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Let $S$ and $T$ are monoids. Prove or disprove if $\varphi: S \to T$ is a Semigroup homomorphism, then $ \varphi$ is monoid homomorphism.

Shaun
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Farshad Hasani
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2 Answers2

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Consider the zero map $f \colon \mathbb{Z} \rightarrow \mathbb{Z}$, $a \mapsto 0$, where we consider $\mathbb{Z} = (\mathbb{Z}, \cdot)$ as a monoid. This is obviously a homomorphism of semigroups as $f(ab) = 0 = 0 \cdot 0 = f(a)f(b)$, but will not send the multiplicative identity $1 \in \mathbb{Z}$ to itself and thus is not a homomorphism of monoids.

Con
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Well, if there is an idempotent element $e$ in the monoid $T$ and $\phi:S\rightarrow T:x\mapsto e$, then $\phi$ is a semi group homomorphism, but the unit element of $S$ is mapped to $e$ and $e$ may not be the unit element of $T$.

About idempotents in semigroups see Is there an idempotent element in a finite semigroup?

Wuestenfux
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  • thank you so much. – Farshad Hasani Aug 01 '19 at 14:56
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    It might be useful for some readers to mention the smallest specific example of what is described in this answer: Let $S=\{0\}$ and $T=\{0,1\}$, both with ordinary multiplication as the operation, and let $\varphi$ send $0$ to $0$. – Andreas Blass Aug 01 '19 at 23:44