$\mathbf {The \ Problem \ is}:$ Let, $A\subset X$ are based topological spaces with basepoint $•\in A.$ Show that the image of $H=\pi_2(X,•)$ in $G=\pi_2(X,A,•)$ lies in the centre of the group $G$.

$\mathbf {My \ approach}:$ If $f\in H$ then $f: I^2\to X, f(\delta I^2)=•.$ So, such an $f$ belongs to $G$ also by definition of $G.$

Thus the map $\phi : H \to G$ is well-defined by $\phi(f)=f.$

Now, for any $g\in G$ we have

$f*g (p,q) = \begin{cases} f(2p,q) , & 0\leq p \leq \frac{1}{2}\\ g(2p-1,q) , & \frac{1}{2}\leq p \leq1 \end{cases}$

$H$ is abelian but $G$ not necessarily.

But, how to show that there's a homotopy of $f*g$ and $g*f ?$

I think it's tricky as there's a switch of $f$ and $g$ in the definition of concatenation .

A hint is required, thanks in adv.


1 Answer 1


The reason why it is true can be explained by hands. Let $I^2$ be a $2$-square, and $K$ be the union of left, bottom, and right sides of $I^2$. Then every element of $\pi_2(X,A,x)$ is represented by the map of triples: $$ \beta: (I^2, \partial I^2, K) \to (X,A,x), $$ while every element of $\pi_2(X,x)$ is represented by some $$ \alpha: (I^2, \partial I^2) \to (X,x). $$ Now if $\beta,\beta'$ are two representatives of elements of $\pi_2(X,A,x)$, then their product $\beta\beta'$ is obtained by "gluing" right side of $\beta$ with left side of $\beta'$, since those sides are mapped into the same point $x$.

Moreover, if $\alpha$ is a representative of some $\pi_2(X,x)$, then one can "attach" $\alpha$ to $\beta$ not only to left or right side, but also to bottom side of $\beta$. It is convenient to denote such map by $\frac{\beta}{\alpha}$, which means not the division, but indicates a "stack". Thus $$ \frac{\beta}{\alpha}(s,t) = \begin{cases} \alpha(s, 2t), & t\in[0;1/2], \\ \beta(s, 2t-1), & t\in[1/2,1]. \end{cases} $$ It remains to note that both $\alpha\beta$ and $\beta\alpha$ are homotopic to $\frac{\beta}{\alpha}$, which means that $\alpha\beta=\beta\alpha$.

In other words, the homotopy between $\alpha\beta$ and $\beta\alpha$ consists of "moving $\alpha$ along $K$ from the left down to the bottom, and then up to the right".

To get a more "conceptual" proof one should use identification $\pi_2(X,x) = \pi_1\Omega(X,x)$ and regard the loop space $\Omega(X,x)$ as an H-space and $\Omega(A,x)$ is its H-subspace.

Recall that a pointed space $(Y,e)$ is called an H-space, whenever there exist a continuous map $$ \mu:Y \times Y \to Y $$ such that $\mu(e,e)=e$ and the restrictions $$ \mu: e \times Y, Y \times e \to Y $$ are homotopic to the identity map $id_{Y}$ via homotopies fixing $e$.

In the H-space one can define multiplication of loops $\alpha,\beta:(I,\partial I) \to (Y,e)$ by $(\alpha\cdot\beta)(t) = \mu(\alpha(t),\beta(t))$, and on the level of homotopy classes this multiplication coincides with the usual product in $\pi_1(Y,e)$. In other words $[\alpha\beta]=[\alpha\cdot\beta]$.

Furthermore, a subset $B\subset Y$ is called an H-subspace if $e\in B$, $\mu(B\times B) \subset B$ and the restrictions $$ \mu:B\times e, e\times B \to B $$ are also homotopic to $id_{B}$ relatively to $e$.

In this case one can define on $\pi_1(Y,B,e)$ the structure of a group by the following rule: if $$ \alpha,\beta: (I,0,1) \to (Y,e,B) $$ two paths started at $e$ and finished in $B$, then their product is given by the same formula as above: $$ (\alpha\cdot\beta)(t) = \mu(\alpha(t),\beta(t)) $$

One can check that then the natural map: $$ j: \pi_2(Y,e) \to \pi_1(Y,B,e) $$ is a homomorphism.

It follows from these formulas that if $\alpha,\alpha':(I,\partial I)\to(Y,e)$ are two loops at $e$ and $\beta,\beta': (I,\partial I, K) \to (Y,B,e)$ are two paths from $e$ into $B$, then it directly follows from the multiplication formulas that $$ (\alpha\cdot\beta)(\alpha'\cdot\beta') = (\alpha\alpha')\cdot(\beta\beta'). $$ In fact both maps are given by $$ \gamma(t) = \begin{cases} \mu(\alpha(2t),\beta(2t)), & t\in [0;1/2], \\ \mu(\alpha'(2t-1),\beta'(2t-1)), & t\in [1/2;1]. \end{cases} $$

Lemma. The image of $j$ is contained in the center of $\pi_1(Y,B,e)$.

Proof. Let $\alpha\in\pi_1(Y,y)$ be a loop at $e$ and $\beta\in\pi_1(Y,B,e)$ be path from $e$ into $B$. Let also $\epsilon:I\to Y$ be a constant map into $e$. Then
$$ [\alpha\cdot\beta] = [(\alpha\epsilon)\cdot(\epsilon\beta)] = [(\epsilon\alpha)\cdot(\beta\epsilon)] = [(\epsilon\cdot\beta)\cdot(\alpha\epsilon)] = [\beta\cdot\alpha]. $$ Lemma is proved.

Corollary. The image of $j:\pi_2(X,x) \to \pi_2(X,A,x)$ is contained in the center of $\pi_2(X,A,x)$.

Proof. As noted above $Y = \Omega(X,x)$ is an H-space and $B = \Omega(A,x)$ is its H-subspace. Just replace $\pi_2(X,x)$ with $\pi_1 Y = \pi_1\Omega(X,x)$ and $\pi_2(X,A,x)$ with $\pi_1(Y,B) = \pi_1(\Omega(X,x), \Omega(A,a))$ and apply Lemma above.


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