does invertibility of product imply invertibility of each term of product?

Question

Suppose $\mathcal{H}$ is a Hilbert space and the product $T_1T_2 \in B(\mathcal{H})$ is invertible. Does this imply that both $T_1, T_2$ are invertible ? I am unable to prove this since, unlike the finite dimensional case, injectivity or surjectivity alone does not imply invertibility. I am of course assuming that both both $T_1,T_2 \in B(\mathcal{H})$

Answer 1

No. Let $H$ be a separable, infinite-dimensional Hilbert space, and let $\{e_k\}_{k \in \mathbb{N}}$ be an orthonormal basis. Let $T_1 \in B(H)$ be given by $T_1 e_k = e_{2k}$, and let $T_2 \in B(H)$ be given by $$ T_2 e_k = \begin{cases} e_{k/2} &\text{if $k$ is even,}\\ 0 &\text{if $k$ is odd;} \end{cases} $$ observe, indeed, that $\|T_1\| = \|T_2\| = 1$. Then $T_1$ is injective but not surjective, $T_2$ is surjective but not injective, and yet $T_2 T_1 = \operatorname{id}_H$. The essential point is that $V := \operatorname{span}\{e_{2k}\}_{k \in \mathbb{N}}$ is a proper closed subspace of $H$ even though $V \cong H$ as Hilbert spaces (via $T_1$ as a unitary isomorphism $H \to V$), all of which is ultimately possible precisely because $H$ is infinite-dimensional.