# (algebraic topology) question about the cellular approximation theorem

I have a question about the proof $$\pi _n(S^k) = 0 ~ for ~ n 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.

Since $$n, $$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$$ is a map to a point.
Since $$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$$.