Borsuk–Ulam theorem in every dimension

We first rule out a continuous odd map $g:S^m\to S^{m-1}$ for $m\ge 1$. Such a map descends to $b:{\mathbb{R}\mathbb{P}}^m\to {\mathbb{R}\mathbb{P}}^{m-1}$, $b([x])=[g(x)]$. This is continuous by the quotient topology, since the composite with the sphere quotient is continuous and $g(-x)=-g(x)$.

In the coordinate subspace ${\mathbb{R}}^{m+1}\oplus {\mathbb{R}}^m\subset {\mathbb{R}}^{\mathrm{\infty }}$, define $H([x],t)=[\cos (\pi t\mathrm{/}2)x,\sin (\pi t\mathrm{/}2)g(x)],0\le t\le 1.$ Its representing vector has norm one. Replacing $x$ by $-x$ negates the whole vector, so the projective point is independent of representative. Continuity can be checked on each chart where a chosen coordinate of $x$ is nonzero: choose the unit representative with that coordinate positive, which varies continuously in the ratios of the remaining coordinates, and use the displayed formula. These charts cover projective space, proving joint continuity in $([x],t)$. Thus $H$ is a homotopy from $i_m$ to $j\circ b$, where $j([y])=[0,y]$ is the inclusion in the second block. The same formula with $y$ in the first $m$ coordinates and $y$ in the second block, namely $[\cos (\pi t\mathrm{/}2)y,0,\sin (\pi t\mathrm{/}2)y]$, is a homotopy from $i_{m-1}$ to $j$. Both homotopies lie in the finite projective subspace ${\mathbb{R}\mathbb{P}}^{2m}$ of the infinite space and therefore are continuous there as well.

Put $a_k=i_k^{\ast }a$. By [F3] and step 2.1, $a_m=b^{\ast }{a}_{m-1}$, including $m=1$ when $a_0=0$. Consequently $a_m^m=b^{\ast }(a_{m-1}^m)=0,$ since $H^m({\mathbb{R}\mathbb{P}}^{m-1};{\mathbb{F}}_2)=0$ by [F2]. On the other hand, [F1] says $a^m\mathrm{\ne }0$ and that $i_m^{\ast }$ is an isomorphism in degree $m$, so $a_m^m=i_m^{\ast }(a^m)\mathrm{\ne }0$. These incompatible conclusions rule out every such odd map.

For $m\ge 1$, if $h(x)-h(-x)$ never vanished, then $g(x)=\frac{h(x)-h(-x)}{\mathrm{\parallel }h(x)-h(-x){\mathrm{\parallel }}_2}$ would be continuous, take values in $S^{m-1}$ and satisfy $g(-x)=-g(x)$. Step 3.1 rules this out, so an antipodal pair has equal images. When $m=0$, the codomain ${\mathbb{R}}^0$ consists of one point and the two points of $S^0$ have equal images directly. AC is inherited exactly as stated in [A1].