I have a question about the example in the image below: the screenshot comes from the notes of an introductory course to topological K-theory, and the annotations (which for this question can be ignored) are made by the teacher; I couldn't find the original textbook, that's why I pasted it directly from the notes.
My doubt is the following. We define $t(x)$ depending on $x\in S^n$, in a way that the map $s\circ p:S^n\to E\subset \mathbb RP^n\times \mathbb R^{n+1}$ is defined, for all $x\in S^n$, by $s\circ p(x)=(p(x), t(x)x)$. Then, since $s\circ p(x)=s\circ p(-x)$, we have $t(x)x=t(-x)(-x)$, implying $t(-x)=-t(x)$. However look at the last three lines in the screenshot: it seems that they set $t:=t(x)$, but then $s\circ p(-x)=(p(-x), t(-x)(-x)=(p(x),tx)$, and not $(p(x),-tx)$; setting $t:=t(-x)$ doesn't change the situation obviously. I'm sure that I'm making confusion somewhere, but I can't find it. Would you please clarify this a bit? Thank you
Those notes are at best confusing and at worst wrong. They seem to be using $t$ to denote $t(x)$ and $t(-x)$ simultaneously which seems dubious to me. Moreover, they seem to be deducing that $t(x)$ has to be zero at all points which is bizarre since there exist sections that are not always zero. The correct conclusion should be that $t(x)=0$ is zero for at least one point $x\in S^n$.
However, I think one can easily patch up the argument. As you say, comparing the last coordinates gives the the equality $t(-x)\cdot (-x)=t(x) \cdot x$, from where one deduces that $t(-x)=-t(x)$ for all $x$ in $S^n$. However $t:S^n \rightarrow \mathbb{R}$ is a continuous function and $S^n$ is connected for $n\geq 1$. So if we have $t(x_0)>0$ for some point $x_0\in S^n$, then $t(-x_0)<0$ and the intermediate value theorem for connected topological spaces implies that there must be a point $y\in S^n$ with $t(y)=0$ which is what we need.