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original poster
hbghlyj
posted 2025-8-11 07:00
Here’s the interchange law in en.wikipedia.org/wiki/Eckmann–Hilton_argument checked by a direct calculation.
Let
$M=\operatorname{Map}\big((I^{n-1},\partial I^{n-1}),(X,A)\big)$ (basepoint = constant map at $x_0$).
For $n\ge 3$ we have two concatenation operations on $M$, by splitting different domain coordinates:
- $\mu_1$: concatenate in $u_1$
$$
(\mu_1(p,q))(u_1,u_2,\ldots)=
\begin{cases}
p(2u_1,u_2,\ldots), & u_1\le \frac12,\\
q(2u_1-1,u_2,\ldots), & u_1\ge \frac12.
\end{cases}
$$ - $\mu_2$: concatenate in $u_2$
$$
(\mu_2(p,q))(u_1,u_2,\ldots)=
\begin{cases}
p(u_1,2u_2,\ldots), & u_2\le \frac12,\\
q(u_1,2u_2-1,\ldots), & u_2\ge \frac12.
\end{cases}
$$
Both respect the boundary-to-$A$ condition.
For any $p,q,r,s\in M$,
$$
\mu_2\big(\mu_1(p,q),\mu_1(r,s)\big)=
\mu_1\big(\mu_2(p,r),\mu_2(q,s)\big)
\quad\text{in }M.
$$Just compute both sides on each quadrant of the $(u_1,u_2)$-square:
- If $u_1\le \frac12,\ u_2\le \frac12$:
$\mu_2(\mu_1(p,q),\mu_1(r,s))(u)=\mu_1(p,q)(u_1,2u_2,\ldots)=p(2u_1,2u_2,\ldots).$
$\mu_1(\mu_2(p,r),\mu_2(q,s))(u)=\mu_2(p,r)(2u_1,u_2,\ldots)=p(2u_1,2u_2,\ldots).$ - If $u_1\ge \frac12,\ u_2\le \frac12$: both sides give $q(2u_1-1,2u_2,\ldots)$.
- If $u_1\le \frac12,\ u_2\ge \frac12$: both give $r(2u_1,2u_2-1,\ldots)$.
- If $u_1\ge \frac12,\ u_2\ge \frac12$: both give $s(2u_1-1,2u_2-1,\ldots)$.
Thus both composites produce the same four-block map
$$
u\mapsto
\begin{cases}
p(2u_1,2u_2,\ldots) & \text{LL quadrant},\\
q(2u_1-1,2u_2,\ldots) & \text{LR quadrant},\\
r(2u_1,2u_2-1,\ldots) & \text{UL quadrant},\\
s(2u_1-1,2u_2-1,\ldots) & \text{UR quadrant},
\end{cases}
$$
so the interchange law holds.
Why $n\ge 3$: we needed two independent domain coordinates $u_1,u_2$ inside $I^{n-1}$ to define $\mu_1$ and $\mu_2$. When $n=2$, $I^{n-1}=I$ has only one such coordinate, so there’s no second product and no interchange law to apply. |
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