I'm studying a bit of de Rham Cohomology and trying to understand the following step in the proof of a lemma:
[...] By definition, $H^{0} = \frac{Ker(d:\Omega^{0}(U) \to \Omega^{1}(U))}{Im(d:\Omega^{-1}(U) \to \Omega^{0}(U))}$ . Since $d:\Omega^{-1}(U) \to \Omega^{0}(U)$ is the zero map, we get simply $H^{0} \cong Ker(d:\Omega^{0}(U) \to \Omega^{1}(U))$
I'm not sure yet how to handle with quotients, so I would like if someone explain me that step by step.
Recall for an (abelian) group $G$, with subgroup $H$, the elements of $G/H$ are exactly the equivalence classes of
$$g_1 \sim_H g_2 \iff g_2 - g_1 \in H$$
Then elements of $G/0$ are exactly equivalence classes of $g_1 \sim g_2 \iff g_2 - g_1 \in 0 \iff g_1 = g_2$. So $G/0$ is just $G$.
This same argument works for modules, and for Cohomology. So
$$ H^0 \cong \text{Ker} d / \text{Im} d \cong \text{Ker} d$$
whenever $\text{Im} d = 0$.
I hope this helps ^_^