$ \newcommand{\nc}{\newcommand} \nc{\SS}{\mathcal{S}} \nc{\TT}{\mathcal{T}} \nc{\ol}{\overline} \nc{\inv}[1]{#1^{-1}} \nc{\int}{\operatorname{int}} \nc{\R}{\mathbb{R}} $ Setup: Suppose we have two topological spaces, $(X,\TT)$ and $(Y,\SS)$. Let $f : X \to Y$ be a closed mapping (i.e. maps closed sets to closed sets).
Suppose that, for some $y \in Y$ and a corresponding $G \in \TT$ ($G$ open in $X$), we have that $\inv f(y) \subseteq G$.
Our goal is to prove or disprove the following claim: $y \in \int (f(\ol{G}))$.
My Attempt: I think I've found a counterexample, but I just wanted to double-check it.
Consider the case of $f : \R \to \R$ with $f(x) \equiv 0$. We will use the usual topology on $\R$.
$f$ is clearly a closed mapping (all closed sets will map to $\{0\}$, which is closed).
Take $y = 0, G = \R$ in the statement of the claim. Then $\inv f(0) = \R \subseteq \R \in \TT$ trivially.
However, $$ \int (f(\ol{G})) = \int(f(\ol{R})) = \int(f(\R)) = \int (\{0\}) = \emptyset $$ and thus cannot contain $0$.
My Question: Is this correct? It just feels a little too easy (especially after struggling for a while to instead prove the claim), so I feel like I must've missed something important.