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Let $f:\mathbb{M}\to\mathbb{M'}$ a morphism between two monoids and denote $N$ the equivalence class of identity element $\epsilon \in \mathbb{M}$ relatively at nuclear relation $Ker f$. Prove that $N$ is submonoid of $\mathbb{M}$.

Can you help me with some hints to start this proof?

Thanks in advance!

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    By "the equivalence class of identity element $\epsilon \in \mathbb{M}$ relatively at nuclear relation $Ker f$", what do you mean, exactly? – 2017-02-02
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    From Wikipedia: Any function $f : X → Y$ itself defines an equivalence relation on $X$ according to which $x_1 ~ x_2$ if and only if $f(x_1) = f(x_2)$. The equivalence class of x is the set of all elements in X which get mapped to $f(x)$, i.e. the class $[x]$ is the inverse image of $f(x)$. This equivalence relation is known as the kernel of f. – 2017-02-02

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Well, $\ker f$ is defined to be the relation $m_1\ker f\; m_2$ if and only if $f(m_1)=f(m_2)$. So the equivalence class of $\epsilon$ is the set of all $m\in\Bbb M$ such that $f(m)=f(\epsilon)=\epsilon'\in\Bbb M'$.

How do you prove something is a submonoid? You need to prove that $\epsilon\in N$ (which we already know to be true), and you need to prove that if $m_1,m_2\in N$ then $m_1m_2\in N$. But this should also be easy by the definition of monoid homomorphism.

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    I got it now! Thank you for your answer! Have a nice day! – 2017-02-02