I read the following answer to a question where the OP clearly intended to use the open mapping theorem from functional analysis here.
I'll add a more general version of the difficult direction, to show that linearity isn't necessary.
Suppose $X$ and $Y$ are complete metric spaces, $f:X\to Y$ is a continuous map with dense range, and there is $\delta>0$ such that $d_Y(f(a),f(b))\ge \delta d_X(a,b)$ for all $a,b\in X$. Then $f$ is a homeomorphism between $X$ and $Y$.
Proof. It suffices to show that $f(X)$ is closed, because being closed and dense, it must coincide with $Y$. Take a sequence $(y_n)$ in $f(X)$ that converges to $y\in Y$. Write $y_n=f(x_n)$. Since $d_X(x_n,x_m)\le \delta^{-1} d_Y(y_n,y_m)$, the sequence $(x_n)$ is Cauchy in $X$. By completeness of $X$, there is a limit $x=\lim x_n$. Since $f$ is continuous, $y=f(x)\in f(X)$. $\quad\Box$
What I reads from this is that if $f $ is such that there is $\delta>0$ such that $d_Y(f(a),f(b))\ge \delta d_X(a,b)$ for all $a,b\in X$ then $f $ is injective, and since we show that the range of $f $ is closed in $Y $ the assumption that the range also is dense in $Y $ leads to that $Y =\overline {f(X) } =f(X)$ so that $f $ also is surjective. If we also assume that $f $ is linear the open mapping theorem from functional analysis applies, but the poster here says that isn't necessary, why is this so and which theorem is then used?
Most grateful for any help!
