If $G$ is generated by $A$, then $f(G)$ is generated by $f(A)$

Question

Let $G$ be a monoid that is generated by the subset $A$. Let $f\colon G\to H$ be a homomorphism between two monoids. Prove that $f(G)$ (the image of $f$) is created by $f(A)$.

I know as a fact that the image of $f$ ($f(G)$) is indeed a submonoid of $H$.

But when we want to prove that $f(G)$, a submonoid of $H$, is generated by $f(A)$, what particular points do we need to prove?

Would appreciate a little direction.

EDIT: Maybe if we can show the fact that $f(G)$ is the smallest submonoid (we already know that it is a submonoid) that contains $f(A)$ then this would prove the theory per definition?

Answer 1

Let $K$ be the submonoid of $H$ generated by $f(A)$ and write the operations on $G$ and $H$ additively.

For every $k \in K$ we have $k = \sum_{i \in I} f(a_i) = f(\sum_{i \in I} a_i)$ for some index set $I$ and $a_i \in A$, thus $k \in f(G)$. Therefore $K \subseteq f(G)$.

On the other hand, for every $g \in G$ we have $g = \sum_{j \in J} a_j$ for some index set $J$ and $a_j \in A$, thus $f(g) = \sum_{j \in J} f(a_j) \in K$. Therefore $f(G) \subseteq K$.

Answer 2

If $K$ is a submonoid of $H$ and $f(A)\subseteq K$ then $A\subseteq f^{-1}(K)$.

Showing that $f^{-1}(K)$ is a submonoid of $G$ is enough (can you do it?).

That allows the conclusion that $\langle A\rangle\subseteq f^{-1}(K)$ or equivalently $f(\langle A\rangle)\subseteq K$.

So in special case $G=\langle A\rangle$ we have $f(G)\subseteq K$.

Proved is now that $f(G)$ is the smallest submonoid of $H$ that contains $f(A)$. Notation: $\langle f(A)\rangle=f(G)$.

In a more general setting: $$f(\langle A\rangle)=\langle f(A)\rangle$$