Does $x \circ y = id_V$ with $x \in GL(V)$ and $y \in L(V)$ implies $y \in GL(V)$?

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Let $V$ be a real Banach space. Denote by $L(V)$ the real Banach space of linear continuous applications from $V$ to $V$ and by $\mathrm{GL}(V)$ the set of all linear bijective bicontinue applications from $V$ to $V$.

Let $x \in \mathrm{GL}(V)$ and $y \in L(V)$. Does $x \circ y = id_V$ implies $y \in \mathrm{GL}(V)$ ? Does $y \circ x = id_V$ implies $y \in \mathrm{GL}(V)$ ?

For the context, the question above arise when I try to apply implicit function theorem to prove $f \in \mathrm{GL}(V) \mapsto f^{-1} \in \mathrm{GL}(V)$ is $\mathcal{C}^{\infty}$. Since I already know one way to prove this $\mathcal{C}^{\infty}$, so this question is not about how to prove this $\mathcal{C}^{\infty}$.

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If $x\in GL(V)$, then $x\circ y=\text{id}_V$ implies $y=x^{-1}$ (and $x^{-1}$ is also trivially an element of $GL(V)$), so yes $y\in GL(V)$.