The monoids $(S,\min)$ and $(T,\min)$, where $S = \{3,4,5,6\}$ and $T = \{1,2,3,4,5,6\}$ and $\min$ is the minimum function of two integers. Let the function $f:S \to T$ be defined by $f(x) = x-1$; which of the following statements are true?
(A) $f$ is an isomorphism.
(B) $f$ is a one-to-one homomorphism, but $f$ is not onto.
(C) $f$ is an onto homomorphism, but $f$ is not onto.
(D) $f$ is bijective, but $f$ is not a homomorphism.
(E) $f$ is one-to-one, but $f$ is neither a homomorphism nor onto.
(F) $f$ is surjective, but $f$ is neither a homomorphism nor one-to-one.
How would I go about this problem? I also get somewhat confused by the $\min$ function
HINT: You should already be familiar with the properties one-to-one and onto for functions in general, so I’ll not say anything about them. The function $f$ is a monoid homomorphism if and only if it has the following two properties:
for all $m,n\in S$, $f\big(\min\{m,n\}\big)=\min\big\{f(m),f(n)\big\}$. By the definition of $f$ this means that $\min\{m,n\}-1=\min\{m-1,n-1\}$; is that always true?
$f(1_S)=1_T$, where $1_S$ is the identity element of $S$, and $1_T$ is the identity element of $T$. Your first step here must be to figure out which element of $S$ is $1_S$, and which element of $T$ is $1_T$. Remember, $1_S$ has the property that $\min\{n,1_S\}=n$ for all $n\in S$, and $1_T$ has a similar property with respect to $T$.
A monoid isomorphism is a homomorphism that is also one-to-one and onto.