I'm having trouble understanding the proof of Borsuk-Ulam theorem ($n=2$) that we did in our class. The only problematic part is the last sentence in the proof of lemma 1.
$\mathbb{S}^1\subseteq\mathbb{C}$. We know that $\mathrm{deg}:\pi_1(\mathbb{S}^1)\rightarrow \mathbb{Z}$, $\mathrm{deg}([\alpha]):=\tilde{\alpha}(1)$, is an isomorphism, where $\tilde{\alpha}:I\rightarrow\mathbb{R}$ is the unique lifting of the loop $\alpha:I\rightarrow\mathbb{S}^1$, i.e. $p\circ\tilde{\alpha}=\alpha$, where $p:\mathbb{R}\rightarrow\mathbb{S}^1$, $p(t)=e^{2\pi it}$.
Lemma 1: continuous $f:\mathbb{S}^1\rightarrow\mathbb{S}^1$ odd (i.e. $f(-x)=-f(x)$) $\Longrightarrow$ $\mathrm{deg}(f)\in\mathbb{Z}$ odd.
Proof: Without loss of generality: $f(1)=1$ (if not, then we compose $f$ with a rotation; the degree and odd-ness is preserved). If $q:I\rightarrow\mathbb{S}^1$, $q(t):=e^{2\pi i t}$, then $q(t+1/2)=-q(t)$. Now $f\circ q:I\rightarrow\mathbb{S}^1$ is a loop at $1$, $\widetilde{f\circ q}$ its lifting, and $\mathrm{deg}(f)=\widetilde{f\circ q}(1)$. Clearly $p\circ\widetilde{f\circ q}(1/2)=f\circ q(1/2)=-1$, so by the definition of $p$, we have $\widetilde{f\circ q}(1/2)=k+1/2$ for some $k\in\mathbb{Z}$. (So far, I understand everything; here is what troubles me.) Since $f$ is odd, we have $\widetilde{f\circ q}(1)=k+1/2+k+1/2=2k+1$. WHY?
Lemma 2: continuous $f:\mathbb{S}^2\rightarrow\mathbb{S^1}$ is not odd.
Proof: $\mathbb{S^1}\overset{i}{\hookrightarrow}\mathbb{S}^2\overset{f}{\rightarrow}\mathbb{S^1}$. If $f$ is odd, then so is $f\circ i$. By Lemma 1, $f\circ i$ has odd degree, so it isn't nullhomotopic. But since $i$ is not surjective, it is nullhomotopic, hence so is $f\circ i$, $\rightarrow\leftarrow$. $\blacksquare$
Theorem (Borsuk-Ulam, $n=2$): $\forall$ continuous $f:\mathbb{S}^2\rightarrow\mathbb{R}^2$ $\exists x\in\mathbb{S}^2$: $f(-x)=f(x)$.
Proof: If the theorem were not true, then $F(x):=\frac{f(x)-f(-x)}{\|f(x)-f(-x)\|}$, would be a well defined continuous odd map $\mathbb{S}^2\rightarrow\mathbb{S}^1$, $\rightarrow\leftarrow$ (Lemma 2). $\blacksquare$