Proof. Let $Q$ be the following filler.

Consider the following cube:

Let its bottom be filled by $Q^{-1}$ and its top be filled by $\textsf{flip}(Q)$ . All sides have their obvious fillers defined by path induction on $p$ , $q$ , $r$ and $s$ . We claim that the cube has a filler. We prove this by first applying path induction on $p$ , $r$ and $s$ , transforming $Q$ to a path $\textsf{refl}_x = q$ . Finally, after applying path induction on $Q$ we are left to fill a cube with $\textsf{refl}_{\textsf{refl}_x}$ on each side, which is trivial.

Specialising the above argument to the case when $p,q,r$ and $s$ are all $\textsf{refl}_x$ and $Q$ is arbitrary, we have our lemma.

Proof. An easy way of proving this is to note that the canonical map

\begin{equation*} i_{\vee } : {\widetilde {K}(G_1,n) \vee \widetilde {K}(G_2,m)} \to {\widetilde {K}(G_1,n) \times \widetilde {K}(G_2,m)} \end{equation*}

is $((n-1) + (m-1))$ -connected, thereby inducing a section via the map

\begin{equation*} f \vee g : (x : {\widetilde {K}(G_1,n) \vee \widetilde {K}(G_1,m)}) \to P (i_\vee (x)) \end{equation*}

From the point of view of computer formalisation, however, this proof leads to poor computational behaviour of subsequent definitions. For this reason, we provide a direct proof.

We proceed by induction on $n$ and $m$ . We first consider the base-case: $n=m=1$ . We are to construct a section ${{f}\tilde {\vee }{g}} : ((x,y): \widetilde {K}(G_1,1) \times \widetilde {K}(G_2,1)) \to P(x,y)$ . Since $P$ is a set, it suffices to define a section ${{f}\tilde {\vee }{g}} : ((x,y): L(G_1) \times L(G_2)) \to P(x,y)$ . We proceed by $L(G_1)$ -induction on $x$ . For the point constructor, we define

\begin{align*} ({{f}\tilde {\vee }{g}})(\star ,y) &:= g(y) \end{align*}

We now need to define $\textsf{ap}_{({{f}\tilde {\vee }{g}})(\!-,y)}({\textsf{loop}_k\,{a}})$ for $a : G_1$ . Since $P$ is set valued, it suffices to do so when $y$ is $\star$ . We define

\begin{align*} \textsf{ap}_{({{f}\tilde {\vee }{g}})(\!-,\star )}({\textsf{loop}_k\,{a}}) &:= p^{-1}\cdot \textsf{ap}_{f}({\textsf{loop}_k\,{a}}) \cdot p \end{align*}

We note that this construction satisfies the additional properties by default: we construct $l : (x : \widetilde {K}(G_1,1)) \to ({{f}\tilde {\vee }{g}})(x,{\textsf{0}_k}) = f(x)$ by induction on $x$ . Since this is a proposition, it suffices to construct $l(0_k) : g({\textsf{0}_k}) = f({\textsf{0}_k})$ , which we do by $l(0_k) := p^{-1}$ . We construct $r : (y : \widetilde {K}(G_2,1)) \to ({{f}\tilde {\vee }{g}})({\textsf{0}_k},x)$ simply by $r(y) := \textsf{refl}$ , since the equality holds by definition. We finally need to show that ${l(0_k)}^{-1} \cdot r(0_k)= p$ which is immediate by construction of $l$ and $r$ .

We proceed by induction on $n$ , letting $n \geq 2$ and omitting the case when $n = 1$ and $m\geq 2$ which is entirely symmetric. Let us define ${{f}\tilde {\vee }{g}} : ((x , y) : {\widetilde {K}(G_1,n) \times \widetilde {K}(G_2,m)}) \to P(x,y)$ . We stress that since $n \geq 2$ , we have that $\widetilde {K}(G_1,n) = \Sigma \widetilde {K}(G_1,n-1)$ by definition. Hence, we may define ${f}\tilde {\vee }{g}$ by suspension induction on $x$ . Let us start with the action on point constructors.

\begin{align*} ({{f}\tilde {\vee }{g}})(\textsf{north},y) &:= g(y)\\ ({{f}\tilde {\vee }{g}})(\textsf{south},y) &:= {\textsf{transport}^{P(\!-,y)}(\textsf{merid}\,{\textsf{0}_k},g(y))} \end{align*}

We now need to define $\textsf{ap}_{{{f}\tilde {\vee }{g}}}(\!-,y)(\textsf{merid}\,a) : {g(y) =_{\textsf{merid}\,a}^{P(\!-,y)} {\textsf{transport}^{P(\!-,y)}(\textsf{merid}\,{\textsf{0}_k},g(y))}}$ for $a :\widetilde {K}(G_1,n-1)$ and $y : \widetilde {K}(G_2,m)$ . The type of $\textsf{ap}_{{{f}\tilde {\vee }{g}}}(\!-,y)(\textsf{merid}\,a)$ is an $(n+m-3)$ -type, and thus, by induction hypothesis, it suffices to construct the following:

\begin{align*} f' &: (a : \widetilde {K}(G_1,n-1))\to {g({\textsf{0}_k}) =_{\textsf{merid}\,a}^{P(\!-,{\textsf{0}_k})} {\textsf{transport}^{P(\!-,{\textsf{0}_k})}(\textsf{merid}\,{\textsf{0}_k},g({\textsf{0}_k}))}} \\ g' &: (y : \widetilde {K}(G_2,m))\to {g(y) =_{\textsf{merid}\,{\textsf{0}_k}}^{P(\!-,y)} {\textsf{transport}^{P(\!-,y)}(\textsf{merid}\,{\textsf{0}_k},g(y))}} \\ p' &: f'({\textsf{0}_k}) = g'({\textsf{0}_k}) \end{align*}

By composition with $p$ , the target type of $f'$ is equivalent to $f(\textsf{north}) =_{\textsf{merid}\,a}^{P(\!-,{\textsf{0}_k})} f(\textsf{south})$ , of which we have a term: $\textsf{ap}_{f}(\textsf{merid}\,a)$ . The target type of $g'$ is equivalent to $g(y) = g(y)$ and we simply give the term $\textsf{refl}$ . The construction of $p'$ is an easy but technical lemma. Thus we have constructed ${f}\tilde {\vee }{g}$ . Now note that $r : (y : \widetilde {K}(G_2,m)) \to ({{f}\tilde {\vee }{g}})({\textsf{0}_k},y) = g(y)$ holds by $\textsf{refl}$ . The path $l : (x : \widetilde {K}(G_1,n-1)) \to ({{f}\tilde {\vee }{g}})(x,{\textsf{0}_k}) = g(y)$ is easily constructed using the corresponding $l$ and $r$ paths from the inductive hypothesis. This can be done so that $l({\textsf{0}_k}) = r({\textsf{0}_k})$ holds almost by definition.

Proof. The statement follows from the group structure on $G$ when $n = 0$ . When $n = 1$ , the target type is a set, since $K(G,1)$ is $1$ -truncated (and hence path types over it are sets). By the set-elimination rule for $K(G,n)$ (see Figure 2B ), it suffices to show that $\lvert \,x\,\rvert +_k 0_k = 0_k = 0_k +_k \lvert \,x\,\rvert$ for $x : L(G)$ . These equalities hold definitionally by $L(G)$ induction. Hence, we also get the filler of the square by $\textsf{refl}$ . When $n \geq 2$ , the statement follows immediately by the defining equations (1) and (2), using the family of paths $l$ and $r$ and the coherence between them, as given in Proposition 12.

Proof. When $n = 0$ , the statement is simply that $G$ is abelian, which we have assumed. Let $n \geq 1$ . By truncation elimination, it suffices to construct paths $\lvert \,x\,\rvert +_k \lvert \,y\,\rvert = \lvert \,y\,\rvert +_k \lvert \,x\,\rvert$ for $x,y:\widetilde {K}(G,n)$ . This path type is $(n-1)$ -truncated which is sufficiently low for Proposition 12 to apply. Hence, it suffices to construct paths $p_x : \lvert \,x\,\rvert + 0_k = 0_k + \lvert \,x\,\rvert$ and $q_y : 0_k +_k \lvert \,y\,\rvert = \lvert \,y\,\rvert + 0_k$ and show that $p_{0_k} = q_{0_k}$ . We construct both paths as compositions of the left- and right-unit laws of $+_k$ in the obvious way. These definitions coincide over $0_k$ by the second part of Proposition 13.

Proof. We have

\begin{equation*} \textsf{ap}^{2}_{+_k}(p,q) = \textsf{ap}_{x\mapsto x +_k 0_k}(p) \cdot \textsf{ap}_{y\mapsto 0_k +_k y}(q) = p \cdot q \end{equation*}

The final step consists in applying the right- and left-unit laws to both components. This is justified since they agree at $0_k$ .

Proof. By commutativity of $+_k$ , it suffices to show that $x -_k x = 0_k$ . When $n = 0$ , this comes from the group structure on $G$ . When $n = 1$ , we note that the goal type is a set, which means that it suffices to show the statement for $x : L(G)$ . We have $\star -_k \star = 0_k$ by $\textsf{refl}$ . For $\textsf{loop}_k\,{}$ , we need to show that

\begin{equation*} \textsf{ap}^{2}_{x,y\mapsto x -_k y}({\textsf{loop}_k\,{g}},{\textsf{loop}_k\,{g}}) = \textsf{refl} \end{equation*}

By Proposition 16, the LHS above is equal to ${\textsf{loop}_k\,{g}} \cdot ({\textsf{loop}_k\,{g}})^{-1}$ which is equal to $\textsf{refl}$ . The case when $n \geq 2$ is entirely analogous, again using Proposition 16 for the path constructors.

Proof. When $n = 0$ , we are automatically done, since $G$ is a set. For $n \geq 1$ , it suffices, by $(n-2)$ -type elimination, to show the statement when $x$ is $0_k$ . Let $p,q: \Omega K(G,n)$ . We have

\begin{align*} p \cdot q = \textsf{ap}^{2}_{+_k}(p,q) = \textsf{ap}^{2}_{+_k}(q,p) = q \cdot p \end{align*}

where the second equality comes from the commutativity of $+_k$ .

Proof. We give the proof for $n = 1$ as the case when $n \gt 1$ is entirely analogous. Let us define a fibration $\textsf{Code}_x: (p : \star = x) \to \textsf{hProp}$ for $x:K(G,1)$ . As $\textsf{hProp}$ is a set, it suffices to define $\textsf{Code}_x(p)$ for $x : L(G)$ . We do this by induction on $x$ and define

\begin{align*} \textsf{Code}_\star (p) := (\textsf{ap}_{-_k}(p) = p^{-1}) \end{align*}

which is a proposition since $K(G,1)$ is $0$ -connected. We need to check that this definition respects the action of $\textsf{Code}$ on $\textsf{loop}_k\,{g}$ for $g:G$ . This amounts to showing that, for any $p : \Omega K(G,1)$ , we have an equivalence

\begin{align*} (\textsf{ap}_{-_k}(p) = p^{-1}) \simeq (\textsf{ap}_{-_k}(p \cdot {\textsf{loop}_k\,{g}}) &= (p \cdot {\textsf{loop}_k\,{g}})^{-1}) \end{align*}

We rewrite the terms on the RHS as follows:

\begin{align*} \textsf{ap}_{-_k}(p \cdot {\textsf{loop}_k\,{g}}) &= \textsf{ap}_{-_k}(p) \cdot \textsf{ap}_{-_k}({\textsf{loop}_k\,{g}}) \\ &= \textsf{ap}_{-_k}(p) \cdot {({\textsf{loop}_k\,{g}})}^{-1} \end{align*}

and

\begin{align*} (p \cdot {\textsf{loop}_k\,{g}})^{-1} &= ({\textsf{loop}_k\,{g}})^{-1} \cdot p^{-1} \\ &= p^{-1} \cdot ({\textsf{loop}_k\,{g}})^{-1} \end{align*}

Thus, we only need to construct an equivalence.

\begin{align*} (\textsf{ap}_{-_k}(p) = p^{-1}) \simeq (\textsf{ap}_{-_k}(p) \cdot ({\textsf{loop}_k\,{g}})^{-1} &= p^{-1} \cdot ({\textsf{loop}_k\,{g}})^{-1}) \end{align*}

This equivalence is given by $\textsf{ap}_{q \mapsto q \cdot ({\textsf{loop}_k\,{g}})^{-1}}$ and thereby the construction of $\textsf{Code}$ is complete.

We now note that we have a section $(p : \star = x) \to \textsf{Code}_x(p)$ for any $x:K(G,1)$ since, by path induction, it suffices to note that $\textsf{refl}$ is an element of $\textsf{Code}_\star (\textsf{refl})$ . Hence, in particular, we have an element of $\textsf{Code}_\star (p)$ for all $p : \Omega K(G,1)$ , which is what we wanted to show.

Proof. For $n = 0$ , we refer to the proof in (Licata and Finster, Reference Licata and Finster2014). For $n \geq 1$ , we choose to provide an encode-decode proof similar to their $n = 1$ case. We define $\textsf{Code} : K(G,n+1) \to {\textsf {{n}-Type}}$ by

\begin{align*} \textsf{Code}(\lvert \,\textsf{north}\,\rvert ) &:= K(G,n) \\ \textsf{Code}(\lvert \,\textsf{south}\,\rvert ) &:= K(G,n) \\ \textsf{ap}_{\textsf{Code}\,\circ \, \lvert \,-\,\rvert }(\textsf{merid}\,\,a) &:= \textsf{ua}(x \mapsto \lvert \,a\,\rvert +_k x) \end{align*}

Here $\textsf{ua}$ is the part of univalence which turns an equivalence into a path. Clearly, the map to which it is applied is an equivalence. For any $t : K(G,n+1)$ , we have a map $\textsf{encode}_t : 0_k = t \to \textsf{Code}(t)$ defined by $\textsf{encode}_t(p):={\textsf{transport}^{\textsf{Code}}(p,0_k)}$ (this is equivalent to path induction on $p$ ). We define $\textsf{decode}_t : \textsf{Code}(t) \to 0_k = t$ by truncation elimination and $K(G,n)$ -induction on $t$ . For the point constructors, we define it by

\begin{align*} \textsf{decode}_{\lvert \,\textsf{north}\,\rvert }(x) &= \sigma _n(x)\\ \textsf{decode}_{\lvert \,\textsf{south}\,\rvert }(x) &= \sigma _n(x)\cdot \textsf{ap}_{\lvert \,-\,\rvert }(\textsf{merid}\,0_k) \end{align*}

For the path constructor, we need to, after some simple rewriting of the goal, provide a path

\begin{equation*} \sigma _n(x-_k\lvert \,a\,\rvert ) = \sigma _n(x)\cdot \sigma _n(\lvert \,a\,\rvert )^{-1} \end{equation*}

for $x : K(G,n)$ and $a : \widetilde {K}(G,n)$ which is easily done using Proposition 20 and Corollary 21. We can now show that, for any $t : K(G,n+1)$ and $p : 0_k = t$ , we have $\textsf{encode}_t(\textsf{decode}_t(p)) = p$ by path induction on $p$ . In particular, we have $\textsf{decode}_{0_k}(\textsf{encode}_{0_k}(p)) = p$ for all $p : \Omega K(G,n+1)$ . We can also show that $\textsf{encode}_{0_k}(\textsf{decode}_{0_k}(x)) = x$ for all $x:K(G,n)$ by simply unfolding the definitions of $\textsf{decode}_{0_k}$ and $\textsf{encode}_{0_k}$ . Indeed, after applying truncation elimination on $x$ , we have:

\begin{align*} \textsf{encode}_{0_k}(\textsf{decode}_{0_k}(\lvert \,a\,\rvert )) &:= \textsf{encode}_{0_k}(\sigma _n(\lvert \,a\,\rvert )) \\ &:= {\textsf{transport}^{\textsf{Code}}(\sigma _n(\lvert \,a\,\rvert ),0_k)}\\ &= {\textsf{transport}^{\textsf{Code}\circ \lvert \,-\,\rvert }(\textsf{merid}\,(0_k)^{-1},{\textsf{transport}^{\textsf{Code}\circ \lvert \,-\,\rvert }(\textsf{merid}\,a,0_k)})} \\ &= {\textsf{transport}^{\textsf{Code}\circ \lvert \,-\,\rvert }(\textsf{merid}\,(0_k)^{-1},\lvert \,a\,\rvert +0_k)}\\ &=(\!-_k 0_k) +_k (\lvert \,a\,\rvert +0_k)\\ &= \lvert \,a\,\rvert \end{align*}

and we are done.

Proof. It is enough to show that $\Omega ^{m+1}(K(G_1,n) \to _\star K(G_2,n+m))$ is contractible. We have

\begin{align*} \Omega ^{m+1} (K(G_1,n) \to _\star K(G_2,n+m)) &\simeq (K(G_1,n) \to _\star \Omega ^{m+1} K(G_2,n+m)) \\ &\simeq (K(G_1,n) \to _\star K(G_2,n-1)) \end{align*}

The fact that $(K(G_1,n) \to _\star K(G_2,n-1))$ is contractible follows easily from the $(n-2)$ -type elimination rule for $K(G_1,n)$ in Figure 2C .

Proof. By assumption, we have a homotopy $h : (x : A) \to f(x) = g(x)$ . We construct $r : \star _B = \star _B$ by

\begin{align*} r := p^{-1} \cdot h(\star _A) \cdot q \end{align*}

and define $P : (B,\star _B) =_{\mathcal{U}_*} (B,\star _B)$ by

\begin{align*} P := \textsf{ap}_{(B,-)}(r) \end{align*}

We get $P = \textsf{refl}$ as an easy consequence of the homogeneity of $B$ . Hence, instead of proving that $(f,p) = (g,q)$ , it is enough to prove that ${\textsf{transport}^{{A \to _\star (\!-\!)}}(P,(f,p))} = (g, q)$ . The transport only acts on $p$ and $q$ , so the identity of the first components holds by $h$ . For the second components, we are reduced to proving that $p \cdot r = h(\!\star _A\!) \cdot q$ . This is true immediately by the construction of $r$ .

Proof. When $n = 0$ , the result follows by a straightforward induction on $m$ . When $n \geq 1$ , we generalise and, for $x,y : K(G_1,n)$ , consider the two functions

\begin{align*} z &\mapsto (x +_k y) \smile _k z \\ z &\mapsto (x \smile _k z) +_k (y \smile _k z) \end{align*}

We claim that these are equal as pointed functions living in $K(G_2,m) \to _\star K(G_1\otimes G_2 , n + m)$ . By Proposition 24, all path types over $K(G_2,m) \to _\star K(G_1\otimes G_2 , n + m)$ are $(n-1)$ -types. This allows us to apply truncation elimination and Proposition 12. In combination with Lemma 26, it suffices to construct, for each $z : K(G_2,m)$ :

\begin{align*} l(y) &: (0_k +_k \lvert \,y\,\rvert ) \smile _k z = (0_k \smile _k z) +_k (\lvert \,y\,\rvert \smile _k z) \\ r(x) &: (\lvert \,x\,\rvert +_k 0_k) \smile _k z = (\lvert \,x\,\rvert \smile _k z) +_k (0_k \smile _k z) \\ q &: l(0_k) = r(0_k) \end{align*}

which is direct using the unit laws for $+_k$ and $\smile _k$ . The obvious construction makes $q$ hold by $\textsf{refl}$ .

Proof. When $n = 0$ , the statement holds by definition. Let us consider the case $n \geq 1$ . The maps we are comparing are of type $K(G_1,n) \to K(G_2,m) \to _\star \Omega (K(G_1 \otimes G_2 , (n + 1) + m))$ . The type

\begin{equation*}K(G_2,m) \to _\star \underset {\simeq K(G_1\otimes G_2,n+m)}{\underbrace {\Omega (K(G_1 \otimes G_2 , (n + 1) + m))}}\end{equation*}

is an $n$ -type by Proposition 24, and hence we may use $K(G,n)$ -elimination on $x$ . Hence, we want to construct an equality of pointed functions

\begin{equation*}(y\mapsto \textsf{ap}_{x\mapsto x \smile _k y}(\sigma _n(a))) = (y \mapsto \sigma _{n+m}(a \smile _k' y))\end{equation*}

for $a : \widetilde {K}(G,n)$ . By Lemma 26, it suffices to show this equality for underlying functions. This follows trivially:

\begin{align*} \textsf{ap}_{x\mapsto x \smile _k y}(\sigma _n(a)) &:= \textsf{ap}_{x\mapsto \lvert \,x\,\rvert \smile _k y}(\textsf{merid}\,a \cdot (\textsf{merid}\,0_k)^{-1}) \\ &= \textsf{ap}_{x\mapsto \lvert \,x\,\rvert \smile _k y}(\textsf{merid}\,a) \cdot (\textsf{ap}_{x\mapsto \lvert \,x\,\rvert \smile _k y}(\textsf{merid}\,0_k))^{-1} \\ &:= \sigma _{n+m}(a \smile _k' y) \cdot \sigma _{n+m}(0_k \smile _k' y)^{-1}\\ &= \sigma _{n+m}(a \smile _k' y) \end{align*}

Proof. We induct on $n$ . Let us first consider the inductive step and save the base case for last. To this end, let $n \geq 1$ . We may consider the two functions we are comparing as doubly pointed functions of type

\begin{equation*}K(G_1,n) \to (K(G_2,m) \to _\star (K(G_3,k) \to _\star K(G_1 \otimes (G_2 \otimes G_3) , n + m + k)))\end{equation*}

Applying an appropriate variant of Proposition 24, the codomain of the above function type is an $n$ -type. When $x$ is a point constructor of $K(G_1,n)$ , the diagram in the statement commutes by $\textsf{refl}$ . For the higher constructor ( $\textsf{merid}\,$ or $\textsf{loop}_k\,{}$ , depending on the value of $n$ ), we are done if we can show, for $x : K(G_1,n-1)$ , that

\begin{align*} \textsf{ap}_{x \mapsto x \smile _k (y \smile _k z)}(\sigma _{n-1}(x)) = \textsf{ap}_{x \mapsto \alpha _{n+m+k} ((x \smile _k y) \smile _k z)}(\sigma _{n-1}(x)) \end{align*}

where $\alpha : (G_1 \otimes G_2) \otimes G_3 \simeq G_1 \otimes (G_2 \otimes G_3)$ and, recall, $\alpha _{p} : K((G_1 \otimes G_2) \otimes G_3,p) \to K(G_1 \otimes (G_2 \otimes G_3),p)$ is the functorial action of $K(\!-,p)$ on $\alpha$ . We have

\begin{align*} \textsf{ap}_{x \mapsto x \smile _k (y \smile _k z)}(\sigma _{n-1}(x)) &= \sigma _{(n-1)+(m+k)}(x \smile _k (y \smile _k z)) & \text{(}\text{Lemma}\,29)\text{} \\ &= \sigma _{(n-1)+(m+k)}(\alpha _{(n-1)+(m+k)} ((x \smile _k y) \smile _k z)) & \text{ (Ind. hyp.)} \\ &= \textsf{ap}_{\alpha _{n+m+k}}(\sigma _{((n-1) + m)+k}((x \smile _k y) \smile _k z)) &\text{(}\text{Lemma}\,11\text{)} \\ &= \textsf{ap}_{\alpha _{n+m+k}}(\textsf{ap}_{x \mapsto x \smile _k z}(\sigma _{((n-1) + m)}(x \smile _k y))) &\text{(}\text{Lemma}\,29\text{)} \\ &= \textsf{ap}_{\alpha _{n+m+k}}(\textsf{ap}_{x \mapsto x \smile _k z}({\textsf{ap}_{x \mapsto x \smile y}(\sigma _{(n-1)}(x))})) &\text{(}\text{Lemma}\,29\text{)} \\ &= \textsf{ap}_{x \mapsto \alpha _{n+m+k} ((x \smile _k y) \smile _k z)}(\sigma _{n-1}(x)) \end{align*}

which concludes the inductive step.

Let us return to the base case, namely, the case $n = 0$ . In this case, we fix $g_1 : G_1$ and instead consider the functions $y,z \mapsto g_1 \smile _k(y\smile _k z)$ and $y,z \mapsto (g_1 \smile _k y)\smile _k z$ to be of type

\begin{equation*}K(G_2,m) \to K(G_3,k) \to _\star K(G_1\otimes (G_2 \otimes G_3) , m + k).\end{equation*}

We induct on $m$ . The inductive step follows in the same way as the inductive step above. For the base case, we fix $y$ to be $g_2 : G_2$ and are left to prove $g_1 \smile _k (g_2 \smile _k z) = (g_1 \smile _k g_2) \smile _k z$ for $z : K(G_3,k)$ . Again, this is followed by induction on $k$ . The base case is given by associativity of $G_1\otimes (G_2 \otimes G_3)$ and the inductive step follows in exactly the same way as before.

Proof. We may, without loss of generality, assume that $n \geq m$ . We induct on the quantity $n+m$ . The base-case when $n = m = 0$ is trivial. Let us consider the two cases $n = m = 1$ and $n,m \geq 2$ and omit the two cases when $n \geq 2$ and $m \lt 2$ as they follow in an entirely analogous manner. In what follows, let $c_p : K(G_2\otimes G_1,p) \to K(G_1 \otimes G_2,p)$ denote the commutator.

We start with the case $n = m = 1$ . In this case, we would like to be able to apply the set-elimination rule for $K(\!-,1)$ . As before, fix $x : K(G_1,1)$ . We strengthen the goal by asking for $x \smile _k (\!-\!)$ and $y \mapsto (\!-1)^{nm}(y \smile _k x)$ to be equal as pointed functions living in $K(G_2,1) \to _\star K(G_1\otimes G_2 , 2)$ . The type of such functions is a $1$ -type, and hence, any path type over it is a set. Thus, we may apply set-elimination and assume that $x : L(G_1)$ . We may now repeat the argument for $y : K(G_2,n)$ : we would like $(\!-\!) \smile _k y$ and $x \mapsto (\!-1)^{nm}(y \smile _k x)$ to be equal as pointed functions living in $L(G_1) \to _\star K(G_1\otimes G_2, 2)$ . By a very similar argument to that of Proposition 24, this pointed function type is again a $1$ -type, which justifies set-elimination being applied to $y$ . Combining this with Lemma 26, it is enough to show that

\begin{align*} x \smile _k y = - c_{2}(y \smile _k x) \end{align*}

for $x : L(G_1)$ and $y : L(G_2)$ . We do this by $L(G_1)$ induction. When either $x$ or $y$ is a point constructor, the identity holds by $\textsf{refl}$ . The remaining case amounts to (by evaluation of $c_2$ ) the following identity

\begin{align*} \textsf{ap}_{x \mapsto \textsf{ap}_{x \smile _k (\!-\!)}({\textsf{loop}_k\,{h}})}({\textsf{loop}_k\,{g}}) = \textsf{flip}(\textsf{ap}_{x \mapsto \textsf{ap}_{x \smile _k (\!-\!)}({\textsf{loop}_k\,{h}})}({\textsf{loop}_k\,{g}})^{-1}) \end{align*}

which holds by Lemma 6.

We give a rough sketch of the case $n,m \geq 2$ . We may apply $K(G_1,n)$ -elimination again, by the same argument as in the previous case. The crucial step this time is verifying

(4) \begin{align} \textsf{ap}_{x \mapsto \textsf{ap}_{x \smile _k (\!-\!)}(\textsf{merid}\,\,h)}(\textsf{merid}\,\,g) = \textsf{ap}_{x \mapsto \textsf{ap}_{(\!-1)^{nm} (x \,\smile _k (\!-\!))}(\textsf{merid}\,\,h)}(\textsf{merid}\,\,g) \end{align}

We note that the above identity and the identities that follow only are well-typed up to some coherences which we brush under the rug for the sake of readability. We notice that (4) follows if we can show that for $x : K(G_1,n-1)$ , $y : K(G_2,m-1)$ , we have

(5) \begin{align} \textsf{ap}_{x \mapsto \textsf{ap}_{x \smile _k (\!-\!)}(\sigma _{m-1}(y))}(\sigma _{n-1}(x)) = \textsf{ap}_{x \mapsto \textsf{ap}_{(\!-1)^{nm} (x \,\smile _k (\!-\!))}(\sigma _{m-1}(y))}(\sigma _{n-1}(x)) \end{align}

Let us, for further readability, omit the equivalence $c_{p}$ in the following equations. Some of the following equalities hold up to multiplication by $(\!-1)^p$ for some $p : \mathbb{N}$ . This will be indicated in the right-hand column.

\begin{align*} \textsf{ap}_{a \mapsto \textsf{ap}_{a \smile _k (\!-\!)}(\sigma _{m-1}(y))}(\sigma _{n-1}(x)) &= \textsf{ap}_{a \mapsto \textsf{ap}_{(\!-\!) \smile _k a}(\sigma _{n-1}(x))}(\sigma _{m-1}(y)) & (\!-1)\\ &= \textsf{ap}_{\sigma _{(n-1)+m}}(\textsf{ap}_{x \smile _k (\!-\!)}(\sigma _{m-1}(y))) \\ &= \textsf{ap}_{\sigma _{m+(n-1)}}(\textsf{ap}_{(\!-\!) \smile _k x}(\sigma _{m-1}(y))) & m(n-1) \\ &= \textsf{ap}_{\sigma _{(m-1)+(n-1)+1}}(\sigma _{(m-1)+(n-1)}(y \smile _k x)) \\ &= \textsf{ap}_{\sigma _{{(n-1)+(m-1)+1}}}(\sigma _{{(n-1)+(m-1)}}(x \smile _k y)) & (n-1)(m-1) \\ &= \textsf{ap}_{\sigma _{n+(m-1)}}(\textsf{ap}_{(\!-\!) \smile _k y}(\sigma _{n-1}(x))) \\ &= \textsf{ap}_{\sigma _{m+(n-1)}}(\textsf{ap}_{y \smile _k (\!-\!)}(\sigma _{n-1}(x))) & (m-1)n \\ &= \textsf{ap}_{a \mapsto \textsf{ap}_{(\!-\!) \,\smile a}(\sigma _{m-1}(y))}(\sigma _{n-1}(x)) \end{align*}

Since $(\!-1)+n(m-1) + (n-1)(m-1) + (n-1)m =_{\mathbb{Z}/2\mathbb{Z}} nm$ , the above identity holds up to multiplication by $(\!-1)^{nm}$ , which gives (5).

Proof. Let $c_p$ denote the commutator $K(G\otimes H,p) \to K(H \otimes G,p)$ . Consider the following diagram.

The statement is that the outer square commutes. By Proposition 31 the left-square commutes. It is easy to verify, using commutativity of $R$ , that $\textsf{mult}_{n+m} \circ c = \textsf{mult}_{m+n}$ . Hence, the right-square commutes, and we are done.

Proof. By a straightforward induction on $n$ , using the recursive definition of the cup product.

Proof. Let $F$ denote the function in question. The fact that it is a homomorphism is trivial. We may construct its inverse $F^{-1} : H^n(X,G) \to \widetilde {H}^n(X,G)$ explicitly by

\begin{align*} F^{-1}\,\lvert \,f\,\rvert := \lvert \, x \mapsto f(x) -_k f(\star _X)\,\rvert \end{align*}

In order to see that $F(F^{-1} \lvert \,f\,\rvert ) = \lvert \,f\,\rvert$ , we note that this is a proposition. By connectedness of $K(G,1)$ , we may assume we have a path $f(\star _X) = 0_k$ . Hence

(6) \begin{align} F(F^{-1}\,\lvert \,f\,\rvert ):= \lvert \, x \mapsto f(x) -_k f(\star _X)\,\rvert = \lvert \,f\,\rvert \end{align}

For the other direction, we need to check that

\begin{align*} F^{-1}(F\,\lvert \,(f,p)\,\rvert ) = \lvert \,(f,p)\,\rvert \end{align*}

By Lemma 26, we only need to verify that the first projections agree, and we are reduced to the second equality in (6), which we have already verified.

Proof. We define $\phi : (X \to _\star G) \times G \to (X \to G)$ by

\begin{equation*} \phi (f,g) := x \mapsto f(x) +_G g \end{equation*}

and $\psi : (X \to G) \to (X \to _\star G) \times G$ by

\begin{equation*} \psi (f) := ((x \mapsto f(x) -_G f(\star _X)) , f(\star _X)) \end{equation*}

The fact that these maps cancel out follows immediately from the group laws on $G$ . This induces, after applying the appropriate set truncations, an equivalence $H^0(X,G) \simeq \widetilde {H}^0(X,G) \oplus G$ . The fact that it is an isomorphism is also trivial using the group laws on $G$ .

Proof. We verify that $\widetilde {H}^n(\!-,G)$ satisfies the four axioms.

• • Suspension: this axiom is trivially satisfied when $n \lt -1$ . When $n = -1$ , we need to verify that $\widetilde {H}^0(\Sigma \,X,G)$ is trivial, which follows immediately by connectedness of $\Sigma \,X$ . For $n \geq 0$ , the suspension axiom follows immediately from the fact that $\Sigma$ and $\Omega$ are adjoint. We have

  \begin{align*} \widetilde {H}^{n+1}(\Sigma \,X,G) &= \lVert \Sigma X \to _\star K(G,n+1)\rVert _0 \\ &\simeq \lVert X \to _\star \Omega \,K(G,n+1)\rVert _0 \\ &\simeq \lVert X \to _\star K(G,n)\rVert _0 \\ &= \widetilde {H}^n(X,G) \end{align*}
  It is easy to see that the above equivalence of types is a homomorphism and is natural in $X$ . This concludes the proof of the suspension axiom.

• • Exactness: Let $f : X \to Y$ . We want to show that the sequence

  \begin{align*} \widetilde {H}^n({\textsf{cofib}\,f},G) \xrightarrow {{{\textsf{cfcod}}}^*} \widetilde {H}^n(Y,G) \xrightarrow {f^*} \widetilde {H}^n(X,G) \end{align*}
  is exact.Footnote ⁶ In other words, we want to show that, given an element $\lvert \,g\,\rvert : \widetilde {H}^n(Y,G)$ , we have $\lvert \,g\,\rvert \in \ker (f^*)$ iff $g \in \textsf{im}({{\textsf{cfcod}}}^*)$ . For the left-to-right implication, the assumption $\lvert \,g\,\rvert \in \ker (f^*)$ gives us a homotopyFootnote ⁷ $p : (x : X) \to 0_k = g(f(x))$ . Using this, we construct a function $g_{\textsf{cf}} : {\textsf{cofib}\,f} \to K(G,n)$ by
  \begin{align*} g_{\textsf{cf}}(\textsf{inl}\,{\star }) &= 0_k \\ g_{\textsf{cf}}(\textsf{inr}\,{y}) &= g(y) \\ \textsf{ap}_{g_{\textsf{cf}}}(\textsf{push}\,x) &= p(x) \end{align*}
  The fact that ${{\textsf{cfcod}}}^*(\lvert \,g_{\textsf{cf}}\,\rvert ) = \lvert \,g\,\rvert$ holds by construction, and hence $\lvert \,g\,\rvert \in \textsf{im}({{\textsf{cfcod}}}^*)$ .

  For the other direction, we simply want to show that the composition $f^* \circ {{\textsf{cfcod}}}^*$ is null-homotopic. This holds by construction, since ${{\textsf{cfcod}}} \circ f : X \to {\textsf{cofib}\,f}$ is null-homotopic by definition of $\textsf{cofib}\,f$ .

• • Dimension: when $n \lt 0$ , the statement is trivial. When $n \gt 0$ , we have

  \begin{align*} \widetilde {H}^n(\mathbb{S}^0,G) = \lVert \mathbb{S}^0 \to _\star K(G,n)\rVert _0 \simeq \lVert K(G,n)\rVert _0 \end{align*}
  Since $K(G,n)$ is $(n-1)$ -connected, $\lVert K(G,n)\rVert _0$ vanishes for $n \gt 0$ .

• • Additivity: Again, the statement is trivial when $n \lt 0$ . For the non-trivial case, the result follows from a simple rewriting of $\widetilde {H}^n(\bigvee _{i:I} X_i,G)$ using the elimination principle for $\bigvee _{i:I} X_i$ :

  \begin{align*} \widetilde {H}^n\left (\bigvee _{i:I} X_i,G\right ) &:= \bigg \Vert \bigvee _{i:I} X_i \to _\star K(G,n) \bigg \Vert _0 \\ &\simeq \Bigg \Vert \sum _{f : (\sum _{i:I} X_i) \to K(G,n)} \sum _{x_0 : K(G,n)} \left (((i : I) \to f(i,\star _{X_i}) = x_0) \times (x_0 = 0_k)\right ) \Bigg \Vert _0\\ &\simeq \Bigg \Vert \sum _{f : (\sum _{i:I} X_i) \to K(G,n)} \sum _{(x_0,p) : \sum _{x : K(G,n)} (x = 0_k)} \left ((i : I) \to f(i,\star _{X_i}) = x_0\right ) \Bigg \Vert _0\\ &\simeq \Bigg \Vert \sum _{f : (\sum _{i:I} X_i) \to K(G,n)} ((i : I) \to f(i,\star _{X_i}) = 0_k) \Bigg \Vert _0 \\ &\simeq \lVert (i : I) \to X_i \to _\star K(G,n)\rVert _0\\ &\simeq (i : I) \to \lVert X_i \to _\star K(G,n)\rVert _0 & (\textsf{AC}_0)\\ &= \Pi _{i:I}\widetilde {H}^n(X_i,G) \end{align*}
  where the step from the third to fourth line is simply a deletion of the (contractible) singleton type $\sum _{x : K(G,n)} (x = 0_k)$ . The fact that this composition of equivalences indeed gives the usual homomorphism $\widetilde {H}^n\left (\bigvee _{i:I} X_i,G\right ) \to \Pi _{i:I}\widetilde {H}^n(X_i,G)$ is immediate by construction.

Proof. Straightforward generalisation of Brunerie (Reference Brunerie2016, Proposition 5.2.2).

Proof. A function $\mathbb{S}^1 \to K(G,1)$ is uniquely determined by its action on the canonical base point and the canonical loop. Hence, it corresponds to choices of a point $x : K(G,1)$ and a loop $x = x$ . In other words, $(\mathbb{S}^1 \to K(G,1)) \simeq \Sigma _{x : K(G,1)} (x = x)$ . But $K(G,1)$ is homogeneous, so $(x = x) \simeq \Omega K(G,1)$ . Thus, we have

\begin{align*} H^1(\mathbb{S}^1,G) &= \lVert \mathbb{S}^1 \to K(G,1)\rVert _0 \\ &\simeq \lVert \Sigma _{x : K(G,1)} (x = x)\rVert _0\\ &\simeq \lVert K(G,1) \times \Omega K(G,1)\rVert _0 \\ &\simeq \lVert K(G,1)\rVert _0 \times G \end{align*}

Now, $K(G,1)$ is $0$ -connected, so $\lVert K(G,1)\rVert _0$ vanishes. Thus, we have $H^1(\mathbb{S}^1,G) \simeq G$ .

We need to verify that this isomorphism is a homomorphism. It is easier to show that its inverse is. Let us call it $\phi$ . By definition, $\phi (g) = \lvert \,f_g\,\rvert$ where $f_g : \mathbb{S}^1 \to K(G,1)$ is the map sending $\textsf{base}$ to $\star$ and $\textsf{loop}$ to $\textsf{loop}_k\,{g}$ . Thus, we only need to verify that for $g_1,g_2 : G$ and $x : \mathbb{S}^1$ , we have $f_{g_1}(x) +_k f_{g_2}(x) = f_{g_1 + g_2}(x)$ . We do this by induction on $x$ . When $x$ is $\textsf{base}$ , the goal holds by $\textsf{refl}$ . For the action on $\textsf{loop}$ , we need to show that

\begin{align*} \textsf{ap}^{2}_{+_k}({\textsf{loop}_k\,{g_1}},{\textsf{loop}_k\,{g_2}}) = {\textsf{loop}_k\,{(g_1 + g_2)}} \end{align*}

which holds by Proposition 16 and functoriality of $\textsf{loop}_k\,{}$ .

Proof. The case $n = 1$ is Proposition 47. For $n \gt 1$ , we simply apply the suspension isomorphism inductively.

Proof. Let us first consider the case $n \gt m$ . We induct on $m$ . When $m = 0$ , we have

\begin{align*} H^n(\mathbb{S}^0) = \lVert \mathbb{S}^0 \to K(G,n)\rVert _0 \simeq \lVert K(G,n) \times K(G,n)\rVert _0 \end{align*}

which vanishes since $K(G,n)$ is $(n-1)$ -connected and $n \gt m = 0$ . For larger $m$ , we know by suspension that $H^n(\mathbb{S}^m) \cong H^{n-1}(\mathbb{S}^{m-1})$ which vanishes by the induction hypothesis.

When $n\lt m$ we have

\begin{align*} H^n(\mathbb{S}^m,G) \simeq H^n(\lVert \mathbb{S}^m\rVert _n,G) \end{align*}

which vanishes because $\mathbb{S}^m$ is $n \leq (m-1)$ -connected and $n \geq 1$ .

Proof. The only non-trivial cohomology groups are $H^0(\mathbb{S}^n,R) \cong H^n(\mathbb{S}^n,R) \cong R$ . Thus, the cup product can only be non-trivial in dimensions $0 \times n$ and $n \times 0$ . In these dimension, the cup product is simply left respectively right $R$ -multiplication. This gives the desired isomorphism by letting constants in $H^*(\mathbb{S}^n,R)$ represent elements coming from $H^0(\mathbb{S}^n,R)$ and elements of degree one represent elements coming from $H^n(\mathbb{S}^n,R)$ .

Proof. The case when $n=0$ follows by Proposition 46 since $\mathbb{T}^2$ is $0$ -connected. For the case $n \geq 1$ , let us inspect the underlying function space of $H^n(\mathbb{T}^2 , G)$ . We have

\begin{align*} (\mathbb{T}^2 \to K(G,n)) &\simeq (\mathbb{S}^1 \times \mathbb{S}^1 \to K(G,n)) \\ &\simeq \mathbb{S}^1 \to (\mathbb{S}^1 \to K(G,n)) \\ &\simeq \sum _{f :\,\mathbb{S}^1\to K(G,n)} (f = f)\\ &\simeq \sum _{f :\,\mathbb{S}^1\to K(G,n)} ((x : \mathbb{S}^1) \to \Omega (K(G,n) , f(x)))\\ &\simeq (\mathbb{S}^1\to K(G,n)) \times (\mathbb{S}^1 \to \Omega (K(G,n)))\\ &\simeq (\mathbb{S}^1\to K(G,n)) \times (\mathbb{S}^1 \to K(G,n-1))\\ \end{align*}

where the step from line 3 to 4 is function extensionality and the step from lines 4 to 5 follows by homogeneity of $K(G,n)$ . Under set truncation, the last type in this chain of equivalences is simply $H^n(S^1,G) \times H^{n-1}(S^1,G)$ , which we know from Section 5.1 is simply $G \times G$ when $n = 1$ , $G$ when $n = 2$ , and trivial for higher $n$ . The fact that this map is a homomorphism follows by a similar argument to that of the corresponding statement in the proof of Proposition 47.

Proof. The elimination principle for $\mathbb{R}P^2$ tells us that

\begin{equation*}({\mathbb{R}P^2} \to K(G,1)) \simeq \sum _{x : K(G,1)}\sum _{p : x = x} (p^{-1} = p) \simeq \sum _{x : K(G,1)}\sum _{p : x = x} (p \cdot p = \textsf{refl}) \end{equation*}

Since $K(G,n)$ is homogeneous, this is equivalent to the type

\begin{equation*}{K(G,1)} \times \sum _{p : \Omega K(G,1)} (p \cdot p = \textsf{refl}) \end{equation*}

which under the equivalence $\Omega K(G,1) \simeq G$ is equivalent to the type

\begin{equation*}{K(G,1)} \times \sum _{g : G} (g + g = 0_G) \end{equation*}

or, in other words, $K(G,1) \times G[2]$ . Thus, we get

\begin{align*} H^1({\mathbb{R}P^2},G) = \lVert {\mathbb{R}P^2} \to K(G,1)\rVert _0 &\simeq \lVert K(G,1) \times G[2] \rVert _0 \\ &\simeq \lVert K(G,1)\rVert _0 \times G[2] \\ &\simeq G[2] \end{align*}

We verify that the inverse of this equivalence is a homomorphism. The inverse $\psi : G[2] \to H^1({\mathbb{R}P^2},G)$ is given by $\psi (g) = f_g$ where $f_g : {\mathbb{R}P^2} \to K(G,1)$ sends $\star$ to $0_k$ and $\ell$ to $\textsf{loop}_k\,{g}$ . We do not need to worry about its action on $\textsf{sq}$ : for truncation reasons, this is not data but a property. We need to verify that $f_{g_1+g_2}(x) = f_{g_1}(x) +_k f_{g_2}(x)$ for $x : {\mathbb{R}P^2}$ and $g_1,g_2 : G$ . When $x$ is $\star$ , this holds by $\textsf{refl}$ . For the action on $\ell$ , we need to verify that

\begin{align*} {\textsf{loop}_k\,{(g_1+g_2)}} = \textsf{ap}^{2}_{+_k}({\textsf{loop}_k\,{g_1}},{\textsf{loop}_k\,{g_2}}) \end{align*}

which we already verified in the proof of Proposition 47. This concludes the proof.

Proof. Like in the proof of Proposition 55, we have

\begin{align*} ({\mathbb{R}P^2} \to K(G,2)) &\simeq \sum _{x : K(G,2)}\sum _{p : x = x} {p \cdot p = \textsf{refl}} \\ &\simeq K(G,2) \times \sum _{p : \Omega K(G,2)} {p \cdot p = \textsf{refl}} \\ &\simeq K(G,2) \times \sum _{x : K(G,1)}{x +_k x = 0_k} \end{align*}

Under set truncation, the $K(G,2)$ -component vanishes which leaves us with the type $\lVert \sum _{x : K(G,1)}{x +_k x = 0_k}\rVert _0$ . Let us construct a map $\phi : \lVert \sum _{x : K(G,1)}{x +_k x = 0_k}\rVert _0 \to G/2$ by means of a curried map $\phi _x : x +_k x = 0_k \to G/2$ depending on $x : K(G,1)$ . We may define it by the set-elimination rule for $K(G,1)$ . When $x$ is $0_k$ , what is desired is a map $\phi _{\star } : \Omega K(G,1) \to G/2$ . This map may simply be defined by $\phi _{\star } := [\!-\!] \circ \sigma _0^{-1}$ . We now need to describe the action on $\textsf{loop}_k\,{g}$ . This amounts to providing a path $[g] = [g' + g + g']$ , which of course trivially exists in $G/2$ . This defines $\phi (\lvert \,x,p\,\rvert ) := \phi _x(p)$ .

The inverse $\psi : G/2 \to \lVert \Sigma _{x : K(G,1)} (x+_k x = 0_k)\rVert _0$ is defined on canonical elements by $\psi [g] = (0_k,{\textsf{loop}_k\,{g}})$ . In order to show that this is well-defined, we need to show that

\begin{align*} \lvert \,(0_k, \textsf{refl})\,\rvert = \lvert \,(0_k,{\textsf{loop}_k\,{(g+_G g)}})\,\rvert \end{align*}

This is done component-wise. For first component, we need to provide a path $0_k = 0_k$ . In fact, the right choice here is not $\textsf{refl}$ but instead $\textsf{loop}_k\,{g}$ . Showing that the second components agree now amounts to providing a filler of the square

which, in turn, is equivalent to proving that ${\textsf{loop}_k\,{(g +_G g)}} = \textsf{ap}^{2}_{+_k}({\textsf{loop}_k\,{g}},{\textsf{loop}_k\,{g}})$ , which we have already done in the proofs of Propositions 47 and 55.

Finally, let us prove that the maps cancel. Verifying that $\phi (\psi (x)) = x$ for $x:G/2$ is very direct: since the statement is a proposition, it suffices to this when $x$ is on the form $[g]$ . In this case, we have $\phi (\psi [g]) := \phi _{0_k}({\textsf{loop}_k\,{g}}) := \sigma _0^{-1}(\sigma _0(g)) = g$ .

For the other direction, the fact that we are proving a proposition means that it suffices to show the cancellation for elements of $\lVert \Sigma _{x : K(G,1)} (x +_k x = 0_k)\rVert _0$ on the form $\lvert \,(0_k,p)\,\rvert$ with $p : \Omega K(G,1)$ . We get $\psi (\phi \,\lvert \,0_k,p\,\rvert ) := \psi (\sigma _0^{-1}(p)) = \lvert \,0_k,\sigma _0(\sigma _0^{-1}(p))\,\rvert = \lvert \,0_k,p\,\rvert$ . Thus, we have shown:

\begin{align*} H^2({\mathbb{R}P^2},G) &\simeq \bigg \Vert K(G,2) \times \sum _{x : K(G,1)}{x +_k x = 0_k} \bigg \Vert _0 \\ &\simeq \bigg \Vert \sum _{x : K(G,1)}{x +_k x = 0_k} \bigg \Vert _0\\ &\simeq G/2 \end{align*}

The argument for why this equivalence is a homomorphism is technical but similar to the argument given in previous proofs.

Proof. As we have seen in the two previous proofs:

\begin{align*} H^n({\mathbb{R}P^2},G) &\simeq \bigg \Vert K(G,n) \times \sum _{x : K(G,n-1)} x +_k x = 0_k \bigg \Vert _0 \\ &\simeq \bigg \Vert \sum _{x : K(G,n-1)} x +_k x = 0_k \bigg \Vert _0 \end{align*}

We may move in the truncation one step further to get

\begin{align*} \bigg \Vert \sum _{x : K(G,n-1)} \lVert x +_k x = 0_k\rVert _0\bigg \Vert _0 \end{align*}

Since $n \gt 2$ , any path space over $K(G,n-1)$ is (at least) $0$ -connected. Thus, the truncation kills the space $(x +_k x = 0_k)$ . Hence, the above type is simply $\lVert K(G,n-1)\rVert _0$ , which also vanishes due to connectedness.

Proof. Consider the function space $(K^2 \to K(G,n))$ :

\begin{align*} (K^2 \to K(G,n)) \simeq \sum _{x : K(G,n)}\sum _{p,q: x = x}(p \cdot q \cdot p = q) \end{align*}

Again, we may use homogeneity of $K(G,n)$ to show that this is equivalent to the type

\begin{equation*}K(G,n) \times \sum _{p,q: \Omega K(G,n)}(p \cdot q \cdot p = q)\end{equation*}

We now consider the path space $p \cdot q \cdot p = q$ for $p,q : \Omega K(G,n)$ . By commutativity of $\Omega \,K(G,n)$ , composing both sides by $q^{-1}$ will cancel out $q$ everywhere. Thus, the type is simply $p\cdot p = \textsf{refl}$ . Hence, we have

\begin{align*} H^n(K^2,G) &:= \lVert K^2 \to K(G,n)\rVert _0 \\ &\simeq \bigg \Vert K(G,n) \times \Omega K(G,n) \times \sum _{p : \Omega K(G,n)} (p \cdot p \equiv \textsf{refl})\bigg \Vert _0 \end{align*}

Excluding the $\Omega K(G,n)$ component, this is precisely the characterisation of $H^n({\mathbb{R}P^2},G)$ . Thus, the above type is equivalent to

\begin{align*} \lVert \Omega K(G,n)\rVert _0 \times H^n({\mathbb{R}P^2},G) \end{align*}

The factor $\lVert \Omega K(G,n)\rVert _0$ vanishes for $n \gt 1$ . When $n = 1$ , we have $\lVert \Omega K(G,1)\rVert _0 \simeq \lVert G\rVert _0 \simeq G$ . Verifying that this is a homomorphism is straightforward but somewhat technical.

Proof. By Proposition 55, we have $H^1({\mathbb{R}P^2},\mathbb{Z}) \cong \mathbb{Z}[2]$ . Since $\mathbb{Z}$ has no non-trivial torsion elements, this tells us that $H^1({\mathbb{R}P^2},\mathbb{Z})$ is trivial. Hence, we only have as non-trivial cohomology groups the groups $H^0({\mathbb{R}P^2},\mathbb{Z}) \cong \mathbb{Z}$ and $H^2({\mathbb{R}P^2},\mathbb{Z}) \cong \mathbb{Z}/2\mathbb{Z}$ . The proof now proceeds just like that of Proposition 50, representing elements from $H^0({\mathbb{R}P^2},\mathbb{Z})$ and elements from $H^2({\mathbb{R}P^2},\mathbb{Z})$ respectively by degree- $0$ and degree- $1$ elements in $H^*({\mathbb{R}P^2},\mathbb{Z})$ . The $2x$ -quotient comes from the $2$ -torsion in $H^2({\mathbb{R}P^2},\mathbb{Z})$ .

Proof. Using Proposition 58 and the preceding results, we know that the only non-trivial cohomology groups of $K^2$ are

\begin{align*} H^0(K^2,\mathbb{Z}) &\cong \mathbb{Z} \\ H^1(K^2,\mathbb{Z}) &\cong \mathbb{Z} \times \mathbb{Z}[2] \cong \mathbb{Z} \\ H^2(K^2,\mathbb{Z}) & \cong \mathbb{Z}/2\mathbb{Z} \end{align*}

Consider first the ring $\mathbb{Z}[x,y]$ . Using the above isomorphisms, we may naively let degree- $0$ elements in $\mathbb{Z}[x,y]$ represent elements from $H^0(K^2,\mathbb{Z})$ , degree- $1$ elements with $x$ -term represent elements from $H^1(K^2,\mathbb{Z})$ and degree- $1$ elements with $y$ term represent elements from $H^2(K^2 ,\mathbb{Z})$ . Under this identification, we see that the relations described by the ideal $(2y,xy,y^2)$ are satisfied. Hence, we have a map $\mathbb{Z}[x,y]/(2y,xy,y^2) \to H^1(K^2,\mathbb{Z})$ . In order to lift this to an isomorphism $\mathbb{Z}[x,y]/(2y,x^2,y^2,xy) \cong H^*(K^2,\mathbb{Z})$ , we need to show that the cup product is trivial in dimension $1 \times 1$ . It suffices to show that it vanishes on generators. We can easily define a map $\alpha : K^2 \to K(1,\mathbb{Z})$ such that $\lvert \,\alpha \,\rvert$ generates $H^1(K^2,\mathbb{Z})$ . We define it by

\begin{align*} \alpha (\star ) &:= 0_k \\ \textsf{ap}_{\alpha }(\ell _1) &:= \textsf{refl} \\ \textsf{ap}_{\alpha }(\ell _2) &:= {\textsf{loop}_k\,{1}} \end{align*}

where, $1 : \mathbb{Z}$ denotes the generator. The final step of the definition – the action of $\alpha$ on $\textsf{sq}$ – corresponds to providing a proof that ${\textsf{loop}_k\,{1}} = {\textsf{loop}_k\,{1}}$ , which we do by $\textsf{refl}$ . Let $x : K^2$ . We show that $\alpha (x) \smile _k \alpha (x) = 0_k$ by induction on $x$ . We prove $\alpha (\star )\smile _k \alpha (\star ) = 0_k$ by $\textsf{refl}$ and $\textsf{ap}^{2}_{\smile _k}(\textsf{ap}_{\alpha }(\ell _2),\textsf{ap}_{\alpha }(\ell _2)) = \textsf{refl}$ by $\textsf{refl}$ . For $\textsf{ap}^{2}_{\smile _k}(\textsf{ap}_{\alpha }(\ell _1),\textsf{ap}_{\alpha }(\ell _1))$ , we need to provide a path $\textsf{ap}^{2}_{\smile _k}({\textsf{loop}_k\,{1}},{\textsf{loop}_k\,{1}}) = \textsf{refl}$ , which we give by ${\textsf{kill-}\Delta }({\textsf{loop}_k\,{1}})$ . Finally, for the action of $\textsf{sq}$ , we need to provide a filler of the square

which is trivial. Thus, we have shown, in particular, that $\lvert \,\alpha \,\rvert \smile \lvert \,\alpha \,\rvert$ is trivial, which means that $\smile$ vanishes everywhere in dimension $1 \times 1$ . This concludes the proof.