Question: Say the topological space $T$ having a basis as the $B=\{(a,b] \vert a,b\in \mathbb{R}, a<b\}$on the real number set, $\mathbb{R}$. Defining the $f\_1, f\_2 : \mathbb{R} \to (\mathbb{R},T)$ by $f\_1(x) = x^2$ and $f\_2(x) = -x^2$. Find the closure of the $(0,1]$ in the $(\mathbb{R}, T\_1)$. Here the $ T\_1$ is the minimal topological space such that both $f\_1, f\_2$ are continuous.

From now, My solution starts.
Try to solve the above question, I find the form of the basis that $f\_1^{-1}(G) \cap f\_2^{-1}(H)$ for the $G,H\in (\mathbb{R},T)$.
Hence, The minimal open set of the each points are $\{\pm a\}, (a\neq 0)$ or $(-\epsilon, \epsilon)$. So my answer is $[-1,1]$ considering the accumulation points. Is my answer right?

Answer: 
I believe the other two answers are wrong: Note that $f\_1^{-1}(]-1,0])=\{0\}$, so that $\{0\}$ is open in the topology $T\_1$. Therefore, $\mathbb R\setminus\{0\}$ is closed and therefore the closure of $]0,1]$ should be
$$[-1, 0[\cup]0,1]$$ and not $$[-1,1].$$
I think your answer is not correct. First of all $[-1,1]$ is closed in your topology because

$\mathbb{R}\setminus[-1,1]=f\_1^{-1}(1,\infty)$

Thus the closure $C$ of $(0,1]$ is contained in $[-1, 1]$. Let us suppose there is $a\in [-1,1]\setminus \{0\}$ that is not in $C$ (so $a<0$).

Any open neighbourhood of $a$ contains $f\_1^{-1}(-\epsilon+a^2, a^2]$ that intersect always $(0,1]$ in $-a$. This means that $a$ is a point in the closure of $(0,1]$ and so $a\in C$, that is a contradiction.

Moreover $\{0\}$ is not in the clousure of $(0,1]$ because $\{0\} =f\_1^{-1}((-1,0]))$ does not intersect $(0,1]$.

Thus $C=[-1,1]\setminus \{0\} $.

You can do a similar reasoning for the map $f\_2$