3.1 Homomorphisms
We start with the official definitions.
Suppose \(M\) and \(N\) are magmas. A magma homomorphism from \(M\) to \(N\) is a map \(f:M\to N\) such that, for all \(x,y\in M\), \(f(xy) = f(x)f(y)\).
Suppose \(M\) and \(N\) are monoids with identity elements \(e_M\) and \(e_N\). A monoid homomorphism from \(M\) to \(N\) is a monoid homomorphism \(f:M\to N\) with the additional property that \(f(e_M) = e_N\).
Suppose \(M\) and \(N\) are groups. A group homomorphism from \(M\) to \(N\) is a monoid homomorphism with the additional property that, for all \(x\in M\), \(f(x^{-1}) = f(x)^{-1}\).
A magma (resp. monoid, resp. group) homomorphism \(f\) is said to be a magma (resp. monoid, resp. group) isomorphism if it is one-one and onto.
A magma (resp. monoid, resp. group) homomorphism \(f:M \to M\) is called a magma (resp. monoid, resp. group) endomorphism.
A magma (resp. monoid, resp. group) isomorphism \(f:M \to M\) is called a magma (resp. monoid, resp. group) automorphism.
We are mostly going to be interested in group homomorphisms, isomorphisms and (sometimes) automorphisms. The following Proposition is helpful in this regard.
Suppose \(G\) and \(H\) are groups and \(f:G\to H\) is a magma homomorphism. Then \(f\) is a group homomorphism as well. In other words, a map \(f:G\to H\) is a group homomorphism if and only if for all \(x,y\in G\), \(f(xy) = f(x) f(y)\).
Suppose \(G\) and \(H\) are groups and \(f:G\to H\) is a magma homomorphism. Write \(e_G\) (resp. \(e_H\)) for the identity element of \(G\) (resp. \(H\)). Then \(f(e_G) = f(e_G) e_H = f(e_G) (f(e_G)f(e_G)^{-1}) = (f(e_G) f(e_g)) f(e_G)^{-1} = f(e_G) f(e_G)^{-1} = e_H\). So \(f:G\to H\) is a monoid homomorphism.
On the other hand, suppose \(g\in G\). Then
So \(f\) is a group homomorphism.
The analogue of Proposition 3.2 is not true for monoids. In other words, there are examples of monoids \(M\) and \(N\) with identity elements \(e_M\) and \(e_N\) respectively and magma homomorhisms \(f:M\to N\) such that \(f(e_M) \neq e_N\).
In fact, here’s an easy example. Let \(N = \{ 0,1\} \) with the binary operation \(*\) given by setting \(0*0 = 0*1 = 1*0 = 0\) and \(1*1 = 1\). It’s not hard to check directly that \(N\) is associative. So it is a monoid with identity element \(1\). The subset \(M =\{ 0\} \) is a submagma. In other words, it is closed under the binary operation. But, according to Definition 2.29, it is not a submonoid because it does not contain the identity element \(1\) of \(M\).
Write \(f:N\to M\) for the inclusion map given by \(f(0) = 0\). Then \(f\) is a magma homomorphism, but it is not a monoid homomorphism.
If \(S\) is any set, we write \(\operatorname{\mathrm{id}}_S\) for the map \(\operatorname{\mathrm{id}}_S: S\to S\) given by \(\operatorname{\mathrm{id}}_S(s) = s\). This map, which is called the identity map is trivial, but it is also fundamental.
Recall that a map of sets \(f:S\to T\) is a bijection if and only if there exists an inverse function \(g:T\to S\) such that \(f\circ g = \operatorname{\mathrm{id}}_T\) and \(g\circ f = \operatorname{\mathrm{id}}_S\). The inverse function \(g\) is then unique and is usually written as \(f^{-1}:T\to S\). It follows directly that, in this case, \(f^{-1}:T\to S\) is also a bijection.
If \(M\) is magma (resp. monoid, resp. group) then \(\operatorname{\mathrm{id}}_M:M\to M\) is a magma (resp. monoid, resp. group) endomorphism.
Suppose \(M\) and \(N\) are magmas and \(f:M\to N\) is a magma isomorphism. Then the inverse function \(f^{-1}:N\to M\) is also a magma isomorphism.
Suppose \(M\) and \(N\) are monoids and \(f:M\to N\) is a monoid isomorphism. Then the inverse function \(f^{-1}:N\to M\) is also a monoid isomorphism.
Suppose \(M\) and \(N\) are groups and \(f:M\to N\) is a group isomorphism. Then the inverse function \(f^{-1}:N\to M\) is also a group isomorphism.
1 Exercise. (It’s basically obvious.)
2 Suppose \(f:M\to N\) is a bijection of magmas, and \(n_1, n_2\in N\). Set \(m_i := f^{-1}(n_i)\) for \(i=1,2\) so that \(f(m_i) = n_i\).
So \(f^{-1}:N\to M\) is a homomorphism of magmas. As it is a bijetion, it is also an isomorphism of magmas.
3 Suppose \(f:M\to N\) is a bijection of monoids with \(e_M\) (resp. \(e_N\)) the identity element of \(M\) (resp. \(N\)). Then \(f^{-1}:N\to M\) is a magma isomorphism by 2. And \(f(e_M) = e_N\), which implies that \(f^{-1}(e_N) = e_M\). So \(f^{-}:N\to M\) is also a monoid homomorphism. As it is a bijection, it is also a monoid isomorphism.
Write \(\mathbb {R}_+ := (0,\infty ) = \{ x\in \mathbb {R}: x {\gt} 0\} \). It’s easy to see that \((\mathbb {R}_+, *)\) is a group where \(*\) is the usual multiplication. Then let \(G = (\mathbb {R}, +)\), and let \(H = (\mathbb {R}_+, *)\). From Calculus, we know that \(\exp (x+y) = \exp (x) \exp (y)\) and \(\exp (x) {\gt} 0\) for all \(x\). So the map \(\exp :G\to H\) is a group homomorphism.
Here I introduced the symbols \(G\) and \(H\) to make it absolutely clear what the binary operations are, but usually, I would express this more directly by saying that \(\exp :\mathbb {R}\to \mathbb {R}_{+}\) is a group homomorphism. (The assumption is that the reader can see that addition is the only obvious binary operation that makes \(\mathbf{R}\) into a group and multiplication is the only obvious binary operation that makes \(\mathbb {R}_+\) into a group.)
In fact, \(\exp :\mathbb {R}\to \mathbb {R}_+\) is an isomorphism of groups with inverse \(\log :\mathbb {R}_+\to \mathbb {R}\).
The composition \(S\circ T:G\to K\) of two group homoorphisms \(T:G\to H\) and \(S:H\to K\) is a group homomorphism. (And similarly for magma and monoid homomorhisms).
Let’s prove it for group homomorphisms. So suppose \(T:G\to H\) and \(S:H\to K\) are group homomorphisms and \(x,y\in G\). Then \((S\circ T)(xy) = S(T(xy)) = S(T(x)T(y)) = S(T(x))S(T(y)) = (S\circ T)(x) (S\circ T)(y)\). So, by Proposition 3.2, we’re done.
The same proof (of course) works for magma homomorphisms, and the proof for monoid homomorphisms is easy (and left to the reader).