(algebraic topology) question about the cellular approximation theorem

Question

I have a question about the proof $\pi _n(S^k) = 0 ~ for ~ n<k~$ by using cellular approximation theorem. According to the proof, (Wikipedia)

$\mathbf{proof} $. Give $S^n, S^k $ their canonical CW-structure, with one 0-cell each, and with one n-cell for $S^n$ and k-cell for $S^k$. Any base-point preserving map $\ f: S^n \to S^k $ is by the cellular approximation theorem homotopic to a constant map, whence $\pi_n(S^k)=0 $

Then, I cannot understand why f is homotopic to a constant map by the cellular approximation theorem.

Answer 1

Since $n<k$, $n$-th skeleton of $S^k$ is just a point, but the $n$-th skeleton of $S^n$ is already $S^n$ itself. Cellular approximation implies that $f$ is homotopic to $g: S^n \to S^k$ with the property $$g(S^n)=g(\text{sk}^n(S^n)) \subset \text{sk}^n(S^k)=\{\text{pt} \},$$ so $g$ is a map to a point.

Answer 2

Since $f$ is homotopic to some cellular map $g$ we only need to consider $g$ for now. We have that $g$ maps the $j$-skeleton of $S^n$, $0 \leq j \leq n$, to the $j$-skeleton of $S^k$. As $n < k$ (and we have these CW-structures), you can only map to the point that gives the $0$-skeleton of $S^k$ as we first add new cells at level $k$.