Image of Bounded Linear Transformation is Everywhere Dense iff Dual Operator is Injective/Proof 2

Proof
From Annihilator of Image of Bounded Linear Transformation is Kernel of Dual Operator, we have:
 * $T \sqbrk X^\bot = \map \ker {T^\ast}$

where $T \sqbrk X^\bot$ denotes the annihilator of $T \sqbrk X$.

From Linear Transformation is Injective iff Kernel Contains Only Zero, we then have that $T^\ast$ is injective :
 * $T \sqbrk X^\bot = \set { {\mathbf 0}_{Y^\ast} }$

From Annihilator of Subspace of Banach Space is Zero iff Subspace is Everywhere Dense, this is equivalent to:
 * $T \sqbrk X$ is everywhere dense in $Y$.

This was the demand.