Degree is the unsigned number of points in a regular fibre

The map is proper. If $K\subseteq \mathbb{R}$ is compact, [F2] makes it closed and bounded, say $\mathrm{\mid }y\mathrm{\mid }\le B$ on $K$. Then $F^{-1}(K)$ is closed by continuity and is contained in $[-\sqrt{B},\sqrt{B}]$ when $B\ge 0$; if $K$ is empty its inverse image is empty. Thus [F2] makes $F^{-1}(K)$ compact.

The value $1$ has exactly the two preimages $-1$ and $1$. Since $F^{\mathrm{\prime }}(x)=2x$, both are regular, but their local orientation signs are $-1$ at $-1$ and $+1$ at $1$. Hence [F1] gives $\mathrm{deg}(F)=(-1)+(+1)=0,$ whereas the unsigned fibre cardinality is $2$.

This explicit witness disproves the claimed equality. A singleton fibre whose local sign is $+1$ would make the two numbers agree, while an empty regular fibre gives both zero; those special cases do not remove the cancellation in step 1.2. The example is one-dimensional, boundaryless, nonempty, and uses no selections or choice principle.