In the textbook that I am currently learning ring theory from, it states the First Isomorphism Theorem in the following manner:
Let $f : R_1 \to R_2$ be a homomorphism of rings. Then the image $\operatorname{im}(f) = \{f (r) \mid r \in R_1\}$ is a subring of $R_2$ and the map $R_1/\ker(f) \to \operatorname{im}(f)$ sending $r + \ker(f)$ to $f(r)$ is an isomorphism.
However, I have been confused by the way that this is stated because how is it guaranteed that the number of $r + \ker(f)$ will be equivalent to $f(r)$ so that is the mapping is surjective?
I then looked into the Modern Algebra lecture on MIT OCW and it states, for the First Isomorphism Theorem, that $f: R_1 \to R_2$ has to be onto (i.e. surjective). This will help to guarantee that the map $R_1/\ker(f) \to \operatorname{im}(f)$ is not only injective but also surjective (i.e. bijective), wouldn't it?
I hope my question makes sense. Thank you!
The isomorphism theorem establishes a bijection (that preserves ring structure) between cosets of the kernel with elements of the image. If the original image happens to be surjective, then that image is the entire codomain.
In more detail, any ring homomorphism $$ f: R_1 \to R_2 $$ factors through $\require{AMScd}$ \begin{CD} R_1 @>{f}>> R_2 \\ @V{\pi}VV @AA{\iota}A \\ R_1/\ker f @>{\smash[t]{\overline{f}}}>> \operatorname{im} f \end{CD} where $\pi$ is the canonical projection, $\iota$ is the canonical inclusion, and $\overline{f}$ is the induced isomorphism.
When $f$ is injective, $\pi$ is the identity map, and when $f$ is surjective, $\iota$ is the identity map.