Surjectivity alone is algebraic, but for Banach spaces it has topological content. The open mapping theorem says that a bounded surjective linear map cannot collapse all small balls into thin sets.
Open mapping theorem
The theorem says that a surjective bounded map between Banach spaces sends balls to sets large enough to contain balls. This is the quantitative content of surjectivity.
Theorem. Let \(X,Y\) be Banach spaces and let \(T\in\mathcal L(X,Y)\) be surjective. Then \(T\) is an open map: for every open \(U\subset X\), \(T(U)\) is open in \(Y\).
Proof. It is enough to prove that \(T(B_X(0,1))\) contains a ball around \(0\). Since \(T\) is surjective, \(Y=\bigcup_{n\ge1} n\,\overline{T(B_X(0,1))}\). By Baire, some \(n\,\overline{T(B_X(0,1))}\) has nonempty interior; by symmetry and convexity, \(\overline{T(B_X(0,1))}\) contains a ball \(B_Y(0,r)\).
We now remove the closure. If \(\|y\|<r/2\), choose \(x_1\) with \(\|x_1\|<1/2\) and \(\|y-Tx_1\|<r/4\). Inductively choose \(x_k\) with \(\|x_k\|<2^{-k}\) and \(\|y-\sum_{j\le k}Tx_j\|<r2^{-(k+1)}\). Completeness gives \(x=\sum_kx_k\) with \(\|x\|<1\), and continuity gives \(Tx=y\). Hence \(B_Y(0,r/2)\subset T(B_X(0,1))\). Translating and scaling proves openness.
Bounded inverse theorem
If a bijective bounded operator connects Banach spaces, its inverse is automatically continuous. Thus algebraic isomorphism and topological isomorphism coincide in this setting.
Corollary. If \(T:X\to Y\) is a bijective bounded linear map between Banach spaces, then \(T^{-1}:Y\to X\) is bounded.
Proof. By the open mapping theorem, \(T(B_X(0,1))\) contains \(B_Y(0,r)\) for some \(r>0\). If \(y\ne0\), then \(r y/(2\|y\|)\in T(B_X(0,1))\), so \(r y/(2\|y\|)=Tx\) with \(\|x\|<1\). Thus \(\|T^{-1}y\|<(2/r)\|y\|\).
Equivalent complete norms
Completeness prevents two comparable norms from defining incompatible analytic structures. This is a common way to identify when a new norm has the same bounded maps.
Corollary. Let \(\|\cdot\|_1,\|\cdot\|_2\) be Banach norms on the same vector space \(X\). If \(\|x\|_2\le C\|x\|_1\), then the norms are equivalent.
Proof. The identity map \(I:(X,\|\cdot\|_1)\to(X,\|\cdot\|_2)\) is bounded and bijective between Banach spaces. The bounded inverse theorem makes \(I^{-1}\) bounded, so \(\|x\|_1\le C'\|x\|_2\).
Quotient formulation
The open mapping theorem is often most transparent through quotients. Surjectivity factors through the kernel, and the induced bijection is an isomorphism.
Proposition. If \(T:X\to Y\) is bounded and surjective between Banach spaces, then the induced map \(\widetilde T:X/\ker T\to Y\), \(\widetilde T(x+\ker T)=Tx\), is a Banach-space isomorphism.
Proof. The kernel is closed because \(T\) is continuous, so \(X/\ker T\) is Banach. The map \(\widetilde T\) is well-defined, linear, bijective, and bounded since \(\|\widetilde T(x+\ker T)\|\le\|T\|\|x\|\), hence \(\|\widetilde T\|\le\|T\|\). The bounded inverse theorem gives boundedness of \(\widetilde T^{-1}\).
A priori estimates
A priori estimates are inverse theorems in analytic language. They say that controlling \(Tx\) is enough to control \(x\), perhaps after quotienting out a kernel.
Remark. The inverse theorem is the abstract form of an a priori estimate. To prove continuous dependence of a solution \(x\) on data \(y=Tx\), it is enough to prove existence and uniqueness in Banach spaces; boundedness of the solution map follows automatically.