Let $I=(0,L)$ be interval in $\mathbb{R}$. Let $N$ be kernel of a differentiating operator $\frac{d}{dx}$ in $L^{2}(I)$ ($N$ is subset of constant functions in $L^{2}(I)$). I have to prove that mapping from $L^{2}(I)/N$ to $H^{-1}(I)$, which sends $u$ to $u'$ is isomophism.
The thing I know is that $$f\in H^{-1}(I)\Leftrightarrow(\exists g\in L^{2}(I)) f=g'$$ where $H^{-1}(I)= (H^{1}(I))'$.
Mapping $u\rightarrow u'$ is obviously linear. Furthermore, upper characterization of elements from $H^{-1}(I)$ yields that $u\rightarrow u'$ is also surjective. Clearly, for every $f\in H^{-1}(I)$ there is $g\in L^{2}(I)$ such that $f=g'$ (of course, there are infinitely many such $g$s, but they're all in same equivalence class of $L^{2}(I)/N$).
Is injectivity of $u\rightarrow u'$ also easy to check? It's problem for me since I don't know how do elements from $H^{-1}(I)$ look like.
I will appreciate any kind of help.
Edit:
Is it just possible to use $V/Ker(A)\simeq Im(A)$, for arbitrary vector space and $V$ linear operator $A$?
Since $N$ is a kernel of $\frac{d}{dx}$, then it suffices to show that $H^{-1}(I)$ is image of $\frac{d}{dx}$ in $L^{2}(I)$, which is obvious from upper equivalence.
Is my reasoning correct?