Why dont we say if $f(U)$ is open for every open set $U$ in $A$, then $f$ is countinous? Definition of continuity in topology;

Let $f :A \to B$.
If $f^{-1}(U)$ is open in A for every open set U in B, we say f is countinous.

Why dont we say if $f(U)$ is open for every open set $U$ in $A$, then $f$ is countinous ? 

Does the image of an open set under a continous mapping need to be open ? 
 Thank you for your help.
No. This would make the constant functions ($f:X\to\mathbb R; \ f(x) = c$) discontinuous, since

$$f(U) = \{c\}$$
for $U\ne\emptyset$ open.
Consider the metric space $T=\left \{ 0,1 \right \}$ with the discrete metric $d(x,y)=1$ if $x \neq y$ and $d(x,y)=0$ if $x=y$. 

It is easy to show that any subset of $T$ is open. 

Take any metric space $X$ and any discontinuous function $f:X \rightarrow T$ to obtain a counterexample.