Equivalent Definitions of Continuity using Open Sets

Question

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.

Answer 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 $[-1,1]$.

Answer 2

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.