Hilbert Space Isomorphism is Bijection

Theorem
Let $H, K$ be Hilbert spaces.

Denote by $\left\langle{\cdot, \cdot}\right\rangle_H$ and $\left\langle{\cdot, \cdot}\right\rangle_K$ their respective inner products.

Let $U: H \to K$ be an isomorphism.

Then $U$ is a bijection.

Proof
As $U$ is an isomorphism, it is necessarily surjective.

Suppose now that $g, h \in H$ are such that $Ug = Uh$.

Then as $U$ is a linear map, it follows that $U \left({g - h}\right) = \mathbf{0}_K$, the zero vector of $K$.

From property $(3)$ of an isomorphism, conclude that:


 * $0 = \left\langle{\mathbf{0}_K, \mathbf{0}_K}\right\rangle_K = \left\langle{U \left({g - h}\right), U \left({g - h}\right)}\right\rangle_K = \left\langle{g - h, g - h}\right\rangle_H$

Property $(5)$ of an inner product ensures us that $g - h = \mathbf{0}_H$, i.e. $g = h$.

Hence $U$ is injective.

As $U$ is both injective and surjective, it is a bijection.