There is a problem in Armstrong's Basic Topology in chapter 9.3 that I just can't seem to solve by myself, neither have I found a solution online. It goes as follows:
Suppose we have a map $f:S^n \to S^n$ that extends over $D^{n+1}$ in the sence that there is a map $F:D^{n+1}\to S^n$ with $F|_{S^n}=f$. Then there exists a $x\in S^n$ such that $f(x)=f(-x)$.
My idea was to glue two $D^{n+1}$ along $S^n$ to obtain the $S^{n+1}$ and a map $\tilde{F}:S^{n+1}\to S^n$ induced by $F$. Then by the Borsuk Ulam theorem I get a $x\in S^{n+1}$ s.t. $\tilde{F}(x)=\tilde{F}(-x)$, but I can't see how to get a $x\in S^n$ with this property.
Armstrong also states that $f$ having even degree is already sufficent, any hint on how to see that would also be much apprechiated.
Best Answer
A map $h:S^n\to S^n$ such that for all $x, h(-x) = -h(x)$ has odd degree (see e.g. here, or here for an easier proof when $n$ is odd - it has been pointed out to me that the even case can be deduced from the odd case by taking suspensions, which makes for an overall easier proof)
So assume an $f$ of even degree exists with no $x$ such that $f(x)=f(-x)$. Then one may define $g(x) := \frac{f(x)-f(-x)}{||f(x)-f(-x)||}$ which is well-defined and continuous, and satisfies $g(-x) = -g(x)$, so $g$ has odd degree.
However, one may also define $H(x,t) := \frac{tf(x)-(1-t)f(-x)}{||tf(x)-(1-t)f(-x)||}$, $t\in [0,1]$
Indeed if $tf(x) = (1-t)f(-x)$ then by taking the norm we get $t=1-t$ so $t=\frac{1}{2}$ so $f(x) = f(-x)$ which contradicts the assumption.
Now $H(-, \frac{1}{2})=g$ so $f\sim g$. Therefore $g$ has even degree. That's absurd.
It follows that there is $x$ such that $f(x)= f(-x)$. In the first special case, if there is an extension $F$ of $f$, then $f$ is nullhomotopic, so has degree $0$, which is even.