Inverse element in associative structure not necessarily unique [duplicate]

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I read from the linkhere proving that the inverse element in associative structure should be unique. However, it's only proving that should there exist a left and a right inverse element, then they should be identical. And I remember reading examples in non-commutative rings (so there should also be a similar case in monoids), an element has multiple multiplicative left inverse elements. But I'm not quite sure anymore. Could anybody reproduce such an example?

Also, interestingly, even for non-commutative groups, the inverse element is unique. Does all of this have to do with the fact that every element in the group is assigned with an inverse element?

marked as duplicate by rschwieb abstract-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Apr 22 '17 at 12:46

• Let $E$ be a set and let $S$ be the semigroup of all maps $f:E\to E.$ If $f$ is injective but not surjective, then $f$ has more than one left inverse but no right inverse. If $f$ is surjective but not injective, then $f$ has more than one right inverse but no left inverse. – bof Apr 22 '17 at 2:20
• However, it's only proving that should there exist a left and a right inverse element, then they should be identical. Well, yes, if you have that then it's obvious there's only one left inverse. If $a'$ is any left inverse of $a$, then $a'a=1$, and right multiplying with $a^{-1}$ you get $a'=a^{-1}$. – rschwieb Apr 22 '17 at 12:48
Or, take the monoid $E$ of functions $\Bbb N\to\Bbb N$ under composition. Now, e.g. $s:n\mapsto n+1$ has infinitely many left inverses, as it has freedom on what to assign to $0\notin{\rm im}(s)$.
If you want, you can make a ring out of this by adding elements (with integer [or whatever] coefficients) freely - denoted by $\Bbb Z[E]\$ [or $R[E]$ for arbitrary coefficient ring $R$].
For another example, along the same line, we can find a linear operator corresponding to $s$ on an infinite dimensional vector space (e.g. separable Hilbert space).