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Having trouble understanding something obvious.

Let $\varphi :S\to S'$ be a surjective homomorphism of finite semigroups. Then for every $\mathcal{J}$-class $J\subset S$ there exists a $\mathcal{J}$-class $J'\subset S'$ s.t $\varphi (J)\subset J'$. Also, for every $J'\subset S'$, there exists $J\subset S$ s.t $\varphi (J)\subset J'$.

A part of Lemma 1.4, p.144 "Semigroups: An introduction to structure theory", P.A Grillet.

The proof asserts that this is clear [which it isn't].

2 Answers 2

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Hint. First prove that if $x \leqslant_{\cal J} y$, then $\varphi(x) \leqslant_{\cal J} \varphi(y)$. Deduce that if $x \mathrel{\cal J} y$, then $\varphi(x) \mathrel{\cal J} \varphi(y)$.

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    indeed, it's obvious :( I hope i proved the Hint correctly. I did use surjectivity, but is it necessary, in general? – 2017-02-18
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    The hint does not require surjectivity, but you need surjectivity for the second part of your question: for every $J' \subset S'$, there exists $J \subset S$ such that $\varphi (J) \subset J'$. – 2017-02-18
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    you're right. I think I improved the solution. – 2017-02-18
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Let $\varphi : S\to T$ be a homomorphism of semigroups, then

$$\forall x,y\in S \left [x\leq _{\mathcal{J}}y \Longrightarrow \varphi (x)\leq _{\mathcal{J}}\varphi (y)\right ] $$

Since $S^1xS^1\subset S^1yS^1$, then for some $s_1,s_2\in S^1$ we get $1x1 = x = s_1ys_2$ and letting $t_1,t_2\in T^1$, we get $$t_1\varphi (x)t_2 = t_1\varphi (s_1)\varphi (y)\varphi (s_2)t_2\in T^1\varphi (y)T^1 $$

Therefore, if $x\mathcal{J}y$, then $\varphi (x)\mathcal{J}\varphi (y)$.

Now let $J\subset S$ be a $\mathcal{J}$-class, then $J=J_x$ for some $x\in S$ and $\varphi (J_x)\subset J_{\varphi (x)}$. Indeed $$t\in \varphi (J_x)\Longrightarrow \exists u\in J_x: \varphi (u)=t $$ since $u\mathcal{J}x$, then $t\mathcal{J}\varphi (x)$.

Secondly, if $\varphi$ is onto, let $J'\subset T$, then $J' = J_y$ for some $y\in T$ and $y = \varphi (x)$ for some $x\in S$ and $\varphi (J_x)\subset J_y$.