Alternate topological definition of continuity The standard topological definition of continuity is as follows:

Definition: Continuity
Let $(X, \mathcal{T}_X)$ and $(Y, \mathcal{T}_Y)$ be topological spaces. A function $f : X \to Y$ is said to be continuous if for each open subset $V$ of $Y$, the set $f^{-1}(V)$ is an open subset of $X$.
In other words $f$ is continuous if for each $V \in \mathcal{T}_Y$, we have $f^{-1}(V) \in \mathcal{T}_X$.


But can we alternatively define continuity in the following way:

Possible Alternate Definition?: $f$ is continuous if for each $W \in \mathcal{T}_X$, we have $f(W) \in \mathcal{T}_Y$

But if we take $W = f^{-1}(V) \in \mathcal{T}_X$, then $f(W) = f(f^{-1}(V)) \subset V \in \mathcal{T}_Y$, but I don't think that $f(f^{-1}(V))$ needs to be open in $Y$ (as $f(f^{-1}(V)) \subset V$) , which leads me to believe that we cannot define continuity in the proposed way above.
Am I correct, or can we define continuity in the following way? If we cannot define continuity in the following way, are there any other (perhaps more intuitive) reasons why we cannot define continuity in arbitrary topological spaces in the proposed way above.
 A: In addition to Noah's example, here is one that shows that open and continuous is not even the same for bijective maps:
Consider $f:[0,2\pi)\to \{\left<x,y\right>\in\mathbb R^2 \mid x^2+y^2 = 1\}$ (both with the subspace topology) defined by
$$ f(t) = \left<\cos t, \sin t\right> $$
This is a bijection, and is continuous but not open. Its inverse is open but not continuous.
A: Such a map is called an open map, and is not the same as a continuous map: not every continuous map is open, and not every open map is continuous.


*

*A continuous map need not be open: consider a constant map. This is continuous, regardless of the topology on the source and target space (exercise), but in many cases is not open. For instance, in $\mathbb{R}$ with the usual topology, $\{0\}$ is not open, but $\mathbb{R}$ is, and the map $x\mapsto 0$ is continuous.

*An open map need not be continuous: consider the indiscrete and discrete topologies, $\mathcal{T}_i$ and $\mathcal{T}_d$, on a set $X$. Then the identity map $x\mapsto x$ is an open map from $(X, \mathcal{T}_i)$ to $(X, \mathcal{T}_d)$, but is not continuous if $X$ has at least two points. For maybe a better example, consider Conway's base thirteen function: this function sends every interval to all of $\mathbb{R}$, so the image of an open set is either $\emptyset$ or $\mathbb{R}$, so it's an open map; but it is clearly not continuous.
Note that we've found counterexamples in each direction in the context of $\mathbb{R}$ with the usual topology, so this isn't a case of the two notions agreeing in "natural" contexts; rather, even in nice, natural spaces, they're fundamentally different.
A: There are other equivalent (& useful) def'ns for the continuity of $f:X\to Y.$
(1). If $S$ is any closed subset of $Y$ then $f^{-1}Y$ is a closed subset of $X.$
(2). If $A\subset X$ and $p\in Cl_X(A)$ (the closure of $A$ in $X$) then $f(p)\in Cl_Y(f''A),$ where $f''A=\{f(a): a\in A\}.$ In other words $f''(Cl_X(A))\subset Cl_Y(f''A).$
(3). $f$ is continuous at every $p\in A.$ What continuity at $p$ means: For every open $V\subset Y,$ with $f(p)\in V,$ there exists open $U \subset X $, with $p\in U$ and $f''U\subset V.$
Number (2), when $X=Y=\mathbb R,$ means that for $f$ to be continuous, if a sequence $(x_n)_n$ converges to $x$ then $(f(x_n))_n$ converges to $f(x).$ Number (3), when $X=Y=\mathbb R,$ is the standard "$\epsilon$ -$\delta$" def'n of continuity at a point.
