How do I prove that there is no map $f:B^2\rightarrow S^1$ such that $f(-x)=-f(x)$ on $S^1$? How do I prove that there is no map $f:B^2\rightarrow S^1$ such that $f(-x)=-f(x)$ on $S^1$?
I'm not familiar with this kind of problems. I'm only comfortable with algebraic relations between functions such as homotopic relation. I have no idea how to tackle this kind of problem asking properties of "elements". (That is, values of functions) Some recommendation please.. 
Thank you in advance
 A: Consider the image over $f$ of the loop $\gamma:[0, 1] \to B^2$ given by $\gamma(t) = e^{2\pi it}$. We have $f(\gamma(t)) = -f(\gamma(t\pm 0.5))$, which means that $\gamma' = f\circ\gamma$, is not nullhomotopic. (Hint: try to lift the image of $\gamma'$ to the covering space $\Bbb R$.)
Now consider any other loop $\tau_r(t) = re^{2\pi it}$ with radius $0 < r < 1$, and its image $\tau'_r$ under $f$. Since there is a canonical homotopy between $\gamma$ and $\tau$ (continuous scaling), there must be a homotopy between $\gamma'$ and $\tau'$, which means that none of the $\tau_r'$ are nullhomotopic. But that means that for any open neighbouthood $U$ of $0$, we have $f(U) = S^1$, which means the function cannot take any value at $0$ and still be continuous.
Edit Some clarifications that should've been there from the beginning: Since I called $\gamma'$ a loop, I assumed that $\gamma'(0) = \gamma'(1)$, and that any homotopy of $\gamma'$ would respect this restriction. About lifting to the covering space, I was talking about a lift from $\gamma':[0, 1] \to S^1$ to $p:[0, 1]\to \Bbb R$, not the other way around.
The reasoning behind $\gamma'$ not being null-homotopic is that for any given loop the following holds: "The loop is null-homotopic" $\Longleftrightarrow$ "Any lift of the loop to any covering space is still a loop" (i.e. the end points meet up). Lift $\gamma'$ to $p$ with, say, $p(0) = 0$. Then the original condition $f(-x) = -f(x)$ means for $p$ that $p(x+0.5) = p(x) + n + \frac{1}{2}$ for some $n \in \Bbb Z$ and $0\leq x \leq \frac{1}{2}$. This condition prevents $p$ from being a loop, and therefore $\gamma'$ is not nullhomotopic.
