Bijective bounded linear operator is invertible

The following is an exercise from Halmos book "A Hilbert space problem book" :

Exercise: If $H$ and $K$ are Hilbert spaces, and if $A$ is a bounded linear transformation that maps $H$ one to one and onto $K$, then $A$ is invertible.

He gives the following solution for this:  I do not know why he consider the operator $0$ in the role of $A^*$. Clearly $0$ is not bijective, so why does he state this part? Please help me. Thanks.

• Note that this is a special case of the (more difficult and important) open mapping theorem. – kahen Mar 24 '15 at 9:24

For operator $A^*=0$ the condition $\Vert A^*g\Vert=1$ is always false. By the fundamental rules of logic False implies anything and in particular that $\Vert g\Vert\leq 1/\delta$.
One could say for example that $\Vert A^*g\Vert=1\implies\pi=10^5$ and would be absolutely correct.
• Operator $0$ is not satisfy in the assumption, so I can not understand why does he state this part in his proof? – niki Mar 24 '15 at 17:05
• This is not a step of the proof, this is an explanation why one had to keep in mind that $\ker A^*\neq\{0\}$. – Norbert Mar 24 '15 at 17:23