I'm brand new to homotopy theory so I'm sure this question is utterly stupid. But anyway I'm trying to understand a proof in the book "Topology and Geometry" by Glen Bredon. This is the proposition:

Let $\partial: \pi_2 (X,A,\ast)\rightarrow \pi_1 (A,\ast)$ be the boundary map from the long exact sequence of homotopy groups. For $\alpha, \beta \in \pi_2 (X,A,\ast)$ we have $(\partial (\alpha))\beta=\alpha \beta \alpha^{-1}$. Where the left hand side means the action of $\pi_1(A,\ast)$ on $\pi_2 (X,A,\ast)$.

My confusion lies in the very first sentence of the proof:

A representative $\mathbb{D}^2\rightarrow X$ of $\beta$ can be taken so that everything maps into the base point except for a small disk and this disk can be placed anywhere. Thus we see that a representative of $\alpha \beta \alpha^{-1}$ can be taken as in the first part of figure VII-7.

But isn't this equivalent to the assumption that the restriction of $\beta$ to $S^1$ is null-homotopic in $\pi_1 (A)$? It seems to me that if $\beta$ lay in the same relative homotopy class as a map which is constant on $S^1$ then any homotopy between these two maps would also provide a contraction of $\beta|_{S^1}$ in $A$.


First of all, this is an important fact, that the boundary map $\partial:\pi_2(X,A,*) \to \pi_1(A,*)$ with the action of $\pi_1(A,*)$ has the structure of crossed module. That is a morphism of groups $\mu: M \to P$ with an action of $P$ on $M$ (on the left, with your conventions) with two rules:

  1. $\partial(^p m )= p(\partial m)p^{-1}$;

  2. $mnm^{-1}={} ^{\mu m} n$.

The second was first given by J.H.C. Whitehead in 1941.

I do not have the book referred to, and prefer a more rectangular approach. Regard $\alpha, \beta$ as maps of squares which map the whole into $X$, the top edge into $A$ and the other three edges to the base point. Divide a rectangle into 6 parts by 1 horizontal and 3 vertical lines. In the middle top put $\beta$, in the bottom left put $\alpha$ and bottom right put $\alpha^{-1}$ (reflected horizontally). The bottom middle is totally constant, and the other two squares are vertically constant.

This diagram looks like

$$\begin{bmatrix} || & \beta & || \\ \alpha & \Box & \alpha^{-1} \end{bmatrix}$$

Reading this diagram in 2 ways (either by rows, or by columns) gives the result. (It is really a result on double groupoids! )

There is full background on crossed modules and double groupoids in the book "Nonabelian algebraic topology: filtered spaces, crossed complexes, cubical homotopy groupoids" of which full details are here, where a pdf is available.


What's happening is that Bredon's sentence is ambiguous and the figure is highly misleading.

When Bredon says that "everything maps into the base point except for a small disk and this disk can be placed anywhere", he means that the disk can be placed anywhere on the boundary of $D^2$, with the boundary of the disk sharing an arc with the boundary of $D^2$.

This ambiguity is exacerbated by the picture, which really shouldn't have a circle surrounding the $\beta$ in the way that it does. The entire right region is a copy of $\beta$, and the portion of the boundary of $D^2$ in the figure between $-\pi/3$ and $\pi/3$ does not map to the basepoint.


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