We consider the follow Dirichlet problem \begin{align*} \displaystyle \begin{cases} -\Delta u =f & \Omega \\ u=0 & \partial \Omega \end{cases} \end{align*} were $\Omega \subset \mathbb{R}^n$ is a bounded open. ($\star$) It is known that for this problem there exists a unique weak solution $u \in H_{0}^1(\Omega)$ for each $f \in L^2(\Omega)$ fixed, and operator \begin{align*} \Delta^{-1} : f \in L^2(\Omega) \longmapsto u \in H_{0}^1(\Omega) \end{align*} is continuous. Now it can be shown that $(-\Delta)^{-1}$ is compact injective self-adjoint operator on $L^2(\Omega)$ and $H_{0}^1(\Omega)$. Ok, I'm not interested to prove that it is compact and self-adjoint. In the book, in the demonstration said that $\Delta^{-1} : f \in L^2(\Omega) \longmapsto u \in H_{0}^1(\Omega)$ is bijective by result in ($\star$).
It seems to me that it should be only injective operator, and not bijective. Is it a book error?
One can show that for $f\in L^2(\Omega)$, we have $\Delta^{-1} f\in H_0^1(\Omega)\cap H^2(\Omega)$, therefore, for any $u\in H_0^1(\Omega)\setminus H^2(\Omega)$, there is no $f\in L^2(\Omega)$ with $\Delta^{-1}f=u$.
On the other hand, if one replace $L^2(\Omega)$ by $H^{-1}(\Omega)$ then, $\Delta^{-1}: H^{-1}(\Omega)\to H_0^1(\Omega)$ is an isomorphism.