= Solution
For $n\ge1$, let $\#$ denote concatenation in the first cube coordinate and let $\cdot$ denote pointwise multiplication of maps into the <topological group>. Both operations descend to the <homotopy group>, and the constant map is their common identity. They satisfy the interchange rule
$$
(\alpha\#\beta)\cdot(\gamma\#\delta)=(\alpha\cdot\gamma)\#(\beta\cdot\delta),
$$
which follows by considering the two halves of the first coordinate. Using the identity class $0$ gives
$$
[\alpha]\cdot[\beta]=([\alpha]\#0)\cdot(0\#[\beta])=([\alpha]\cdot0)\#(0\cdot[\beta])=[\alpha]\#[\beta].
$$
Therefore \b[$[\alpha\beta]=[\alpha]+[\beta]$]. The equality is on based <homotopy classes>; the usual reparametrization homotopies justify the unit identities for concatenation. This is the <Eckmann-Hilton argument>, and proves <homotopy-group addition in a topological group> even for $n=1$.
For the <unit quaternions>, the explicit inverse is
$$
\boxed{\Phi_{k,l}^{-1}(r,s)=(s,s^{-k}rs^{-l}).}
$$
Both compositions cancel in the displayed order, without commuting the <quaternions>. Multiplication, inversion and all integer powers are continuous, so this proves that $\Phi_{k,l}$ is a <homeomorphism> for all integers $k,l$.
Put $m=k+l$. Let $a,b$ be the standard generators of $H_3(S^3\times S^3;\mathbb Z)$, represented by the first and second factors. The previous pointwise-multiplication result and the <Hurewicz theorem> imply that the <mapping degree> of $q\mapsto q^m$ on $S^3$ is $m$, including negative integers. Restricting $\Phi$ to the two factors therefore gives
$$
\boxed{\Phi_*a=ma+b,\qquad\Phi_*b=a,\qquad [\Phi_*]_{H_3}=\begin{pmatrix}m&1\\1&0\end{pmatrix}.}
$$
The map on $H_0$ is the identity. For the top <homology group>, take the dual degree-three <cohomology classes> $x,y$. We have $\Phi^*x=mx+y$, $\Phi^*y=x$. Since these classes have odd degree, <graded commutativity of the cup product> gives
$$
\Phi^*(xy)=(mx+y)x=yx=-xy.
$$
Thus \b[$\Phi_*$ is multiplication by $-1$ on $H_6$], and all other <homology groups> of the product are zero.
For the gluing, regard $\Phi$ as the attaching identification from $\partial(D^4\times S^3)$ to $\partial(S^3\times D^4)$. In these coordinates its first input is the boundary coordinate of $D^4$, its second input the fibre coordinate; its second output is the boundary coordinate of the other $D^4$. This makes the two pieces the usual two trivializations of a <three-sphere bundle over the four-sphere>.
Let $A=S^3\times D^4$ and $B=D^4\times S^3$, and parametrize their common boundary $T=S^3\times S^3$ using the $B$ coordinates. Collar neighbourhoods give an open cover with the same <homotopy types>, so the <Mayer–Vietoris sequence> applies. Both pieces retract onto $S^3$. In degree three, the map into the <homology> of the pieces is
$$
L=(i_{A*}\Phi_*,-i_{B*}):\mathbb Z^2\longrightarrow\mathbb Z^2,\qquad L=\begin{pmatrix}m&1\\0&-1\end{pmatrix}.
$$
Indeed, inclusion into $A$ retains the first output coordinate, while inclusion into $B$ retains the second input coordinate. Exactness now gives
$$
0\longrightarrow H_4(X_{k,l})\longrightarrow\mathbb Z^2\xrightarrow{L}\mathbb Z^2\longrightarrow H_3(X_{k,l})\longrightarrow0.
$$
The second relation eliminates the second generator of the cokernel, leaving a single generator with relation $m$ times that generator equal to zero. Thus $\operatorname{coker}L=\mathbb Z/m\mathbb Z$. The kernel is zero if $m\ne0$ and is generated by $(1,0)$ if $m=0$.
The degree-six boundary class gives $H_7(X_{k,l})\cong H_6(T)=\mathbb Z$. All remaining positive-degree groups outside degrees $3,4,7$ vanish by the same <Mayer–Vietoris sequence>; connectedness gives $H_0=\mathbb Z$. Consequently the complete integral answer is
$$
\boxed{H_i(X_{k,l};\mathbb Z)\cong\begin{cases}
\mathbb Z,&i=0,7,\\
\mathbb Z/m\mathbb Z,&i=3,\\
\mathbb Z,&i=4\text{ and }m=0,\\
0,&\text{otherwise},
\end{cases}\qquad m=k+l.}
$$
Here $\mathbb Z/0\mathbb Z$ means $\mathbb Z$, and a negative $m$ gives the same cyclic group as $|m|$. In particular, $m=\pm1$ gives the integral <homology> of a seven-sphere, while $m=0$ gives the integral <homology> of $S^3\times S^4$.
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