a generalization of unit elements
If I have a monoid
, and an element
, then the following equivalence holds:
is a unit iff
and
are bijections onto
.
Proof. Suppose
is a unit. Let
Then we have
1. ;)
2. if
then
so 
Conversely, if
and
are bijections onto
then there exists
such that
and
such that
Thus
has a left inverse and a right inverse, which is known to imply that
is a unit.
Now let's say
is a semigroup without an identity element. Can there be an element
such that
and
are bijections? Is it possible when
is finite? (Then it's equivalent to asking whether a semigroup without ideantity can have a cancellable element.) Is it possible when
is infinite? Is it possible when we only demand that at least one of
be a bijection onto 
Re: a generalization of unit elements
a partial answer:
if such bijections x→xu and x→ux exists for some element u, then for some element x:
xu = u, and since for ANY y in S we have y = ua (for some a): xy = x(ua) = (xu)a = ua = y. thus x is a left-identity for S.
similarly, there must be some z in S with uz = u, and thus (writing y = bu), yz = (bu)z = b(uz) = bu = y, so z is a right-identity for S.
but then x = xz = z, so S possesses an identity element, which we can call e.
so in THAT case, we have xu = e and uy = e for some x,y in S, so u is indeed a unit (and x = y: x = xe = x(uy) = (xu)y = ey = y).
note that above, we only require that x→ux, x→xu be surjective to show that a right- or left-identity exists, uniqueness of this identity follows (for one of left- or right-) if one of those maps is injective as well (if S is finite, then surjective = bijective) (if x→ux is injective, then we have a unique right-identity, if x→xu is injective, we have a unique left-identity).
i am unsure of how you would define a unit without the presence of a two-sided identity.
Re: a generalization of unit elements