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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. Let $\beta':S^1 \rightarrow I \vee S^1$ be defined by $(2t,_\ast)$ if $t \leq 1/2$ and $(\ast,2t-1)$ 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 $t \leq 1/4$, $\alpha(t) = (\ast,2t-1/2)$ if $1/4 \leq t \leq 3/4$ and $\alpha(t) = (4(1-t),\ast)$ if $3/4 \leq t \leq 1$. However, I get that the composition is $(4s,\ast)$ for $t \leq 1/2$ and $(\ast, 2t-1)$ for $t \geq 1/2$. Is Switzer wrong, or am I misunderstnading something?

Dmitry
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Helle Karlsson
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  • For the formulas you've written to make sense (and to coincide with those in Switzer's book) $\wedge$ must be replaced by $\vee$ (\vee). The problem is, $\wedge$ in the context of homotopy theory always means smash product. What's more, $X \wedge S^1$ is a suspension over $X$, so your post looks as if you are doing something concrete with suspensions of the segment and of the circle. Surely that's not even close to the truth, all these spaces are wedges, not suspensions. – Dmitry Aug 20 '13 at 19:00
  • Also $\beta'$ as you've written is not a continuous map from $S^1$. Look what happens about $t = 1/2$: when $t$ is less than $1/2$, you are at the end of the segment $I$, quite far from $S^1$ (which I assume is attached to $I$ at $0$). When $t$ is a bit larger than $1/2$, you suddenly jump into the circle, not good. Even if $S^1$ is attached to $I$ not to zero, but to $1$, the formula doesn't work (it maps $[0, 1]$ continuously, but the ends do not meet) – Dmitry Aug 20 '13 at 19:15

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$\beta'$ in Switzer's book is different. Look at the general definition, there is one more variable. For a circle embedded as $x^2 + y^2 = 1$, it works like that: emphasize points $(1, 0)$ and $(0, 1)$, they are $S^0$ over which $S^1$ is suspended. The $t$ of Switzer's book in $\beta'(s, x, t)$ is more or less vertical coordinate in the plane (zero when $y = -1$, increasing twice slower than $y$ so $t = 1$ means $y = 1$). So you "fold" lower half of the circle into the segment $I$ and then map upper half to $S^1$. Using single real coordinate on $S^1$ the map $\beta'\colon S^1 \to I\vee S^1$ sends $[0, 1/4]$ and $[3/4, 1]$ to $I$, $[1/4, 3/4]$ to $S^1$.

With this in mind, it should be possible to write down $\beta'$ as a formula of $t$. Though it's not necessary: things happening in that chapter in Switzer's book are not so complicated to imagine geometrically. It is way easier and more pleasant than manipulating with formulas, and you wouldn't spend your time dealing with meaningless errors like forgetting which semicircle is parametrized by $[0, 1/2]$ and which by $[1/4, 3/4]$.

Generally, I would not recommend using explicit formulaic parametrizations for circles and segments like Switzer does in the book unless the situation is too complicated to be imagined in detail. Hard to understand, easy to make mistakes, and no room for geometric intuition. I would suggest reading about higher homotopy groups in some other source, for example in (freely available online) book by A. Hatcher: it has beautiful (meaningful!) pictures and explains what's going on conceptually, not just by writing messy formulas.

Dmitry
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