The first isomorphism theorem states that if $f$ is a homomorphism between two groups $G$ and $H$, then $G/\textrm{Ker}(f) \cong H$.
We know that $f$ is a homomorphism, but do we also get to assume that it is surjective in order to prove the theorem?
Let $\textrm{Ker}(f) = K$. Let $\phi: G/K \mapsto H$ such that for any $a \in G$, $\phi(aK) = f(a)$. To prove the first isomorphism theorem, we will need to show that $\phi$ is well defined, injective, and surjective.
If $\textrm{Im}(f) = H$ (i.e. $f$ is surjective), then we can write any element $h \in H$, as $f(a)$ (for some $a \in G$), since surjectivity of $f$ guarantees the existence of $a$. Thus, $\phi$ being surjective naturally follows from how we use $f$ to define it.
It seems that Rotman is making this assumption:
But according to an answer on this website, it is not necessarily the case that $f$ must be surjective: https://math.stackexchange.com/a/2201729/115703 ?
Is Rotman actually assuming that $f$ is surjective, or am I misunderstanding? If he is assuming it, how can he do so given what the answer above discusses?
Best Answer
The first isomorphism theorem actually states that if $f$ is a homomorphism between two groups $G$ and $H$, then $G/\ker(f)\cong \operatorname{im}(f)$.
Note that every map is surjective onto it's image.
If you want to state that $G/\ker(f) \cong H$ then you do have to assume that $f$ is surjective.