Suppose $f : X → \mathbb{S}^k$ is smooth, where $X$ is compact and $0 < \text{dim } X < k$. Then for all closed $Z \subset \mathbb{S}^k$ of dimension complementary to $X$, $I_2(X,Z) = 0$.
The hint is to use Sard's theorem to see that $\exists p \notin f(X)\cap Z$ and the stereographic projection.
My idea:
As $\dim X < k = \dim \mathbb{S}^k$ we have that $\exists p \notin f(X)\cap \mathbb{S}^k$ and as $Z \subset \mathbb{S}^k$ then $ p \notin f(X)\cap Z$. Now if we take the stereographic projection $P : \mathbb{S}^k / \{p\} \to \mathbb{R}^k$ we have that $P$ is a diffeomorphism and injective. So $P(Z)$ is a closed manifold in $\mathbb{R}^k$.
$I_2(X,Z) = I_2(i_X,Z) = I_2(f,Z) = I_2(P \circ f, P(Z))$
But as $P \circ f : X \to \mathbb{R}^k$ and $\mathbb{R}^k$ îs contractible we have that $I_2(P \circ f, P(Z)) = 0 = I_2(X,Z)$.
So my question is, is Sard's theorem necessary to prove this?