These ideas are being used a lot, but I cannot justify why they are correct:
If M is a matroid and $T$ a subset of $E(M).$ Then $$(a)\ cl(T) = T \cup \{e \in E(M) - T: e \text{ is a loop of M/T}\}.$$ and $$(b)\ cl^* (T) = T \cup \{e \in E(M) - T: e \text{ is a coloop of M\T}\}.$$
I know the following:
$(i)\ cl_{M\setminus T}(X) = cl_M(X) - T.$
$(ii)\ cl_M(X) = \{ x \in E : r(X \cup x) = r(X)\}.$
$(iii)\ cl_{M/T}(X) = cl_M(X \cup T) - T \text{ for all X \subseteq E - T.}$
And I know the properties of the closure operator. I also know that $M \setminus e = M/e$ iff $e$ is a loop or a coloop of $M.$
Any explanation will be greatly appreciated.
First, $\operatorname{cl}_M(X)=\{x∈E:r(X∪x)=r(X)\}$ is usually taken to be the definition of the closure operator. Recall that a loop in a matroid is an element $e$ such that $r(\{e\})=0$.
Let $T\subseteq E(M)$. For part $(a)$ we want to show that the elements $e\in E(M)-T$ such that $e$ is a loop in $M/T$ are exactly the elements outside $T$ we can add to $T$ without increasing the rank of $T$. Recall that for any $S\subseteq E(M)-T$, $$r_{M/T}(S) = r_M(S\cup T)-r_M(T).$$ Letting $S=\{e\}$, this gives
$$r_{M/T}(\{e\}) = r_M(\{e\}\cup T)-r_M(T).$$
We are now ready to prove the assertion. It is clear that $T\subseteq \operatorname{cl}_M(T)$. Now, if $e$ is a loop in $M/T$, then $r_{M/T}(\{e\}) = 0$, so the above equation yields $ r_M(\{e\}\cup T) = r_M(T)$, i.e., $e\in \operatorname{cl}_M(T)$. Similarly, if $e\in \operatorname{cl}_M(T)$ either $e\in T$, or $e\in E(M)-T$ and $r_M(\{e\}\cup T) = r_M(T)$. This implies $r_{M/T}(\{e\}) = 0$ by the above equation, i.e., $e$ is a loop in $M/T$.
Now, $(b)$ is exactly the dual statement of $(a)$, since $(M/T)^* = M^*\backslash T$, and loops and coloops are dual notions. This occurrence of two statements dual to each other is a common occurrence in the theory of matroids.