Continuous function mapping non-open sets to open sets

Question

While studying the product topology, I've come across an example that I can't shake.

Consider the product topology on $\mathbb{R}^2$. We know that

1) This product topology is the same as the "usual" topology on $\mathbb{R}^2$, i.e. that induced by the Euclidean metric

2) that this topology is the coarsest topology for which all projection maps are continuous.

Consider the set $(0,1)$ in one of the factor spaces $\mathbb{R}$. This is an open set whose preimage under the canonical projection map from $\mathbb{R}^2$ is $(0,1) $ X $ (-\infty, \infty)$, easily seen as open in the topology from (1).

Yet my problem is that the set $(0,1)$ X $[1,0]$ also maps to (0,1) under this projection map, and this set is not open. So while the preimage condition for continuity is satisfied, if someone gives you the question:

"$f$ is a continuous function between topological spaces $X$ and $Y$. Let $U \subset X$ and $V \subset Y$, with $V$ open. If $f(U) = V$, what can we say about $U$?"

The answer is really "nothing"?

Answer 1

That's correct, the answer is "nothing".

Maybe the following trivial example will help guide your intuition: Let $X$ be any topological space, and let $Y$ be the singleton space $\{*\}$. There is a unique function $f\colon X\to Y$ defined by $f(x) = *$ for all $x\in X$, and this map is automatically continuous. Now for any subset $X'\subseteq X$, we have $f(X') = Y$ if $X'$ is nonempty, and $f(X') = \emptyset$ if $X' = \emptyset$. In either case, $f(X')$ is open in $Y$. So the fact that $f(X')$ is open doesn't tell us anything at all about $X'$.