If I have the following sequence:
$$J \xrightarrow{v} H \xrightarrow{u} G \rightarrow 0 \qquad (1)$$
Where $v$ is the zero map.
Does that mean $G \cong H$ i.e., $G$ isomorphic to $H$?if so why?
Or, do I need the extra assumption that $(1)$ is an exact sequence for $G$ to be isomorphic to $H$? If so, what is also the justification for the isomorphism between $G$ and $H$?
For me, it is clear that $u$ is onto.
Any elaboration will be greatly appreciated!
If the sequence is exact, we have $\operatorname{im}(v) = \ker(u)$. But $\operatorname{im}(u) = G $ since it coincides with the kernel of the zero map, therefore $u$ is surjective. Since you mentioned that $v$ is the zero map, it follows that $u$ is injective. Hence $u$ is an isomorphism between $H$ and $G$. If the sequence is not exact then you can easily find a counterexample to show that it does not imply $G$ and $H$ are isomorphic.