 Solution

Let $B_X$ be the open unit ball. Since $T(X)={\bigcup }_{n\ge 1}T(n{B}_X)$ is not meagre in $Y$, at least one closure $\overline{T(n{B}_X)}$ has nonempty interior. A ball contained in that closure, subtracted from itself, gives a ball about zero contained in $\overline{T(2n{B}_X)}$. Scaling yields some $\delta >0$ withNo completeness of $Y$ is used here; the hypothesis is second category in $Y$, not merely in the relative image topology.

If $\Vert y\Vert <\delta /2$, choose $x_1\in \frac{1}{2}{B}_X$ with $\Vert y-T{x}_1\Vert <\delta /4$. Then choose $x_2\in \frac{1}{4}{B}_X$ so the new residual is below $\delta /8$, and continue. The series ${\sum }_j{x}_j$ converges in the Banach space $X$, has norm less than one, and continuity of $T$ gives $T({\sum }_j{x}_j)=y$. Thus $B_Y(0,\delta /2)\subseteq T(B_X)$.

The linear subspace $T(X)$ contains a ball about zero, so it is all of $Y$: $T$ is surjective. Translated and scaled unit-ball images show that every open-set image is open. If $T$ is injective, its inverse is linear and the ball inclusion gives $\Vert T^{-1}y\Vert \le (2/\delta )\Vert y\Vert$, by scaling and taking a limiting bound. Thus $T^{-1}$ exists and is bounded.

For the first nonlinear counterexample use $f(x)=e^x$: its image $(0,\infty )$ is of second category but is not all of $\mathbb{R}$. For the second use the continuous surjection $f(x)=x+1$ for $x\le -1$, $f(x)=0$ for $-1\le x\le 1$, and $f(x)=x-1$ for $x\ge 1$. The open interval $(-1/2,1/2)$ maps to the nonopen singleton $\{0\}$. Linearity is essential to the conclusions.