A proof that $T^*$ injective with closed range implies $T$ is surjective Given $T\in B(H)$ for some Hilbert space $H$, if $T^*$ is injective and has closed range, then $T$ is surjective. 
My professor sketched a proof by saying that, since $T^*$ has an inverse on its range (by the open mapping theorem), then $T$ maps $T^*(H)$ onto $H$. 
Can anyone explain why this implication holds?
On my separate attempt, I know that $\text{ker}(T^*)^\perp=\overline{T(H)}$, but I'm not sure where to employ the fact that $T^*(H)$ is closed. 
 A: Let $A = T^\ast(H)$. Then $A$ is a Hilbert space, and viewing $T^\ast$ as a map $H\to A$, we have an isomorphism $S \colon H\to A;\; S a = T^\ast a$, and hence an isomorphism $S^\ast \colon A \to H$. Furthermore, since $\ker T = T^\ast(H)^\perp$, we have an injective map $T\lvert_A \colon A \to H$. These maps satisfy
$$\langle T\lvert_A a, y\rangle = \langle a, T^\ast y\rangle = \langle a, S y\rangle = \langle S^\ast a, y\rangle$$
for all $a \in A$ and $y \in H$, and therefore $T\lvert_A = S^\ast$. Since $S^\ast$ is an isomorphism, $T\lvert_A(A) = S^\ast(A) = H$.
A: Your claim is that $\mathcal{R}(T)=H$ under the stated conditions, which is equivalent to showing that the range of $T$ is closed and $\mathcal{N}(T^{\star})=\{0\}$. This equivalence holds because $\overline{\mathcal{R}(T)}=\mathcal{R}(T)^{\perp\perp}=\mathcal{N}(T^{\star})^{\perp}$. You were given $\mathcal{N}(T^{\star})=\{0\}$; all that remains is to show that the range of $T$ is closed.
You were given that $T^{\star}$ has a closed range, which means that the inverse $(T^{\star})^{-1} : \mathcal{R}(T^{\star})\rightarrow H$ is a closed linear operator from one Hilbert space to another ($\mathcal{R}(T^{\star})$ is a Hilbert space because it is a closed subspace of a Hilbert space.) Therefore $(T^{\star})^{-1}$ is bounded by the closed graph theorem. You can easily extend $(T^{\star})^{-1}$ to a continuous linear operator $M$ on $H$ by setting $M=0$ on $\mathcal{R}(T^{\star})^{\perp}=\mathcal{N}(T)$. This extension $M\in B(H)$ satisfies $MT^{\star}=I$. So $TM^{\star}=I^{\star}=I$, which proves that $T$ is surjective.
