TITLE: $X=\{a,b,c\}$ and $\mathcal{T}=\{X, \emptyset, \{a\}, \{b\}, \{a,b\}\}$. Determine if $f:X \rightarrow X$ is $\mathcal{T}-\mathcal{T}$ Continuous
QUESTION [5 upvotes]: $X=\{a,b,c\}$ and $\mathcal{T}=\{X, \emptyset, \{a\}, \{b\}, \{a,b\}\}$. 
Assume $f: X \rightarrow X$ is given by $f(a)=a, f(b)=c,$ and $f(c)=b.$ 
Determine if $f:X \to X$ is $\mathcal{T}-\mathcal{T}$ continuous
What I got: 
For $f$ to be continuous, then for each $\mathcal{T}$-open subset $V$ of $X,$ $f^{-1}(V)$ is a $\mathcal{T}$-open subset of $X$.
Since $f:(X,\mathcal{T}) \to (X,\mathcal{T})$, then $\{b\}$ is an open set but $f^{-1}(\{b\})=\{c\}$ which is not an open since $\{c\}$ $\notin \mathcal{T}$
Thus $f$ is not continuous.
Does that sound correct?

REPLY [2 votes]: Yes, your proof works.
Alternatively, you could argue that $c\in\overline{\{b\}}$ but $f(c)=b\notin\overline{f[\{b\}]}=\overline{\{c\}}$.
Note that $f\circ f=1_X$, so $f$ is continuous if and only if $f^{-1}=f$ is open.