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[几何/拓扑] $\pi_n(X)$ is abelian for $n≥2$

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hbghlyj posted 2025-8-11 07:03 |Read mode

Let’s walk through the answer.
Use $J=[-1,1]$. For $f,g:(J^n,\partial J^n)\to (X,x_0)$, the sum is concatenation in the first coordinate
$$\tag1\label1
(f+g)(s_1,\ldots,s_n)=
\begin{cases}
f(2s_1+1,s_2,\ldots,s_n) & s_1\le 0,\\
g(2s_1-1,s_2,\ldots,s_n) & s_1\ge 0.
\end{cases}
$$
Idea. Find a map $R:J^n\to J^n$ that is homotopic (rel $\partial J^n$) to the identity and swaps the two halves used by the sum. Then
$$
[f+g]=[(f+g)\circ R]=[g\circ R+f\circ R]=[g+f].
$$
Constructing $R$. Split $J^n=J^2\times J^{n-2}$. We’ll 180°-rotate the first two coordinates and leave the rest alone.
  • Build a rotation on the square $J^2$:
    Define $h:J^2\to D^2$ by sending each point straight out/in along the line from the origin so that the boundary $\partial J^2$ goes to the unit circle. (This is a homeomorphism—think “radial projection from the square to the disk”.)
  • Rotate the disk: $G:D^2\times I\to D^2,\ G(z,t)=e^{\pi it}z$. This rotates by angle $\pi$. At $t=1$, $G_1(z)=-z$.
  • Bring that rotation back to the square: $H=h^{-1}\circ G\circ(h\times \mathrm{id}_I)$. Then $H$ is a homotopy of pairs $(J^2,\partial J^2)$ from the identity to $H_1$, which is the “$\pi$-rotation” of the square fixing the boundary setwise.
  • Extend to $J^n$ by ignoring the last $n-2$ coordinates:
    $$
    R = H_1\times \mathrm{id}_{J^{n-2}},\qquad
    R(s_1,s_2,s_3,\ldots,s_n)=(-s_1,-s_2,s_3,\ldots,s_n).
    $$
    Because $H$ is a homotopy rel boundary, precomposing any $\phi:(J^n,\partial J^n)\to(X,x_0)$ with $R$ does not change its homotopy class: $[\phi]=[\phi\circ R]$.

Why $R$ swaps the order in the sum. Compute using \eqref{1}
  • Case $s_1\le 0$
    $$
    (g\circ R+f\circ R)(s)= (g\circ R)(2s_1+1,\ldots)
    = g(-2s_1-1,-s_2,s_3,\ldots,s_n)
    $$
    But since $-s_1\ge 0$,
    $$
    ((f+g)\circ R)(s)=(f+g)(-s_1,-s_2,\ldots)
    = g(2(-s_1)-1,-s_2,\ldots)
    = g(-2s_1-1,-s_2,\ldots)
    $$
  • Case $s_1\ge 0$
    $$
    (g\circ R+f\circ R)(s)= (f\circ R)(2s_1-1,\ldots)
    = f(-2s_1+1,-s_2,s_3,\ldots,s_n)
    $$
    But since $-s_1\le 0$,
    $$
    ((f+g)\circ R)(s)=(f+g)(-s_1,-s_2,\ldots)
    = f(2(-s_1)+1,-s_2,\ldots)
    = f(-2s_1+1,-s_2,\ldots)
    $$
so the two expressions match.

Why does “precompose with a map homotopic to the identity rel boundary” guarantee the same class in $\pi_n(X,x_0)$?
We have a homotopy of pairs $H_t:(J^n,\partial J^n)\to(J^n,\partial J^n)$ with $H_0=\mathrm{id}$ and $H_1=R$.
Then for any $\phi:(J^n,\partial J^n)\to(X,x_0)$, the composite $\phi\circ H_t$ is a homotopy rel $\partial J^n$ from $\phi$ to $\phi\circ R$, because $H_t(\partial J^n)\subseteq \partial J^n$ and $\phi|_{\partial J^n}=x_0$.

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original poster hbghlyj posted 2025-8-11 06:35

$\pi_n(X,A)$ is abelian for $n\ge 3$

Here’s the same rotation trick, but done in the relative cube model so you can see exactly where $A$ enters and why we need $n\ge 3$.

Model for $\pi_n(X,A,x_0)$

Use $J=[-1,1]$ and write $J^n=J^{n-1}\times J$ with coordinates $(u,s)$, where $u=(u_1,\dots,u_{n-1})\in J^{n-1}$ and $s\in J$.
Represent $\pi_n(X,A,x_0)$ by maps of triples
$$
f:(J^n,B,S)\longrightarrow (X,A,x_0),
$$
where
$$
B=\partial J^{n-1}\times J\ \cup\ J^{n-1}\times\{-1\},\qquad
S=J^{n-1}\times\{-1\}.
$$
(So the whole boundary goes to $A$, and the “bottom face” $S$ goes to $x_0$.)

Define $f+g$ by concatenating in the first tangential coordinate $u_1$ (so we never touch the “relative” direction $s$):
$$
(f+g)(u,s)=
\begin{cases}
f(2u_1+1,u_2,\dots,u_{n-1},s), & u_1\le 0,\\
g(2u_1-1,u_2,\dots,u_{n-1},s), & u_1\ge 0.
\end{cases}
$$
This respects the triple $(J^n,B,S)$: on $B$ we still land in $A$, and on $S$ we land at $x_0$.

The rotation $R$ (works when $n\ge 3$)

Now assume $n\ge 3$. Rotate inside the tangential block $J^{n-1}$ using the first two coordinates and leave $s$ fixed:
  • As before, build a $180^\circ$ rotation on $J^2$ via $h:J^2\to D^2$, $G(z,t)=e^{\pi it}z$, and $H_t=h^{-1}\circ G_t\circ h$.
  • Extend by the identity on the remaining coordinates to get
    $$
    R(u_1,u_2,u_3,\dots,u_{n-1};s)
    =(-u_1,-u_2,u_3,\dots,u_{n-1};s).
    $$
  • $R$ is homotopic to $\mathrm{id}$ by a homotopy $R_t$ of triples $(J^n,B,S)$: it preserves $\partial J^{n-1}\times J$ setwise and fixes $S=J^{n-1}\times\{-1\}$ pointwise (since $s$ is untouched).
  • Hence for any $\phi$ we have $[\phi]=[\phi\circ R]$ in the relative group.

Swapping $f$ and $g$

Because the sum uses $u_1$ and $R$ flips $u_1$,
$$
(f+g)\circ R = g\circ R + f\circ R,
$$
by the same calculation you did in the absolute case (now $s$ is just carried along). Therefore
$$
[f+g]=[(f+g)\circ R]=[g\circ R+f\circ R]=[g+f],
$$
so $\pi_n(X,A)$ is abelian for $n\ge 3$.

Why this fails for $n=2$

When $n=2$, the tangential block $J^{n-1}=J$ has only one coordinate $u_1$. There is no second tangential direction to perform a $180^\circ$ rotation while keeping the relative face $S=J\times\{-1\}$ fixed. Any map that flips $u_1$ necessarily disturbs $S$ unless you also move $s$, which breaks the “rel $S$” condition. That’s the obstruction—hence $\pi_2(X,A)$ need not be abelian.

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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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