Invertible Continuous Linear Operator is Bijective

Theorem
Let $\struct {X, \norm {\, \cdot \,} }$ be the normed vector space.

Let $\map {CL} X := \map {CL} {X, X}$ be a continuous linear transformation space.

Let $I \in \map {CL} X$ be the identity element.

Suppose $A \in \map {CL} X$ is invertible.

Then $A$ is bijective.

$A$ is injective
Let $x, y \in X$ be such that $\map A x = \map A y$.

Then:


 * $A^{-1} \circ \map A x = A^{-1} \circ \map A y$

where $A^{-1}$ is the inverse of $A$.

By definition:


 * $A^{-1} \circ A = I$

Hence:


 * $x = y$

By definition, $A$ is injective.

$A$ is surjective
Let $y \in X$

Then:


 * $x := \map {A^{-1} } y \in X$

Moreover:

Hence:


 * $\forall y \in X : \exists x \in X : \map A x = y$

By definition, $A$ is surjective.

By definition, $A$ is bijective.