About the range of an operator and its adjoint I need a hand with the proof of this result:
If we have an operator between Banach spaces $T:X\to Y$, with closed range, then the adjoint operator $T^*:Y^*\to X^*$ has also closed range.
Thanks in advance for any help.
 A: This is how it's proved in Dunford and Schwartz:
Using the Open Mapping Theorem, you can prove the following:
Lemma: If $T:X\rightarrow Y$ is a bounded linear operator between the Banach spaces $X$ and $Y$ with closed range, there is a constant $K$ so that to each $y\in TX$ there is an $x\in X$ with $Tx=y$ and $\Vert x\Vert\le K\Vert y\Vert$.
Using the Lemma, you can prove the:
Theorem:
For $T$ as in the Lemma, the range of $T^*$ is the set of $x^*\in X^*$ such that $Tx=0$ implies $x^* x=0$.
This in particular implies $T^*$ has closed range. 
To prove the non-trivial inclusion of the Theorem: 
Suppose $x^*$ satisfies the given condition. Then one uniquely defines $y_0^*$ on $TX$ via $y_0^*(Tx)=x^* x$. Use the Lemma to show $y_0^*$ is continuous. Then extend by Hahn-Banach to a $y^*\in Y^*$ with $T^*y^*=x^*$.
A: If you factor $T$,
$$X \xrightarrow{\pi} X/\ker T \xrightarrow{\tilde{T}} \mathcal{R}(T) \xrightarrow{\iota} Y,$$
and then take the dual of that sequence,
$$Y^\ast \xrightarrow{\iota^\ast} \mathcal{R}(T)^\ast \xrightarrow{\tilde{T}^\ast} (X/\ker T)^\ast \xrightarrow{\pi^\ast} X^\ast,$$
can you see how that quickly leads to the result?
