I am trying to get a deeper understanding of the First Isomorphism Theorem, but am having trouble seeing the "natural mapping" that my textbook says exists. Upon looking it up more online I came across this which said that "The map that describes the isomorphism is the real power of the result, rather than the fact that the groups happen to be isomorphic", however, like I said, I am having trouble seeing what mapping this is. I would appreciate it if you could define the mapping and show that it is a ring homomorphism (both additive and multiplicative). Thanks!
Note: I do see how the two groups R/K --> Img (R) have the same number of elements, I just can't see the natural bijective homomorphism between the two.
Let $\phi:R\to S$ be a homomorphism. Then $K=\mathrm{ker}(\phi)$ is an ideal of $R$ and $R/K$ is a ring. The elements of $R/K$ are classes of equivalence: $[x]=[y]\iff x-y\in \mathrm{ker}(\phi).$ Note that then $\phi(x)=\phi(y),$ that is, $\phi$ takes a unique value on each class of equivalence. Moreover, $[x]\ne [y]\iff x-y\notin \mathrm{ker}(\phi)\iff \phi(x)\ne \phi(y).$ That is, $\phi$ takes different values on different classes of equivalence.
Thus the map that defines in a natural way the isomorphism is
$$R/\mathrm{ker}(\phi)\to \mathrm{im}(\phi), [x]\to \phi(x).$$
It follows from above that it is well-defined. Can you prove that it is an isomorphim?