The thing about isomorphism is that given (M, *, 1) and (N, +, 0) and a bijectoin f : M -> N which satisfies the equation, f is a monoid isomorphism. Proof: Let’s assume that f w = 0, for w \in M. We calculate for any m:

w * m =

g (f w) * g (f m) =

g (f (g (f w) * g (f m))) =

g (f (g (f w)) + f (g (f m))) =

g (f w + f m) =

g (0 + f m) =

g (f m) =

m

so w is an identity, so w = e, so f e = 1, so f is a monoid homomorphism (and a bijection, hence it is an isomorphism). (QED)

I dont’t think they want to say anything more than that.

]]>[blockquote]φ(m) = φ(m + 0) = φ(m) × φ(0).[/blockquote]

And indeed, in this case the characterisation of neutral element is given in terms of semiring-based equations only, much like the inverses for groups. That said, I could have made some embarrassing mistake above, so feel free to point it out.

]]>Wonder if you could give some (category-theoretical?) presentation of functors for which this sort of morphism “lifting” holds?

Also, first! 😉

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