Definition talk:Inverse of Subset
This is confusing. Is it really supposed to be cancellable elements or only invertible ones? If just cancellable, the set of left inverses of elements of $X$ can be different from the set of right inverses, no? --Dfeuer (talk) 19:42, 4 February 2013 (UTC)
PM, do you have any sources for the monoid definition other than the one exercise in Warner? For the reason above, it strikes me as shaky. To compound the shakiness, not every cancellable element even has a left or right inverse, so cancellability only implies that no element has more than one inverse in the inverse subset. --Dfeuer (talk) 20:35, 4 February 2013 (UTC)
- Invertible Element of Monoid is Cancellable, Inverse in Monoid is Unique --prime mover (talk) 21:06, 4 February 2013 (UTC)
- You don't need it. a) The only invertible elements of $S$ are going to be found in $C$. Therefore $X \subseteq C$. b) Then, if any element of $X$ actually does have an inverse, that element will be found in $X^{-1}$. Nothing in this definition to stop $X^{-1} = \varnothing$. --prime mover (talk) 21:17, 4 February 2013 (UTC)
- Under such conditions, right inverses are left inverses, or so I believe. --prime mover (talk) 21:50, 4 February 2013 (UTC)
- Not true, e.g. consider the left and right shift operators on sequences ($(a_n)_n \mapsto (a_{n \pm 1})_n$ with the convention $a_{-1} = 0$), considered as (set) endomorphisms of $\C^\N$ which evidently form a monoid. --Lord_Farin (talk) 22:06, 4 February 2013 (UTC)
PM has convinced me about that, actually. Cancellability in the set should get you left inverse is right inverse. If $a$ is cancellable and $ab = e$ then $aba = a = ae$ so $ba = e$. Also if $ba=e$ then $aba=a=ea$ so $ab=e$. --Dfeuer (talk) 22:16, 4 February 2013 (UTC)
- Ah forgot about cancellability. Obviously left shift isn't cancellable. --Lord_Farin (talk) 22:18, 4 February 2013 (UTC)