# Contraction of loops in matroids

If $M=(E,I)$ is a matroid, and $e$ is not a loop (a loop is an element of the matroid which is not an element of any independent set), we may define the matroid obtained by contracting $e$ to be the matroid whose edges are $E\smallsetminus\{e\}$ and whose independent sets are $\{A\smallsetminus\{e\}\mid A\cup\{e\}\in I\}$.

The graphic intuition behind this works well with the graphic notation of contraction; in case of graphic matroids, both definitions coincide. One other important observation is that a contraction in $M$ is equivalent to a removal in the dual matroid $M^*$.

I've taken the graphic intuition and checked the case in which $e$ is a loop. In that case, "contracting" $e$ would mean removing it. Taking a simple example of a lemniscate graph (one vertex, two loops), the dual matroid in this case is the graphic matroid of a $V$-shaped tree (3 vertices, 2 edges). A removal in the dual would leave us with a single edge, and the dual of that is a single loop. That is, even when $e$ is a loop, at least in this example, the "contraction as a removal in the dual" definition coincides with the graphic one. In this case it also coincides with the original definition.

My question is this: why are loops omitted from the original definition? The dual question would be: why are coloops (bridges) omitted from the definition of removal?

There are no independent sets containing the loop $e$. So $\{A\smallsetminus\{e\}\mid A\cup\{e\}\in I\} = \phi$.
But this means $M^*$ has no dependent sets, which is not true (this could have been true if we consider isolated vertices after deleting edges, but we don't. We simply ignore the isolated vertices.)