set of all monoid homomorphisms $(\mathbb N_0, +) \to M$ Let $M$ be a monoid. Determine the set of all monoid homomorphisms $(\mathbb N_0, +) \to M$.
I know that for a monoid homomorphism the following has to be true:

*

*$f(a\circ b) = f(a) \ast f(b)$

*$f(e) = e$
Here, $\circ = +$ on the natural numbers and $\ast$ is not explicitly given. $e$ is the identity element.
Now, if for example $\ast = \cdot$, so the basic multiplication, then the function $f: x \mapsto e^x$ would be such a homomorphism. $f(a+b) = f(a) \cdot f(b)$ would be $e^{a+b} = e^a \cdot e^b$ with $e^0 = 1$ which is true because $0$ is the identity element for $+$ and $1$ is the identity element for $\cdot$ .
But because $*$ is not explicitly given, how to find the monoid homomorphisms?
 A: It is sufficient to know where $1$ is sent to. After that, one has $f(0)=e, f(n) = f(1+1+\cdots+1) = f(1)^n$ for any monoid homomorphism $f$. This shows that $f(1)$ determines $f$ totally. Next, given any $m\in M$, let $f_m\colon\mathbb{N}\to M$ be given by $f_m(0)=0, f_m(n) = m^n$. Then there is a function $\varphi$ from the monoid $M$ to the set $\text{Hom}_{\text{Mon}}(\mathbb{N}, M)$ of monoid homomorphisms from the naturals to $M$, given by $\varphi(m) = f_m$. Conversely, there is the evaluation function $\varepsilon$ from $\text{Hom}_{\text{Mon}}(\mathbb{N}, M)$ to $M$, given by $\varepsilon(f) = f(1)$. Starting with any $m\in M$, we have $(\varepsilon\circ\varphi)(m) = f_m(1) = m$, and similarly $(\varphi\circ\varepsilon)(f)=\varphi(f(1))=f_{f(1)}=f$ for any $f\in\text{Hom}_{\text{Mon}}(\mathbb{N}, M)$. Thus, there is a bijection between those. If you want to do some extra, you can show that $\text{Hom}_{\text{Mon}}(\mathbb{N}, M)$ is actually a monoid under $f_m\cdot f_n = f_{m\cdot n}$, and then $\varphi$ is an isomorphism of monoids.
A: Such a homomorphism is completely characterised by $f(1)$. Using a multiplicative notation for the monoid $M$, one gets
$f(0) = e$ by definition and $f(n) = f(1)^n$ since $n = \underbrace{1 + \dotsm + 1}_{\text{$n$ times}}$.
