I am currently reading Switzer's book "Algebraic Topology: Homotopy and Homology". On page 50, the proof of 3.30 c), he claims that a certian composition is something I can't see how it possibly can be what he states it to be. Let $\beta':S^1 \rightarrow I \vee S^1$ be defined by $(2t,\ast)$ if $t \leq 1/2$ and $(\ast,2t1)$ if $t > 1/2$. Consider the quotient map $q:I \rightarrow S^1$ given by $q(t) = e^{2\pi t}$. Switzer then claims that the composition $\alpha= (q \vee 1) \circ \beta': S^1 \rightarrow S^1 \vee S^1$ is given by $\alpha(t) = (4t,\ast)$ it $t \leq 1/4$, $\alpha(t) = (\ast,2t1/2)$ if $1/4 \leq t \leq 3/4$ and $\alpha(t) = (4(1t),\ast)$ if $3/4 \leq t \leq 1$. However, I get that the composition is $(2s,\ast)$ for $t \leq 1/2$ and $(\ast, 2t1)$ for $t \geq 1/2$. Is Switzer wrong, or am I misunderstanding something? Any help would be very appreciated, I don't seem to be able to fix this detail.

$\begingroup$ For one thing, those should be wedge sums ($\vee$), which you can get using either of the LaTeX commands \vee or \lor. $\endgroup$ – Dan Kneezel Aug 20 '13 at 15:50

$\begingroup$ Dan Kneezel: Thank you! Fixed. Do you see how the composition works out as Switzer claims? $\endgroup$ – Helle Karlsson Aug 20 '13 at 16:06

1$\begingroup$ Check this link for further perspectives on the action of $\pi_1$ on higher homotopy groups. mathoverflow.net/questions/19775/… $\endgroup$ – Dan Kneezel Aug 21 '13 at 5:09
I think that what you wrote is not precisely what Switzer defined, but perhaps you made some simplifications that I did not notice. I'm going to give up for now on combing through the details of the formulas and just say what the big picture is.
Look at the schematic on page 47. We see that what $\beta':(D^{n+1},s_0)\to ((I,1) \lor (D^{n+1},s_0),0)$ does is it squashes some contractible neighborhood of $s_0 \in D^{n+1}$  say the right half of the disk in the picture, up to homotopy  down to an interval, and handles the rest of the disk in such a way that everything is nice and continuous. Why did I write my target as $((I,1) \lor (D^{n+1},s_0),0)$? Notice that Switzer says he wants the basepoint of that space (and related subspaces) to be the point he labeled as 0. He's abusing the wedge sum notation for convenience, but what he's really doing is squashing.
To get $\beta:(S^n,s_0)\to ((I,1) \lor (S^n,s_0),0)$, you just restrict $\beta'$ to the boundary $\partial D^{n+1} = S^n$. Finally, to get $\alpha:(S^n,s_0)\to (S^1,*) \lor (S^n,s_0)$, you just glue together the endpoints of the interval you got from squashing. (To our relief, now the basepoint suggested by that notation is finally back where it's supposed to be.)
So, with this understanding, what does $\alpha:S^1\to S^1\vee S^1$ do? Think of each $S^1$ as $(I\sqcup*)/(0\sim *, 1\sim * )$ with basepoint $*$. For the neighborhood of $*$ we want to squash, we can take $N = [3/4,1]\cup[0,1/4]$. To squash $N$ to an interval, all you really do is just glue $t \in [0,1/4]$ to $(1t) \in [3/4,1]$.
(In physical terms, you're doing the following in the n=1 case: Take a piece of string and knot it into a loop; treat the knot as the basepoint. Pinch the loop together in the middle (away from the basepoint) to make two loops. Finally, pull taut the "subloop" which has the knot in it so that you have a loop on one side and an interval on the other side. You could imagine doing something similar with a balloon to visualize $\alpha: S^2 \to S^1\vee S^2$. When I want to think of the case for general $n$, I imagine the picture for $n=2$ and then squint my eyes so the 2's look like n's.)
Now we just implement my description of $\alpha:(S^n,s_0)\to (S^1,*) \lor (S^n,s_0)$ for the case $n=1$, taking care to choose parametrizations appropriately. We need to send the squashed $N$ to the first copy of $S^1$ in $S^1\lor S^1$. We can achieve this by sending $t \in [0,1/4]$ to $4t$. (Please forgive my imprecise notation. I think it'll be easier to see what's going on if I leave out some of the extra bells and whistles.) Correspondingly, since in the squashed $N$ we have that $t \in [0,1/4]$ is identified with $s = (1t) \in [3/4,1]$, you also have to send $s$ to $4t = 4(1s)$. These account for the first and third lines of the definition of $\alpha$. The second line just sends your remaining loop (the boundary circle you see on the right hand side of the schematic on page 47) to the second $S^1$.