8.3 ${\pi }_{k\le n}$ of an n-connected space and

Let $(A,a)$ be a pointed type and $n:\mathbb{N}$. Recall from thm:homotopy-groups that if $n>0$ the set ${\pi }_n(A,a)$ has a group structure, and if $n>1$ the group is abelian.

We can now say something about homotopy groups of $n$-truncated and $n$-connected types.

Lemma 8.3.1.

If $A$ is $n$-truncated and $a\mathrm{:}A$, then ${\pi }_k\mathit{}\mathrm{(}A\mathrm{,}a\mathrm{)}\mathrm{=}\mathrm{1}$ for all $k\mathrm{>}n$.

Proof.

The loop space of an $n$-type is an $(n-1)$-type, hence ${\mathrm{\Omega }}^k(A,a)$ is an $(n-k)$-type, and we have $(n-k)\le -1$ so ${\mathrm{\Omega }}^k(A,a)$ is a mere proposition. But ${\mathrm{\Omega }}^k(A,a)$ is inhabited, so it is actually contractible and ${\pi }_k(A,a)=\parallel {\mathrm{\Omega }}^k(A,a){\parallel }_0=\parallel \mathrm{𝟏}{\parallel }_0=\mathrm{𝟏}$. ∎

Lemma 8.3.2.

If $A$ is $n$-connected and $a\mathrm{:}A$, then ${\pi }_k\mathit{}\mathrm{(}A\mathrm{,}a\mathrm{)}\mathrm{=}\mathrm{1}$ for all $k\mathrm{\le }n$.

Proof.

We have the following sequence of equalities:

$${\pi }_k(A,a)=\parallel {\mathrm{\Omega }}^k(A,a){\parallel }_0={\mathrm{\Omega }}^k(\parallel (A,a){\parallel }_k)={\mathrm{\Omega }}^k(\parallel \parallel (A,a){\parallel }_n{\parallel }_k)={\mathrm{\Omega }}^k(\parallel \mathrm{𝟏}{\parallel }_k)={\mathrm{\Omega }}^k(\mathrm{𝟏})=\mathrm{𝟏}.$$

The third equality uses the fact that $k\le n$ in order to use that ${\parallel \text{–}\parallel }_k\circ {\parallel \text{–}\parallel }_n={\parallel \text{–}\parallel }_k$ and the fourth equality uses the fact that $A$ is $n$-connected. ∎

Corollary 8.3.3.

${\pi }_k({𝕊}^n)=\mathrm{𝟏}$ for .

Proof.

The sphere ${𝕊}^n$ is $(n-1)$-connected by cor:sn-connected, so we can apply lem:pik-nconnected. ∎

Title	8.3 ${\pi }_{k\le n}$ of an n-connected space and