Continuous maps in topology; the definition? I am just wondering, given the definition of continuous maps as follows,

A functionn $f:X \to Y$ is continuous if for every open subset $U $ of $Y$ the preimage $f^{-1}U$ is open in $X$.

I guess mathematically, this doesn't necessarily mean that "an open subset of $X$ is mapped to an open subset in $Y$"?
It's only that the open subset of $Y$ must originate from an open subset in $X$, but not necessarily that every open $V$ of $X$ will be mapped to some open $U$ of $Y$.
Is this understanding correct?
 A: Yes, that is correct.
A function that maps open sets to open sets is called an open map,
i.e a function $f : X \rightarrow Y$ is open if for any open set $U$ in $X$,
 the image $f(U)$ is open in $Y$.
Open maps are not necessarily continuous.
Then there is the concept of closed maps which maps closed sets to
closed sets. A map may be open, closed, both, or neither and
continuity is independent of openness and closedness.
A continuous function may have one, both, 
or neither property.
A: Continuous maps don't have to map open sets to open sets. An example is the map $f:\mathbb{R}\rightarrow\mathbb{R}$ given by $f(x)=x^2$ which maps $(-1,1)$ to $[0,1)$ which is not open and not closed.
A: Yes it is. Consider, for example, the continuous function $$f(x) = x^2$$ What is the image of the open set $(-1,1)$ ?
A: Yes.  I tend to think of this (quickly) as "continuous = open sets come from open sets" and more slowly as "the preimage of an open set is an open set".
There is another term:  A map is open if it takes open sets to open sets.  A map may be open, continuous, neither, or both.  This is the other idea that a person first learning this definition of continuous may conflate with continuity.
To help set this idea.  The sine map on $\Bbb{R}$ is clearly continuous (using the ideas you had before topology).  However, the interval $(-100,100)$ (large enough easily to hold an entire period) is mapped by sine to $[-1,1]$, so continuous maps are not automatically open maps.
A: Yes that is correct. 
If a function maps open sets to open sets, then it is said to be an open map.
A continuous map is not necessarily open. For example the $\sin$ function is continuous but not open since it maps the open interval $(0,\pi)$ to $(0,1]$, which is not open.  
However, note that if a continuous map $f$ has an inverse $f^{-1}$, then $f^{-1}$ is an open map.
