Here is the question:
Having answered this question:
by the following bijection:
$$\pi(f) = f^{-1}({a})$$ where $a,b$ are the points in the Sierpinski Space. and its bijection is for an open set $U$ in $X$ we have $f_{U}(x) = a$ if $x \in U$ and $f_{U}(x) = b$ if $x \notin U$
My question is:
How can I answer my new question? is there a way to get a solution depending on modifying the solution of the problem I already solved?
Similarly let $Z=\{0,1\}$ with discrete topology, $C(X,Z)$ be the set of all continuous functions and $clo(X)$ be the set of all clopen subsets of $X$. Define
$$\chi:C(X,Z)\to clo(X)$$ $$\chi(f)=f^{-1}(\{1\})$$
First of all $\chi$ is well defined, since $\{1\}$ is clopen in $Z$ and thus its preimage is clopen in $X$. We will now construct its inverse:
$$\psi:clo(X)\to C(X,Z)$$ $$\psi(A):X\to Z$$ $$\psi(A)(x):=\begin{cases}1 & \text{if }x\in A \\ 0 & \text{otherwise}\end{cases}$$
Note that $\psi(A)$ is a continuous function because $A$ is clopen, i.e. $\psi$ is well defined. Can you verify that it is the inverse of $\chi$?
EDIT: So how to prove that these are inverses of each other? First we need to show that $\psi(\chi(f))=f$. Two functions are equal if they are equal on each value, so lets evaluate:
$$\psi(\chi(f))(x)=\begin{cases}1 &\text{if }x\in\chi(f) \\ 0&\text{otherwise}\end{cases}$$
And what does it mean for $x$ to belong to $\chi(f)$? By the definition of $\chi$ this means that $x\in f^{-1}(\{1\})$ or equivalently $f(x)=1$. So we obtain that if $\psi(\chi(f))(x)=1$ then $f(x)=1$ and so $\psi(\chi(f))(x)=1=f(x)$. Anlogously we show that if $\psi(\chi(f))(x)=0$ then $f(x)=0$. And so $\psi(\chi(f))=f$.
Now the other way around we need to show that $\chi(\psi(A))=A$ for a clopen subset $A$. This we can show by two inclusions. So let $x\in \chi(\psi(A))$. This means (by definition of $\chi$) that $x\in(\psi(A))^{-1}(\{1\})$. But if you look at $\psi(A)$ then you'll notice that $\psi(A)^{-1}(\{1\})=A$. And so $x\in A$. This shows "$\subseteq$" inclusion. For "$\supseteq$" consider $x\in A$. Then $\psi(A)(x)=1$ by definition of $\psi$. And thus by definition of $\chi$ we have $x\in\chi(\psi(A))$.