3 ms·
I think you're right. Your version of the definition is more elegantly stated, and I may want to use it going forward, as long as a, b, and c still make intuiti
by tylerneylon 9y ago
I think you're right. Your version of the definition is more elegantly stated, and I may want to use it going forward, as long as a, b, and c still make intuitive sense (thinking about that part).
- contravariant 9y agoFeel free to use it. Using your notation you have a=t1, b=t1∘s, c=t2^-1. Inversely you have s=a^-1∘b, t1=a, t2=c^-1.