Why is $\phi$ an epimorphism of rings?

Question

$$\phi:\mathbb{Z} \rightarrow \mathbb{Z}_m$$ $$\phi(a)=[a]_m$$

Why is $\phi$ an epimorphism of rings ? A surjective function is when $$\forall y, \exists x : f(x)=y$$ Isn't it?? Is $\phi$ surjective because both $x$ and $y$ are related with $a$??

Answer 1

Because for any $[a] \in \mathbb{Z}_m$, the same $a$ in $\mathbb{Z}$ is mapped to it.

In fact all numbers $b$ of the form $b=mq+a$ for some $q \in \mathbb{Z}$ are mapped to $[a]$ in $\mathbb{Z}_m$.

Moreover, this shows that $\phi$ is not injective. Since $b=mq$ is mapped to $[0] \in \mathbb{Z}_m$ then the kernel is $\ker{\phi}=m \mathbb{Z}$.

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More generally, if $R$ is an equivalence relation on the set $X$, i.e. $R \times R \subseteq X$ is a reflexive, symmetric and transitive relation then the map that sends each element in $X$ to its equivalence class in $X/R$ is a surjective map.

Please review elementary set theory if you have not mastered this stuff yet. You'll need to know these things well to study abstract algebra.

Answer 2

Think $Z_m=\{0,1,2,...,m-1\}$ if $a\in Z_m$ then $a\in Z$ and $a\cong a \mod (m)$