Originally Posted by

**topsquark** Here is the question:

Just in case:

I have constructed a couple of monoids with homomorphisms, but I'm not getting anywhere. In addition I have a proof of $\displaystyle f(e_G) = e_H$ that does not use the inverses relationship $\displaystyle f \left ( a^{-1} \right ) = f(a)^{-1}$, so I don't see where this would enter the problem. Additionally the proof does not seem to give me any clues as to how $\displaystyle f(e_G)$ might not be equal to $\displaystyle e_H$.

The proof is short so I will include it. Perhaps someone may be able to see a loophole that is escaping me.

Proof:

Let G be a set with a binary relation * and let there be an element of G, $\displaystyle e_G$ such that $\displaystyle e_G*g = g*e_G = g$ for all g in G. Let H be another set with a binary relation $\displaystyle \times$ and let there be an element $\displaystyle e_H$ such that $\displaystyle e_H \times h = h \times e_H = h$ for all h in H. Further let f be a homeomorphism $\displaystyle f: G \to H$.

Then

$\displaystyle f(a) \times f(e_G) = f(a*e_G) = f(a)$. Since by definition $\displaystyle h \times e_H = h$ for all h, $\displaystyle f(e_G)$ must be $\displaystyle e_H$.

-Dan