TITLE: Very basic homotopy question
QUESTION [8 upvotes]: 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 $A$.

REPLY [2 votes]: 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.