identity arrow on $n$ is ${\langle [\mj_0],n,n\rangle}$. We can prove the following.

The functor from $\cal S$ to ${\cal S}^{\ast t}$ giving this isomorphism is identity on objects and an arrow $f\!:n\str m$ is mapped to ${\langle [f], n, m\rangle}$. To prove the proposition, we rely on the fact that for every element $[f]$ of ${\cal
S}^\ast$, and every arrow ${g:\!n\str m}$ in $[f]$, the arrow $g$ is the only arrow in $[f]$ of the type ${n\str m}$. To establish this fact we need to establish the following.

An analogous lemma (called the $\cal L$ Cancellation Lemma) was proved in [@DP03], but with a restriction on the type of $f$ and $g$. It was stated there that the restriction can be lifted, and it was suggested how to achieve that. The suggestion envisaged two ways, one of which is rather straightforward (both of these ways are however lengthy, and for lack of space a detailed exposition was omitted.) As a matter of fact, there is a direct way to prove the $\cal S$ Cancellation Lemma along the lines of the ${\cal K}_c$ Cancellation Lemma of [@DP03]. Here are indications for this direct proof (which presuppose an acquaintance with [@DP03]).

[Proof 