# Continuous function mapping non-open sets to open sets

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 $$U$$?"

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 $$X'$$.