The plan is to observe that $T^{-1}$ exists as a [linear map](/page/Linear%20Map) by abstract linear algebra, and then use the [Open Mapping Theorem](/theorems/631) to establish its [continuity](/page/Continuity). The completeness of both spaces is essential. **Step 1: Existence and linearity of $T^{-1}$.** Since $T$ is bijective, the [set](/page/Set)-theoretic inverse $T^{-1}: Y \to X$ exists. Linearity of $T^{-1}$ follows from the linearity of $T$: if $T(x_1) = y_1$ and $T(x_2) = y_2$, then $T(\lambda x_1 + \mu x_2) = \lambda y_1 + \mu y_2$, so $T^{-1}(\lambda y_1 + \mu y_2) = \lambda T^{-1}(y_1) + \mu T^{-1}(y_2)$. **Step 2: Boundedness of $T^{-1}$ via the Open Mapping Theorem.** Since $T$ is a bounded, surjective linear map between [Banach spaces](/page/Banach%20Space), the [Open Mapping Theorem](/theorems/631) implies that $T$ is an open map. Therefore, for any [open set](/page/Open%20Set) $U \subset X$, $T(U)$ is open in $Y$. But $T(U) = (T^{-1})^{-1}(U)$, which means that the preimage of every open set under $T^{-1}$ is open. Hence $T^{-1}: Y \to X$ is continuous. **Step 3: Conclude.** Since $T^{-1}$ is a continuous linear map, it is bounded by the [Equivalence of Continuity and Boundedness](/theorems/873).