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I'm trying to prove this isomorphism. I defined this function

$ \psi: M \rightarrow Hom(\mathbb{N}^{+}, M) \\ m \mapsto \phi(n) $ where $ \phi(n) = \begin{cases} e_M, & \text{if }n\text{ is even} \\ m, & \text{if }n\text{ is odd} \end{cases} $

$\psi$ is obviously injective, and this shows that $|Hom(\mathbb{N^{+}}, M)| \ge |M|$. I have yet to show surjectivity, I've been told to use right inverse definition of surjectivity but I don't quite understand what to do.

edit- $\phi$ is definitely not a homomorphism, oops.

So the question is how would one define this homomorphism and then prove bijectivity.

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    @HerpDerpington Actually, the notation $\mathbb N^+$ would suggest (to me at least) rather that $0\notin\mathbb N^+$, whereas $\mathbb N_0$ would sucggest $0\in \mathbb N_0$. In the context of (additive) mopnoids, however, it should clearly be $\mathbb N\cap\{0\}$.2012-12-22

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The map $\psi$ cannot possibly be surjective. When $M = \mathbb{N}^+$, what element of $M$ would map to the identity of $\mathbb{N}^+$?

There is a standard isomorphism for these structures. Let $Hom_{mon}(A,B)$ stand for the set of monoid homomorphisms $A \to B$ where $A$ and $B$ are monoids, and let $\mathbb{N}^+ = \{0, 1, 2, 3, \ldots\}$ be the natural numbers as an additive monoid.

$ \psi : M \to Hom_{mon}(\mathbb{N}^+, M) $ where $\psi(m) : \mathbb{N}^+ \to M$ is the map defined by $\psi(m)(n) = n\cdot m = \underbrace{m + m + m +\cdots + m}_n$.

It should be fairly straightforward to show both injectivity and surjectivity of this mapping.

Hope this helps!

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    @ShaunAult Ah right, thank you.2012-12-23