# Regarding graphs of continuous functions between zero dimensional spaces

Background for the question: Let for any topological space $B$, $I(B)$ denote the topological space which has the same set of points as of $B$, and the topology is generated by closed and open sets of $B$.

Let $X$ be a zero dimensional topological space, $B$ be any topological space, $f : X \rightarrow I(B)$ be a continuous function, $G$ be the graph of $f$, i.e., $G = \{(x, f(x)): x \in X\}$ as a subspace of $X \times I(B)$, and $P$ be the topological space with the same underlying set as $G$ but considered as a subspace of $X \times B$.

Question: When shall $I(P)$ and $G$ be homeomorphic?

An Equivalent Formulation (using the same terminology as above) (added on April 04, 2014)

It is easy to see that $P$ is none else than the same underlying set of $X$ with the topology $\tau$ being generated by the open subsets of $X$ and inverse images of open subsets of $B$ under $f$, i.e., the new topology $\tau$ has basic open sets of the form $V \cap f^{-1}[U]$, where $V$ is (cl)open in $X$ and $U$ is open in $B$.

In this set up:

Question 1 Is it possible to characterize all the clopen sets of $\tau$?

Question 2 (reformulation of the old question): Under what conditions do the clopen subsets of $\tau$ become (cl)open in $X$ (i.e., the old topology of $X$)?

• If $f$ is constant then both $P$ and $G$ are homeomorphic to $X$. Thus $I(P)$ and $G$ are homeomorphic only if $X$ is such that every open set is closed. In the $T_0$ case this means that $X$ is discrete. – Ramiro de la Vega Apr 30 '14 at 16:53
• Thanks Ramiro, however, if one takes the identity map on $I(B)$, then also $P$ and $G$ are homeomorphic; in particular, if $B$ were zero dimensional then its identity map is such a case. But then $f$ is not constant, and neither does every open subset become closed. – Partha Pratim Ghosh Apr 30 '14 at 17:43
• Also, I could not follow the proof of your statement; could you please provide a proof? – Partha Pratim Ghosh Apr 30 '14 at 17:44