0

The Topological Definition of Continuity was stated by my Differential Geometry Professor as:

A function $f:X\rightarrow Y$ is continous if given any open set $V\subset Y$ its pre image $f^{-1}(V)$ is also an open subset of $X$.

My question is: can the definition be stated backwards? In every example I can think of this definition should also work

A function $f:X\rightarrow Y$ is continous if given any open set $V\subset X$ its image $f(V)$ is also an open subset of $Y$.

I haven't found this other statement in any book and I wanted to know if it is equivalent or if there is some case where it just doesn't work.

P. C. Spaniel
  • 169
  • 1
  • 11
  • 1
    A map satisfying that property is called open. See

    https://en.wikipedia.org/wiki/Open_and_closed_maps

     –  Jan 26 '18 at 17:58
  • 2
    As a counterexample, the constant function $0 : \mathbb{R} \to \mathbb{R}$ is continuous but not open. And similarly, for any subspace $X$ of $Y$, the inclusion $X \hookrightarrow Y$ is always continuous, but it is open if and only if $X$ is an open subset of $Y$. – Daniel Schepler Jan 26 '18 at 18:00
  • @PaulK thanks! knowing that my question actually becomes the same as this ones: https://math.stackexchange.com/questions/68468/example-of-a-non-continuous-function-that-is-open-and-not-closed?rq=1 , https://math.stackexchange.com/questions/285462/open-and-closed-mapping-are-not-necessarily-continuous , https://math.stackexchange.com/questions/75589/open-maps-which-are-not-continuous. – P. C. Spaniel Jan 26 '18 at 18:04
  • @DanielSchepler great example, thanks! – P. C. Spaniel Jan 26 '18 at 18:04

2 Answers2

1

It's not equivalent. Consider the map $y=\sin x$ as a map from $\mathbb{R}$ to $\mathbb{R}$. It's continuous by the conventional definition. But the image of the open set $(-100,100)$ (or the open set $\mathbb{R}$) is the non-open set $[-1,1]$.

BallBoy
  • 14,472
1

A function that maps open sets to open sets is called "open".

Consider the set $A=\{x,y\}$ and the topologies $\alpha=\mathcal P(A)$ and $\beta=\{\emptyset, A\}$.

The identity $i:(A,\alpha)\to(A,\beta)$ is continuous but not open and "the other identity" $i^{-1}:(A,\beta)\to(A,\alpha)$ is open but not continuous.

ajotatxe
  • 65,084