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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 $\operatorname{deg}(f)=-1?$

Kalashinikov
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    You didn't make a mistake. Assuming $I$ denotes an interval in $\mathbb{R}$, it's contractible, so the degree of any self-map is $0$, to the extent that this definition makes sense. – hunter Nov 05 '21 at 19:21
  • @hunter thank you sir, but what's the difference from what I did for the complex case? – Kalashinikov Nov 05 '21 at 19:44
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    You have to be very careful working with degree on non-compact manifolds or manifolds with boundary. The answer to your question, however, is that holomorphic maps are always orientation-preserving away from critical points. The same is not true of real smooth mappings. – Ted Shifrin Nov 05 '21 at 20:48
  • @TedShifrin ooh, thank you sir! – Kalashinikov Nov 05 '21 at 21:09

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