Steve Vickers wrote:
Dear David,
There are more complicated examples. Here's one.
Take A to be the semigroup {0,a,b,c} in which x0=0x=0 for all x, aa=a, cc=c, ab = bc = b, and all other products are 0. (This of this as being derived from a category with two objects and three morphisms, so a and c represent the two identities and b is a morphism between the two objects. 0 takes care of products of non-composable pairs.)
Take B = A u {d}, with cd = da = d, bd=a, db=c and all other binary products involving d give 0. (Think of adjoining an inverse to b in the category.)
Now the inclusion A -> B is a semigroup epi. To see this, suppose f: A -> C is a semigroup homomorphism, and x in C satisfies
xf(a) = f(c)x = x f(b)x = f(a) xf(b) = f(c)
Then x is the unique such. For if x' is another then
x' = x'f(a) = x'f(b)x = f(c)x = x
If g: B -> C is a semigroup homomorphism agreeing with f on A, then g(d) does satisfy those equations for x, and so any two such g's must be equal.
This semigroup example can be easily turned into monoids by adjoining a unit element.
There is still the same idea that B is got by adjoining inverses to elements of A, but they are not inverses in the monoid sense and it is not clear to me in general how one would formalize the idea that they are inverses when embedded in some category.
There is a related epi in rings: the inclusion of upper triangular 2x2 matrices (over any ring) into all 2x2 matrices.
Regards,
Steve Vickers.
i like the example of the semigroups. i think in some sense the addition of d does not add enough information to our monoid so if we have two semigroup homomorphisms that agree on A then by using the properties of semigroup homomorphisms we are forced to define how they behave on d in only one way. i'm still trying to incorporate the inclusion of the 2x2 upper triangular matrices into all 2x2 matrices and Michael Barr's example into this framework.