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Theorem Let $(X,x)$ be a pointed Kan complex and let $n$ be a positive integer. Then there is a unique group structure on the set $\pi _{n}(X,x)$ with the following properties:


Let $e: \Delta ^{n} \rightarrow \{ x\} \rightarrow X$ be the constant map. Then the homotopy class $[e]$ is the identity element of $\pi _{n}(X,x)$.


Let $f: \operatorname{\partial \Delta }^{n+1} \rightarrow X$ be a morphism of simplicial sets, corresponding to a tuple $(\sigma _0, \sigma _1, \ldots , \sigma _{n+1})$ of $n$-simplices of $X$ (see Proposition Assume that each restriction $\sigma _{i}|_{ \operatorname{\partial \Delta }^ n }$ is equal to the constant map $\operatorname{\partial \Delta }^ n \rightarrow \{ x\} \subseteq X$. Then $f$ extends to a map $\Delta ^{n+1} \rightarrow X$ if and only if the product

\[ [ \sigma _0 ]^{-1} [ \sigma _1 ] [ \sigma _2 ]^{-1} [ \sigma _3 ] \cdots [ \sigma _{n+1} ]^{ (-1)^{n} } \]

is equal to the identity element of $\pi _{n}(X,x)$.

Moreover, if $n \geq 2$, then the group $\pi _{n}(X,x)$ is abelian.

Proof of Theorem For every pair of elements $\alpha , \beta \in \pi _{n}(X,x)$, let $\alpha \beta $ denote the homotopy class $m_1(\alpha ,\beta )$, where $m_1: \pi _{n}(X,x) \times \pi _{n}(X,x) \rightarrow \pi _{n}(X,x)$ is the multiplication map of Lemma We first note that this multiplication is associative: for every triple of elements $\alpha , \beta , \gamma \in \pi _{n}(X,x)$, Lemmas and yield identities

\begin{eqnarray*} \alpha (\beta \gamma ) & = & m_1( \alpha , m_1(\beta , \gamma )) \\ & = & m_1( \alpha , m_2( \gamma , \beta ) ) \\ & = & m_2( \gamma , m_1( \alpha , \beta ) ) \\ & = & m_1( m_1(\alpha , \beta ), \gamma ) \\ & = & (\alpha \beta ) \gamma . \end{eqnarray*}

Example shows that $[e]$ is a two-sided identity with respect to multiplication. For every element $\alpha \in \pi _{n}(X,x)$, Lemma implies that we can choose an element $\beta \in \pi _{n}(X,x)$ for which the tuple $( \alpha , [e], \beta , [e], [e], \ldots , [e])$ bounds, so that $\alpha \beta = m_1( \alpha , \beta ) = [e]$. This shows that $\alpha $ has a right inverse, and a similar argument shows that $\alpha $ has a left inverse. It follows that multiplication determines a group structure on the set $\pi _{n}(X,x)$, having $[e]$ as the identity element.

We now verify that the multiplication on $\pi _{n}(X,x)$ satisfies condition $(b)$ of Theorem Suppose we are given an $(n+1)$-tuple $\vec{\eta } = ( \eta _0, \eta _1, \ldots , \eta _{n+1} )$ of elements of $\pi _{n}(X,x)$. We wish to show that $\vec{\eta }$ bounds if and only if the product $\eta _0^{-1} \eta _1 \eta _2^{-1} \cdots \eta _{n+1}^{(-1)^{n}}$ is equal to the identity element of $\pi _{n}(X,x)$. If $\vec{\eta } = ( [e], [e], \ldots , [e] )$, there is nothing to prove. Otherwise, there exists some smallest positive integer $i$ such that $\eta _{i-1} \neq [e]$. We proceed by descending induction on $i$. If $i > n$, we must show that $([e], [e], \ldots , [e], \eta _{n}, \eta _{n+1} )$ bounds if and only if $\eta _ n = \eta _{n+1}$, which follows from Example Let us therefore assume that $1 \leq i \leq n$. Define $\vec{\eta }' = ( \eta '_0, \eta '_1, \ldots , \eta '_{n+1} )$ by the formula

\[ \eta '_{j} = \begin{cases} m_ i( \eta _{i-1}^{-1}, \eta _{j} ) & \text{ if $j = i-1$ or $j=i$ } \\ \eta _ j & \text{ otherwise. } \end{cases} \]

Invoking Lemma repeatedly, we obtain

\[ \eta '_{i-1} = m_ i( \eta _{i-1}^{-1}, \eta _{i-1} ) = \begin{cases} \eta _{i-1}^{-1} \eta _{i-1} & \text{ if $i$ is odd } \\ \eta _{i-1} \eta _{i-1}^{-1} & \text{ if $i$ is even } \end{cases} = [e] \]

\[ \eta '_{i} = m_ i( \eta _{i-1}^{-1}, \eta _{i} ) = \begin{cases} \eta _{i-1}^{-1} \eta _{i} & \text{ if $i$ is odd } \\ \eta _{i} \eta _{i-1}^{-1} & \text{ if $i$ is even }\end{cases}. \]

We therefore have an equality

\[ \eta _0^{-1} \eta _1 \eta _{2}^{-1} \cdots \eta _{n+1}^{(-1)^{n} } = \eta '^{-1}_{0} \eta '_{1} \eta '^{-1}_{2} \cdots \eta '^{(-1)^{n}}_{n+1}. \]

Invoking our inductive hypothesis, we conclude that this product vanishes if and only if the tuple $\vec{\eta }'$ bounds. By virtue of Lemma, this is equivalent to the assertion that $\vec{\eta }$ bounds.

We now complete the proof of Theorem by showing that the multiplication on $\pi _{n}(X,x)$ is commutative. Fix a pair of elements $\sigma , \sigma ' \in \Sigma $. Then the tuples of $n$-simplices $( \sigma , e, \sigma ', \bullet , e, e, \ldots , e)$ and $(\sigma ', e, \sigma , \bullet , e, e, \ldots , e)$ determine maps of simplicial sets $f, f': \Lambda ^{n+1}_{3} \rightarrow X$ (Proposition Since $X$ is a Kan complex, we can extend $f$ and $f'$ to $(n+1)$-simplices of $X$, which we will denote by $\tau $ and $\tau '$, respectively. It follows from the preceding arguments that the faces $d^{n+1}_3(\tau )$ and $d^{n+1}_3(\tau ')$ are representatives of the products $[\sigma '] [\sigma ]$ and $[\sigma ] [\sigma ']$ in $\pi _ n(X,x)$, respectively. Let $\overline{e}: \Delta ^{n+1} \rightarrow X$ denote the constant map taking the value $x$. Then the tuple of $(n+1)$-simplices $(\tau , s^{n}_0(\sigma ), s^{n}_1(\sigma ), \tau ', \bullet , \overline{e}, \overline{e}, \ldots , \overline{e} )$ determines a map of simplicial sets $g: \Lambda ^{n+2}_{4} \rightarrow X$ (Proposition Since $X$ is a Kan complex, we can extend $g$ to an $(n+2)$-simplex of $X$. Then the fourth face of this extension witnesses that the tuple of $n$-simplices $( d^{n+1}_3( \tau ), e, e, d^{n+1}_3(\tau '), e, \ldots , e)$ bounds, so that we have an equality $[\sigma '] [\sigma ] = [ d^{n+1}_3(\tau ) ] = [ d^{n+1}_3(\tau ') ] = [ \sigma ] [\sigma ']$ in the homotopy group $\pi _{n}(X,x)$. $\square$