Let $X,Y$ be real Banach spaces. Let $f:X\to Y$ be a continuous surjective function (not necessarily linear), and let $g:Y\to \mathbb R$ be an arbitrary function. Suppose that $g\circ f$ is continuous. Suppose further that $\|x_n\|_X\to \infty$ if and only if $\|f(x)\|_Y\to \infty$. Does it follow that $g$ is continuous?
I have tried to investigate this question via contradiction. Suppose that $y_n\to y$ in $Y$, as $f$ is surjective we can define $f(x_n)=y_n$ and $f(x)=y$. Now if $x_n\to x$ then the continuity of the composition would guarantee that the answer to my question is in the affirmative. So suppose for sake of contradiction that $x_n\not \to x$. Then either $x_n$ is unbounded or $\|x_n\|\leq M\in \mathbb R$ but $x_n$ either does not converge or converges to some other element. Clearly our assumption on $f$ forbids the first case, so we must have that $x_n$ is bounded. However, I cannot see what prevents $x_n\to z\neq x$ where $f(z)=y$. So I do not believe that we must have $x_n\to x$. Obviously this does not imply that $g$ is not continuous, but I cannot think of another way to investigate the matter right now. Unfortunately the only counter examples I have found either do not have the extra condition about unbounded sequences or are functions on non-linear spaces.
I encountered this question via a lemma given without proof in some notes on Markov processes, which stated that the composition of a Feller process with a map satisfying the properties of $f$ is also a Feller process. The only way I can currently see how this lemma can be true is if the answer to my question is affirmative, which is obviously no reason to assume that the question has an affirmative answer, but I'd be glad if it did. A counter example putting me out of my misery would be grand, and a proof of its veracity even grander.
If f is a quotient map such as an open (or closed) continuous
surjection and g $\circ$ f is continuous then g is continuous.
That is the general topological view.
Perhaps it could helpful with your perplexment.