Let's say I have two sets $A$ and $B$, which are power sets of natural numbers less than or equal to $1$ and $2$ respectively. So $A$ = {$\emptyset$, {$0$}, {$1$}, {$0,1$}} and B = {$\emptyset$, {$0$}, {$1$}, {$2$}, {$0,1$}, {$0,2$}, {$1,2$},{$0,1,2$}}. Let's consider a map $g: (A, \cap) \to (B,\cap )$ defined as $g(a) = a$ where $a \in A$. Now, I know that the sets $A$ and $B$ only form semigroup under $\cap$ and do not form monoid. So, is it correct to assume that the map $g$ is not a monoid homomorphism?
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asked Mar 26, 2019 at 22:42
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$A$ and $B$ are monoids under $\cap$ (with $\{0, 1\}$ as the identity element for $A$ and $\{0, 1, 2\}$ as the identity element for $B$). $g$ is not a monoid homomorphism because it does not respect the identity elements: $g(\{0, 1\}) \neq \{0, 1, 2\}$.
answered Mar 26, 2019 at 23:05
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$\begingroup$ Oh yes. Completely forgot about it. Thank you. $\endgroup$ Mar 26, 2019 at 23:29