I'm studying the trace space on a smooth boundary $\Gamma$, namely $H^{\frac{1}{2}}(\Gamma)$. Let $\gamma$ be the classical trace operator $\gamma: H^1(\Omega) \rightarrow H^{\frac{1}{2}}(\Gamma)$. It is linear, surjective, and bounded.
Then, my professor said that its restriction to the orthogonal complement $\tilde{\gamma}:H_0^1(\Omega)^{\perp} \rightarrow H^{\frac{1}{2}}(\Gamma)$ is a linear and bounded bijection. This is still clear to me. Then, he wrote:
It follows that $\tilde{\gamma}^{-1}: H^{\frac{1}{2}}(\Gamma) \rightarrow H_0^1(\Omega)^{\perp}$ is a linear bijection as well and $\tilde{\gamma}^{-1}$ is a right inverse of $\gamma$.
I have two questions about the above quote.
Of course, the inverse of a bijective linear map is linear and by definition that map is injective, but why is $\tilde{\gamma}$ surjective? I should show that if I take an element $w \in H_0^1(\Omega)^{\perp}$, there exists a $v$ s.t. $$\tilde{\gamma}^{-1}(v) = w$$ I would say that it suffices to choose $v := \tilde{\gamma}(w)$. Is that correct?
Also, I don't see why $\tilde{\gamma}^{-1}$ is a right inverse of $\gamma$.
The inverse of a bijection is always a bijection ... yes the proof can be done as your first point.
First, $\tilde \gamma^{-1}$ is not a left inverse of $\gamma$ because $\tilde \gamma^{-1}\gamma : H^1(\Omega) \to H^1_0(\Omega)^\perp$.
Contrarily, Since $\tilde \gamma^{-1} : H^{1/2}(\Gamma) \to H^1_0(\Omega)^\perp$ and $\gamma = \tilde\gamma$ on $H^1_0(\Omega)^\perp$, you deduce that $\gamma\tilde \gamma^{-1} : H^{1/2}(\Gamma) \to H^{1/2}(\Gamma)$ is well defined and $\gamma\tilde \gamma^{-1} = \tilde\gamma\tilde \gamma^{-1} = \mathrm{Id}_{H^{1/2}(\Gamma)}$. So $\tilde \gamma^{-1}$ is a right inverse of $\gamma$.