generalizes duality for planar graphs. That is, if $G$ is a planar graph, and $M(G)$ is the cycle matroid of $G$, then $M^d(G)=M(G^d)$ (here, $G^d$ is the planar dual of $G$).

Finally, here is a proof for free of Euler's Formula via matroid duality. Let $G$ be a connected planar graph with edge set $E$. Suppose that $G$ has $v$ vertices, $e$ edges and $f$ faces. Let $r$ be the rank function of the cycle matroid of $G$ and let $r_d$ be the rank function of the dual of the cycle matroid of $G$. Then

\[ e=r(E)+r_d(E)=(v-1)+(f-1). \]

I am personally fond of matroid duality. Let $M$ be a matroid with ground set $E$. The dual of $M$ is the matroid with ground set $E$ and whose bases are the complements of bases of $M$. It is easy to verify that the dual of $M$ is indeed a matroid and we immediately have that $M^{dd}=M$.

Matroid duality illustrates that deletion and contraction are actually dual operations. That is, deletion corresponds to contraction in the dual and vice versa.

It also nicely generalizes duality for planar graphs. That is, if $G$ is a planar graph, and $M(G)$ is the cycle matroid of $G$, then $M^d(G)=M(G^d)$ (here, $G^d$ is the planar dual of $G$).

Finally, here is a proof for free of Euler's Formula via matroid duality. Let $G$ be a connected planar graph with edge set $E$. Suppose that $G$ has $v$ vertices, $e$ edges and $f$ faces. Let $r$ be the rank function of the cycle matroid of $G$ and let $r_d$ be the rank function of the dual of the cycle matroid of $G$. Then

\[ e=r(E)+r_d(E)=(v-1)+(f-1). \]

I am personally fond of matroid duality. Let $M$ be a matroid with ground set $E$. The dual of $M$ is the matroid with ground set $E$ and whose bases are the complements of bases of $M$. It is easy to verify that the dual of $M$ is indeed a matroid and we immediately have that $M^{dd}=M$.

Matroid duality illustrates that deletion and contraction are actually dual operations. That is, deletion corresponds to contraction in the dual and vice versa.

It also nicely generalizes duality for planar graphs. That is, if $G$ is a planar graph, and $M(G)$ is the cycle matroid of $G$, then $M^d(G)=M(G^d)$ (here, $G^d$ is the planar dual of $G$).

Finally, here is a proof for free of Euler's Formula via matroid duality. Let $G$ be a connected planar graph with edge set $E$. Suppose that $G$ has $v$ vertices, $e$ edges and $f$ faces. Let $r$ be the rank function of the cycle matroid of $G$ and let $r_d$ be the rank function of the dual of the cycle matroid of $G$. Then

\[ e=r(E)+r_d(E)=(v-1)+(f-1). \]

I am personally fond of matroid duality. Let $M$ be a matroid with ground set $E$. The dual of $M$ is the matroid with ground set $E$ and whose bases are the complements of bases of $M$. It is easy to verify that the dual of $M$ is indeed a matroid and we immediately have that $M^{dd}=M$.

Matroid duality illustrates that deletion and contraction are actually dual operations. That is, deletion corresponds to contraction in the dual and vice versa.

It also nicely generalizes duality for planar graphs. That is, if $G$ is a planar graph, and $M(G)$ is the cycle matroid of $G$, then $M^d(G)=M(G^d)$ (here, $G^d$ is the planar dual of $G$).

Finally, here is a proof for free of Euler's Formula via matroid duality. Let $G$ be a connected planar graph with edge set $E$. Suppose that $G$ has $v$ vertices, $e$ edges and $f$ faces. Let $r$ be the rank function of the cycle matroid of $G$ and let $r_d$ be the rank function of the dual of the cycle matroid of $G$. Then

\[ e=r(E)+r_d(E)=(v-1)+(f-1). \]