If $M$ is a matroid, and $e \in M$, than what is the closure function of $M/e$, and what is the closure function of $M\setminus e$? If $M$ is a matroid, and $e \in M$, than what is the closure function of $M/e$, and what is the closure function of $M\setminus e$?
What I find confusing to start with, is what elements does the closure of matroid $M$ contain. Isn't it all of the elements?
Thanks!   
 A: Closures of sets in a matroid are flats. I forget the textbook definition of the closure operation, but intuitively, you can think of $\mathrm{cl}(X)$ as the union of $X$ and the set $\{e \in M : X \cup e$ has a new circuit that's not entirely in $X \}$.
So the question is, what does deletion and contraction do to circuits? If $C$ is a circuit and $e \in C$, then contracting $e$ leaves $C$ intact, just with one less element. So if $C \subseteq \mathrm{cl}(X)$ in $M$, then $C-e \subseteq \mathrm{cl}(X-e)$ in $M/e$. However, if $e \notin C$, then in $M/e$, $C$ is more accurately a union of circuits. Consider a graph that is just a cycle with a chord through it. If you contract the chord, the cycle becomes two cycles joined at a vertex. Therefore it's possible that $\mathrm{cl}_{M}(X) - e \subseteq \mathrm{cl}_{M/e}(X)$.
Deleting works the other way: instead of possibly having more elements in the closure, you have less. If $e \notin C$, then deleting $e$ does nothing to $C$. But if $e \in C$, then in $M\backslash e$, $C$ is no longer a cycle. Consider a graph that's just one big cycle and let $X$ be the whole thing except an edge $e$. Then $\mathrm{cl}_{M}(X) = M$. But if you delete any edge in the graph, then there are no cycles so every set is closed, and $\mathrm{cl}_{M\backslash e}(X) = X$. In general, $\mathrm{cl}_{M\backslash e}(X) \subseteq \mathrm{cl}_{M}(X) - e$.
And to answer your other question, yes, $\mathrm{cl}(M) = M$. You know this because the closure of a set always contains the set, and there's nothing else in the matroid to add, so it has to be just $M$.
For a longer explanation, if $M$ has rank $r$, take any basis $B$. Then $B$ has $r$ elements, and one property of a basis is that if you add any element $e$, then $B \cup e$ has a circuit. So $\mathrm{cl}(B) = M$, thus proving that $M$ is the only rank-$r$ flat. And the closure of a flat is just the flat.
