# A bijective function $f$ is continuous iff it is open(?)

I'm reading Matsuzaka Topology and it says a bijective function $f$ is continuous iff it is open. Then how can I prove this? The book also says it is obvious but it's not obvious to me. I'm so confused, how can I prove this?

• This is false. $f$ is continuous if and only if $f^{-1}$ is open. Can you see why now? It should be much easier, it's just the condition $$\mbox{ for all open U, f^{-1}(U) is open}$$ – Crostul Jun 25 '17 at 13:09
• In general, this claim cannot be correct, since a bijective, continuous and open function is a homeomorphism, which is (much) stronger than bijective+continuous. In which context does Matsuaka make this claim? – haemi Jun 25 '17 at 13:13
• Oh, I'm so sorry...I misread my textbook. Thank you very much Crostul and Haemi! – lacm Jun 25 '17 at 13:17
• What did it say then? – Henno Brandsma Jun 25 '17 at 14:37
• not $f$ but $f^{-1}$ – lacm Jun 25 '17 at 16:41

This isn't true: consider the function $f:[0,1) \rightarrow S^1$ given by $f(x) = e^{2\pi ix},$ where the notation $S^1$ refers to the $1$-sphere, defined by: $$S^1 = \{z \in \mathbb{C} : |z|=1\}.$$ Then:
• $f$ is a bijection
• $f$ is continuous
• $f$ isn't open; for instance, the set $f([0,1/2))$ isn't open in $S^1$.
The author probably means that a bijective function $f:X \to Y$ is open iff its inverse is continuous. And $f$ is continuous iff its inverse is open. This is just the observation that if $g: Y \to X$ is the inverse of $f$: for any (open) subset of $Y$: $f^{-1}[O] = g[O]$.
Also true for bijections: $f$ is open iff $f$ is closed, because $f[X\setminus A] = Y \setminus f[A]$ for all $A$ (so if $f$ is open, the image of a closed set, which is of the form $f[X\setminus O]$ is again the complement of an open set $f[O]$, hence closed, and vice versa).