About a condition for a continuous mapping to be open. The text (Foundations of General Topology, by Pervin, Second edition) says 

a (continuous) mapping $f$ of $X$ into $X^*$ is open iff $f(i(E))\subseteq i^*(f(E))$ for every $E\subseteq X$.

EDIT: $i(E)$ is the interior of set $E$ in $X$. Similarly, $i^*$ is the interior operator in $X^*$. 
However, shouldn't $f(i(E))=i^*(f(E))$?
Motivation:


*

*$i^*(f(E))$ is an open set. Hence $f^{-1}[i^*(f(E))]$ will be an open set in $X$. This open set $\subseteq i(E)$. Hence, $i^*(f(E))\subseteq f(i(E))$

*$i(E)$ is an open set in $X$. Hence, as $f$ is an open mapping, $f(i(E))$ will be an open set in $X^*$. This open set will be $\subseteq i^*(f(E))$. Hence, $f(i(E))\subseteq i^*(f(E))$

*From $(1)$ and $(2)$ we conclude that $f(i(E))=i^*(f(E))$.  
 A: Let $\newcommand{\Int}{\operatorname{Int}}X = \{ 0 , 1 \}$ be given the anti-discrete (trivial) topology, and let $Y = \{ y \}$ be given its only topology.  It is easy to check that the constant mapping $f : x \mapsto y$ is open continuous.  However $$f [ \operatorname{Int} ( \{ 0 \} ) ] = f [ \varnothing ] = \varnothing \subsetneq \{ y \} = \operatorname{Int} ( \{ y \} ) = \operatorname{Int} ( f [ \{ 0 \} ] ).$$

As far as your justification goes, while true that $f^{-1} [ \Int ( f [ E ] ) ]$ is open, it may not be a subset of $E$, because $f$ may not be one-to-one.  Therefore the claim that $f^{-1} [ \Int ( f [ E ] ) ] \subseteq \Int (E)$ may be false (and it is in the above example).
A: Arthur’s minimal example really exposes what’s going on, but you may want to see another example in a more familiar setting.
Let $X=\Bbb R$, let $Y=[0,\to)$ with the subspace topology inherited from $\Bbb R$, and let $$f:X\to Y:x\mapsto x^2\;.$$ Certainly $f$ is continuous. Suppose that $I=(a,b)$ is an open interval in $X$. Then
$$f[I]=\begin{cases}
\left(a^2,b^2\right),&\text{if }0\le a<b\\\\
\left(b^2,a^2\right),&\text{if }a<b\le 0\\\\
\left[0,\max\{a^2,b^2\}\right),&\text{if }a<0<b\;.
\end{cases}$$
so in every case $f[I]$ is open in $Y$, and $f$ is an open map. Let 
$$A=\Big(\Bbb Q\cap[0,\to)\Big)\cup\Big((\leftarrow,0)\setminus\Bbb Q\Big)\;,$$
the set of non-negative rationals and negative irrationals. Clearly $A$ is not open in $X$, but $f[A]=Y$ is open in $Y$. Moreover, 
$$f[\operatorname{int}A]=f[\varnothing]=\varnothing\subsetneqq Y=\operatorname{int}f[A]\;.$$
A: I'll attempt to answer my own question. This will be a little theoretical, as compared to Arthur and Brian's brilliant example-illustrated answers. 
I stated that if set $E\in X$ is mapped to $f(E)$, then a subset of $E$ will be mapped to a subset of $f(E)$.
This is not necessarily true if this is a many-one mapping. A set $\supset E$ may also be mapped to the same mapping as the subset of $E$. Hence, if $f$ is an open continuous mapping, it is NOT necessary that $f^{-1}[Int(f(E))]\subseteq Int(E)$. The condition I have stated is valid only when $f$ is a one-one or one-many mapping. 
As for $f(Int(E))\subseteq Int^*(f(E))$, 


*

*the relation is $f(Int(E))= Int^*(f(E))$ for one-one and one-many mappings, and 

*the relation is $f(Int(E))\subseteq Int^*(f(E))$ for many-one mapping. 
