$\DeclareMathOperator{\im}{im}$The First Isomorphism Theorem for rings is usually stated as follows:
Let $f\colon R\to S$ be a surjective homomorphism of rings with kernel $K$. Then the quotient ring $R/K$ is isomorphic to $S$.
In some other textbook, the theorem would be stated slightly differently as:
Let $f\colon R\to S$ be a surjective homomorphism of rings with kernel $K$. Then the quotient ring $R/K$ is isomorphic to $\im(f)$ where $\im(f)$ denotes the direct image of $f$.
In the second form, if $R/K \cong\im(f)$ and $\im(f)$ is a subring of $S$, then how would I show that $R/K\cong S$. I can prove the theorem in both formulations, but from the second formulation of the theorem, I don't know how to how to show $R/K \cong S$ from $R/K \cong\im(f)$. I know that $f$ is a surjective homomorphic mapping from $R$ to $S$ and that if $R/K\cong\im(f)$, can Is that enough to show that $\im(f)=S.$
Thank you in advance.
Best Answer
If $f$ is surjective then $\operatorname{im}f=S$ by definition. Hence the second formulation implies the first one when we consider a surjective map. Conversly, any map $f\colon R\to S$ "can be made surjective" by restricting the codomain $S$ to the subring $\operatorname{im}f$ onto which $f$ is trivially surjective. In this case the first formulation implies the second one.
Your argument in the comments is correct. Alternatively, if we write $\pi\colon R\to R/I$ for the canonical projection (defined by $r\mapsto r+I$) we have that $\phi\circ\pi=f$. By elementary set theory the surjectivity of $f$ then implies the surjectivity of $\phi$ (this is a nice standard fact to keep in mind).