Suppose a function $\phi:[0,1] \rightarrow [-1,1]$. Assume that the function $\phi$ has discontinuity at $x=1$ and $\phi(1)=0$.
Question: Is it possible to construct a bijection and continuous function $\phi$ which satisfies all the conditions above?
EDIT: Sorry, the question should be: Construct a bijection function whereby it is continuous everywhere in $[0,1]$ except $x=1$.
Suppose such a function $f$ exists that is continuous on $[0,1)$, bijective and $f(1)=0$. Then $f$ must be surjective, so it must attain all values in $[-1,1]$. $f$ is also injective.
In particular, $f$ is continuous and injective on $[0,1)$, so $f$ is monotone on $[0,1)$: https://proofwiki.org/wiki/Continuous_Injection_of_Interval_is_Strictly_Monotone
So $f(0)= -1 $ or $f(0)= 1$. This is because if $f$ is strictly decreasing on $[0,1)$ then $f(0)=1$ and if $f$ is strictly increasing, $f(0)=-1$. WLOG assume $f(0)=-1$.
$f$ cannot attain the value $0$ anywhere in $[0,1)$ since that would make it non-injective since $f(1)=0$.
To be surjective $f(x) = 1$ for some $x \in [0,1)$. Now take the compact interval $[0,x]$. Then by the intermediate value theorem there exists some $y \in [0,x]$ such that $f(y) = 0$. But $y \in [0,x] \subseteq [0,1)$. Contradiction.
For every compact interval $[0,a]$ (where $0<a<1$) the restriction of $\phi$ to $[0,a]$ is monotonic and reaches its maximum or its minimum at $a$.
Assume, for simplicity, that $\phi$ is increasing. Then $$\lim_{x\to1^-}\phi(x)=1\text{ (Why?)}$$ and since $\phi([0,1))=[-1,1)$, $\phi(1)$ must be $1$ and hence $\phi$ is continuous.
