Continuous image of a F-space

Question

Let $X$ be a $F$-space, i.e. there exists complete metric $d$ on $X$ such that $d(x,y)=d(x+z,y+z), \ x,y,z\in X$.

Suppose that we have a normable vector space $Y$, which is of first Baire category. Is is true that there is no continuous linear map $\phi:X\rightarrow Y$ such that $\phi(X)=Y$ ?

Answer 1

No, that's not true. Consider any Banach space, e.g., $X=\ell^1$ with the usual norm and a strictly coarser coarser norm on $Y=X$, e.g. $\|\cdot\|_\infty$.

However, if $\phi$ is open or almost open (i.e. the closure of $\phi(U)$ is a $0$-neighborhood in $Y$ for each $0$-neighborhood $U$ in $X$) then $Y$ is again an $F$-space and hence of second category.