# Question on image of the inclusion of $2$nd fundanental group of a space inside a relative fundamental group .

$$\mathbf {The \ Problem \ is}:$$ 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.

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.