Maps from even projective space to the sphere use mod-two degree

By [F1], and because $n$ is even, ${\mathbb{R}\mathbb{P}}^n$ is a compact connected nonorientable smooth $n$-manifold with the affine-chart structure. Parametrize its top-cell attachment by $\pi (x)=[x:\sqrt{1-\mathrm{\mid }x{\mathrm{\mid }}^2}]$ for $x\in D^n$. On the interior, the last-coordinate affine chart gives $u=x\mathrm{/}\sqrt{1-\mathrm{\mid }x{\mathrm{\mid }}^2}$, with smooth inverse $x=u\mathrm{/}\sqrt{1+\mathrm{\mid }u{\mathrm{\mid }}^2}$, so $\pi$ restricts to a diffeomorphism onto the open top cell. On the boundary $\pi (x)=[x:0]$ is the antipodal attachment onto ${\mathbb{R}\mathbb{P}}^{n-1}$. In particular ${\mathbb{R}\mathbb{P}}^n$ is nonempty and has no boundary.

The continuous collapse map $q:=\bar{c}\circ Q$ is well defined and continuous, where $Q:{\mathbb{R}\mathbb{P}}^n\to {\mathbb{R}\mathbb{P}}^n\mathrm{/}{\mathbb{R}\mathbb{P}}^{n-1}$ is the quotient map and $\bar{c}$ is the homeomorphism ${\mathbb{R}\mathbb{P}}^n\mathrm{/}{\mathbb{R}\mathbb{P}}^{n-1}\to S^n$ induced by $c$ through [F2]; equivalently $q\circ \pi =c$ on $D^n$, and $\bar{c}$ is the homeomorphism of [F2], so the identification of the quotient with $S^n$ is exactly the one exhibited by the classical pinch.

The model $F$ is constant $N$ for $\mathrm{\mid }x{\mathrm{\mid }}^2\ge 3\mathrm{/}4$. Thus $\tilde{q}(\pi (x))=F(x)$ is well defined: only boundary points of $D^n$ are identified, and $F$ has the same value on them. It is smooth on the open cell because $\pi$ is a local diffeomorphism there; near the lower skeleton it is constant, since the closed smaller disk $\mathrm{\mid }x{\mathrm{\mid }}^2\le 3\mathrm{/}4$ has image disjoint from that skeleton. To exhibit a homotopy from $c$ to $F$ relative to $\mathrm{\partial }{D}^n$, write $s=\mathrm{\mid }x{\mathrm{\mid }}^2$, let $W(s)=1-\chi (s{)}^2(1-s)$ be the profile of [F3], and set $W_{\tau }(s)=(1-\tau )s+\tau W(s)$. For $x\mathrm{\ne }0$ define $C_{\tau }(x)=(2\sqrt{W_{\tau }(s)(1-W_{\tau }(s))\mathrm{/}s}x,2W_{\tau }(s)-1)$, and set $C_{\tau }(0)=y_{-}$. This is continuous jointly in $x,\tau$: near $s=0$ one has $W_{\tau }(s)=s$, and elsewhere $s>0$ the displayed square root is continuous and nonnegative. Its norm is one, $C_0=c$, $C_1=F$ because $\chi \ge 0$, and for $s=1$ it is always $N$. The map $\pi \times {\mathrm{i}\mathrm{d}}_I:D^n\times I\to {\mathbb{R}\mathbb{P}}^n\times I$ is a quotient map by [F2], since its source is compact and its target Hausdorff. Thus this family, constant on its fibres, descends continuously to a homotopy $q\simeq \tilde{q}$. No radial diffeomorphism or homeomorphism assertion about $F$ is needed.

The point $y_{-}=F(0)$ is a regular value of $\tilde{q}$ with the single preimage $\pi (0)$: $\tilde{q}(\pi (x))=F(x)=y_{-}$ forces $x=0$ by [F3], the differential satisfies $d{\tilde{q}}_{\pi (0)}=d{F}_0\circ (d{\pi }_0{)}^{-1}$ by step 1.1, hence is invertible, and points of ${\mathbb{R}\mathbb{P}}^n$ outside the image of the interior of the disk have value $N\mathrm{\ne }{y}_{-}$; therefore ${\mathrm{deg}}_2(\tilde{q})=1$ by [F4].

Since $\tilde{q}$ is homotopic to $q$, [F4] gives ${\mathrm{deg}}_2(q)={\mathrm{deg}}_2(\tilde{q})=1$; a constant map has an empty regular fibre over any other value, so its mod-two degree is $0$, and the two values of $\mathbb{Z}\mathrm{/}2$ are realized. By [F5] applied to the nonorientable manifold ${\mathbb{R}\mathbb{P}}^n$, the mod-two degree is a bijection $[{\mathbb{R}\mathbb{P}}^n,S^n]\to \mathbb{Z}\mathrm{/}2$, so the classes of $q$ and of the constant map are the two classes and no integer degree is available for ${\mathbb{R}\mathbb{P}}^n$.