Re: Equational correspondence and equational embedding

[Janus <janus@rtfm.org.ar>]

Dear Prof. Waddler,

   I don't know if the following proof is correct, please, let me know
if I made any mistake:

   By hypothesis there exists k:S->T and h:k(S)->S such that h(k(S))
=_S s for all s in S and s =_S s' implies k(s) =_T k(s') for all s, s'
in S
   On the other hand, there exists f:T->S and g:f(T)->S such that
g(f(t)) =_T t for all t in T and t=_T t' implies f(t) =_S f(t') for
all t, t' in T

   Assume that there exists t in T\k(S) => |T| > |S| because k is
injective and S = dom(k). Hence, there exists t' not equal to t such
that f(t) =_S f(t') which is absurdum.

   So, there not exists t in T\k(S), then T = k(S). So, k and k^-1
constitutes an equational correspondence.

   Yours,
     Alejandro

-- 
Alejandro Díaz-Caro
Homepage: http://www.fceia.unr.edu.ar/~diazcaro
Weblog: http://computacioncuantica.exactas.org