I am reading a proof of the above-mentioned statement from the book Characteristic Classes by Milnor and Stasheff. There, in the proof they used the Intermediate value theorem to prove the result by constructing a continuous, odd function $t$ from $S^n$ to $E(\gamma_n^1)$. I am facing problem in understanding the construction of $t$.
Proof from the book:
The total space
$E(\gamma_n^1) := \{(\{\pm x\}, \lambda x) : \{\pm x\} \in \mathbb{R}P^n ~~\&~~ \lambda \in \mathbb{R}\}$. Now we take any section
$$
s:\mathbb{R}P^n \to E(\gamma_n^1)
$$
and show that it is not nowhere-zero. Consider the composition
$$
S^n \xrightarrow[]{q} \mathbb{R}P^n
\xrightarrow[]{s} E(\gamma_n^1)
$$
which carries each $x \in S^n$ to some pair $(\{\pm x\},t(x)x) \in E(\gamma_n^1)$.
Evidently $t:S^n \to \mathbb{R}$ is a continuous function with $t(-x)=-t(x)$
Since $S^n$ is connected it follows from IVT that $t(x_0)=0$ for some $x_0 \in S^n$. Hence $s(\{\pm x_0\}) = (\{\pm x_0,0\})$. This completes the proof.
My question is regarding the construction of $~t~$:
Since no explicit definition of $t$ is given in the book, I have tried to construct a definition of $t$: As for each $x \in S^n$ we have $$ q(x)=\{\pm x\} \in \mathbb{R}P^n ~~\&~~ s(\{\pm x\}) = (\{\pm x\},\lambda_0 x) ~\text{for some} ~\lambda_0 \in \mathbb{R} ~\text{we define} ~t(x) ~\text{as:} $$ $$ t(x) := \lambda_0 $$ Defining $t$ this way we get $t(-x) = t(x)$. Which is not the same as what is stated in the book, i.e., $t(-x) = -t(x)$. Is the definition of $t$ different from what I understood?
The point is simply that for each $x \in S^n$, the element $q(s(x)) \in E(\gamma^1_n)$ has first coordinate $\{\pm x\}$; and therefore, like every element of $E(\gamma^1_n)$ with first coordinate $\{\pm x\}$, $q(s(x))$ has second coordinate $\lambda x$ for a uniquely determined value of $\lambda \in \mathbb R$. That value $\lambda$, uniquely associated to each $x \in S^n$, therefore defines a function $t(x)$, $$t : S^n \to \mathbb R $$ Of course you must still convince yourself that $t$ is continuous.
Regarding the identity $t(-x)=-t(x)$, it follows from this calculation: $$\bigl(\{\pm x\},t(x) x \bigr) = sq(x)=sq(-x) = \bigl(\{\pm(- x)\},t(-x)(-x)\bigr)=\bigl(\{\pm x\},-t(-x)x\bigr) $$