Let $f:I\to I,$ where $I$ is compact set, given by $f(x)=x^2$
First, if $I$ is centered in $0$ then deg$(f)=0,$ cause $f$ isn't onto. But, if we consider $f:I\to f(I),$ then $\operatorname{deg}(f)=2?$ Is true?
I thought this way: if $q\in f(I)$ is a regular value, then exist $p_{1}=\sqrt{q}$ and $p_{2}=-\sqrt{q}.$ Now, $\operatorname{sign}df(p_{1})=1$ and sign$ \ df(p_{2})=-1$, therefore $\operatorname{deg}(f)=0.$
Where is my mistake? I know that in complexes, the degree of $z^{m}$ is $m$.So it's natural to think that the degree of $x^2$ is $2.$
Other question, if $I=[a,b],$ if $a\geq 0,$ then $\operatorname{deg}(f)=1?$ If $b\leq 0,$ then $\operatorname{deg}(f)=-1?$