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Suppose we have two monoids $N_1=\{0,1,2,3,\ldots\}$,$N_2=\{0,-1,-2,-3,\ldots\}$ under addition.

Its easy to see that $-(a+b)=(-a)+(-b)$, so there exists a function $h(x)=-x,\ x\in N_1\cup N_2$ such that it is an isomorphism between them. But it is an involution, meaning that it is indeed an automorphism, saying that $N_1=N_2$, which is false.

We defined a valid "structure preserving map" between those monoids that is not an isomorphism. What is happening here? Why it is not an isomorphism?

Garmekain
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  • The isomorphism between $N_1$ and $N_2$ is a function from $N_1$ to $N_2$, not from the union to itself. The inverse map is given by the same formula, but is defined on a different set. – Wojowu Nov 02 '17 at 11:49

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No, the isomorphism is $$ h: N_1 \to N_2 : h(x) = -x $$ This is not an involution -- it is not even a permutation -- because its domain and codomain are not the same set (they barely even intersect).

This $h$ can be viewed as a restriction of $$ g: \mathbb Z\to\mathbb Z: g(x) = -x $$ which does happen to be an involution. But neither the domain nor the codomain of $g$ is one of the two algebraic structures you're considering here. So it is not a map between them (structure-preserving or not).

hmakholm left over Monica
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