I've been working through an old example sheet I found online at Cambridge geometry 2011, to fill in gaps in my topology.
The question #14 asks:
Suppose $f\colon X\to S^k$ is smooth where $X$ is compact and $0<\dim X< k$. Let $Z\subseteq S^k$ be a closed submanifold of dimension $k-\dim X$. Show that $I_2(f,Z)=0$.
$I_2(f,Z)$ is the mod 2 intersection number, which is the number of points in $f^{-1}(Z)$ modulo $2$.
I scoured a somewhat handwavy explanation by Knutson. Since $\dim X<k$, there exists a regular value $p\notin f(X)\cap Z$. The idea is to then stereographically project away from $p$, and "pretend" $f\colon X\to\mathbb{R}^k$. Then homotope to a constant by shrinking in $\mathbb{R}^k$ to get $I_2(f,Z)=0$.
My intuition is not strong enough to understand this explanantion. Can anyone more rigourously elucidate this explanation? That would be helpful, thank you.
If $p$ is a regular value, then $p\notin f(X)$ because of dimensions. We can moreover choose such a $p\notin Z$ (why?). You can stereographically project $\pi\colon S^k-\{p\}\to\Bbb R^k$. But the fact that $\Bbb R^k$ is contractible means that $I_2(\pi\circ f,\pi(Z))=0$. Note that $\pi$ is a diffeomorphism and mod 2 intersection numbers are invariant under diffeomorphisms.