If there exists a bounded operator $T:X \to Y$ with $T^{-1}$ bounded then X is Banach iff Y is Banach I have been asked to show that if there exists a bounded operator $T:X \to Y$  with $T^{-1}$ bounded then X is Banach iff Y is Banach.
I have shown it for $T$ a linear operator.  But I can't use the same trick for a bounded operator which is not linear.
The trick is to use the fact that $||T^{-1}(y_n - y_m)||=||T^{-1}(y_n)-T^{-1}(y_m)||$ to show that $(T^{-1}(y_n))$ is a Cauchy sequence. And the same for $(T(y_n))$ is a Cauchy sequence.
 A: Supposing that the definition of a bounded operator is

An operator $A \colon X \to Y$ is bounded if for every bounded $S\subset X$ the image $A(S)$ is a bounded subset of $Y$,

the linearity of $T$ must be implicitly assumed in that exercise, since we can give counterexamples with nonlinear bounded operators.
Let $1 \leqslant p < q \leqslant +\infty$, and consider $X = \ell^p(\mathbb{N})$ endowed with the $\ell^p$-norm, and $Y$ the same space endowed with the $\ell^q$-norm. Presumably it is known that then $X$ is a Banach space, and $Y$ an incomplete normed space.
However, for every $\rho > 0$, the cardinality of the spheres $S_{\rho}^r = \{ x : \lVert x\rVert_r = \rho\}$ is $2^{\aleph_0}$ for either $r\in \{p,q\}$, so there a bijection $t \colon S_1^p \to S_1^q$. Defining $T\colon X \to Y$ by $T(x) = \lVert x\rVert\cdot t(x/\lVert x\rVert)$ for $x\neq 0$ and $T(0) = 0$, we thus have a bounded bijection with bounded inverse.
If a different definition of a nonlinear bounded operator is used, the assertion might hold without the linearity assumption.
