$f:\mathbb{S}^1\rightarrow\mathbb{S}^1$ odd $\Rightarrow$ $\mathrm{deg}(f)$ odd (Borsuk-Ulam theorem) 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). $\blacksquare$
 A: To keep more to formulas, consider the following: As you already pointed out, by the lifting property there is a unique continuous function $q: \mathbb R \to \mathbb R$ such that 
$$f(e^{2\pi i t}) = e^{2\pi i q(t)}$$
Since $f$ is odd we have $$e^{2\pi i q(t+1/2)} = f(e^{2\pi i (t+1/2)}) = f(-e^{2\pi it}) = - f(e^{2\pi it}) = e^{2\pi i (q(t)+1/2)}$$ so
$$q(t+1/2) \equiv q(t) + 1/2 \pmod {\mathbb Z}$$
Therefore the continuous function $t \mapsto q(t+1/2) - q(t) - 1/2$ only takes values in $\mathbb Z$ and thus has to be constant.
So there is $n \in \mathbb Z$ such that $n = q(t+1/2) - q(t) - 1/2$ for all $t$.
Now
\begin{align*}\deg(f) &= q(1) - q(0)\\ &= [q(1) - q(1/2) - 1/2] + [q(1/2) - q(0) - 1/2] + 1 \\ &= 2n + 1\end{align*}
and we are done. Maybe you can find out what you've been missing by comparing this to your version of the proof (they are both more or less the same).
A: Finally figured it out, using the comment of Miha.
It is enough to prove $\widetilde{f\circ q}(1/2+t)=\widetilde{f\circ q}(1/2)+\widetilde{f\circ q}(t)$, since $t:=1/2$ gives the desired formula.
In general, for a given $\alpha:I\rightarrow\mathbb{S}^1$, $\alpha:\partial I\mapsto1$, by the definition and uniqueness of liftings, if $\beta:I\rightarrow\mathbb{R}$, $\beta:0\mapsto0$, and $p\circ\beta=\alpha$, then $\beta=\widetilde{\alpha}$.
In our case, $\alpha:=f\circ q(1/2+t)$ and $\beta:=\widetilde{f\circ q}(1/2)+\widetilde{f\circ q}(t)$. We have
$$\widetilde{f\circ q}(1/2)+\widetilde{f\circ q}(0)=(k+1/2)+0=\widetilde{f\circ q}(1/2+0)$$ 
and 
$$p\big(\widetilde{f\circ q}(1/2)+\widetilde{f\circ q}(t)\big)=e^{2\pi i ...}=p(\widetilde{f\circ q}(1/2))\cdot p(\widetilde{f\circ q}(t))$$
$$=-1\cdot f\circ q(t)=f(-q(t))=f\circ q(1/2+t),$$
which proves the claim.
Intuitively, when $t$ moves from $0$ to $1/2$, $q(t)$ moves in the upper half of the circle, $f\circ q(t)$ wraps around $\mathbb{S}^1$, and $\widetilde{f\circ q}(t)$ moves up and down on the spiral $\mathbb{R}$. But when $t$ moves from $1/2$ to $1$, $q(t)$ moves in the lower half of the circle, on which $f$ has the same values as on the upper half ($f(-x)=-f(x)$), so $f\circ q(t)$ is the same, and therefore the lift $\widetilde{f\circ q}(t)$ is also the same, just translated in $\mathbb{R}$ by $\widetilde{f\circ q}(1/2)$.
