A doubt in Hatcher's Algebraic Topology.

I refer to pg. 27 of Hatcher's Algebraic Topology.

I refer to the part where Hatcher proves that $f.(g.h)\cong (f.g).h$

For the life of me, I cannot figure out how the diagram on the right proves this. What does the diagram even mean?

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Have you read the preceding paragraph about reparametrizations and how it leads to homotopy? –  Martin Brandenburg Jul 26 '14 at 8:21
@MartinBrandenburg- I did. I don't see how the part that confuses me follows from the first paragraph. I am probably making a grave understanding error somewhere. I am looking for a straight answer, and not for hints, as I have spent far too much time thinking about these concepts. –  Ayush Khaitan Jul 26 '14 at 8:27
@MartinBrandenburg- I suppose one of the reasons why the second paragraph does not necessarily follow from the first is that we are only talking about one path in the first paragraph, while we are talking about three in the second. Hence, there is a bit of a logical gap which the author has not filled in. I see that the graph on the right has three 'components' (or three straight lines joined to one another). However, composing the three parts separately with the help of the three separate paths does not make the picture any clearer. –  Ayush Khaitan Jul 26 '14 at 8:37
It's the same method, except the "one path" is $(f.g).h$, which simply happens to be the concatenation of three paths. If you compose this one path with the reparametrization, you get $f.(g.h)$ (or the reverse, you get $(f.g).h$ from $f.(g.h)$). –  Najib Idrissi Jul 26 '14 at 8:47
I am not sure why texts use only paths or length 1 instead of following the book by Crowell and Fox "Introduction to knot theory" –  Ronnie Brown Jul 26 '14 at 14:38

Remember that $f\cdot(g\cdot h)$ is the function from $[0,1]$ to $X$ such that $$[f\cdot(g\cdot h)](t) = \begin{cases} f(2t) &t\in[0,\frac1{2}]\\ g(4t-2) & t\in[\frac1{2}, \frac{3}{4}]\\ h(4t-3) & t\in[\frac3{4}, 1]\end{cases}$$

while $(f\cdot g)\cdot h$ is the function from $[0,1]$ to $X$ such that $$[(f\cdot g)\cdot h](t) = \begin{cases} f(4t) &t\in[0,\frac1{4}]\\ g(4t-1) & t\in[\frac1{4}, \frac{1}{2}]\\ h(2t-1) & t\in[\frac1{2}, 1]\end{cases}.$$

Let us denote those two functions by $p_0$ and $p_1$, respectively.

So we want to find a homotopy between $p_0$ and $p_1$. Note that if $f$ is a path, and $\gamma\colon [0,1]\rightarrow [0,1]$ is an increasing homeomorphism, then $f$ and $f\circ\gamma$ are homotopic: the homotopy is $\varphi:[0,1]\times [0,1]\rightarrow X$ such that $\varphi(s,t) = f(t\gamma(s) + (1-t)s)$. Indeed:

• we have $\varphi(s,0) = f(s)$ for all $s\in[0,1]$,
• $\varphi(s,1) = f(\gamma(s))$ for all $s\in[0,1]$,
• $\varphi(0,t) = f(t\gamma(0) + (1-t)\gamma(0)) = f(\gamma(0)) = f(0)$, and
• $\varphi(1,t) = f(t\gamma(1) + (1-t)\gamma(1)) = f(\gamma(1)) = f(1)$.

Now it remains to note that there exists $\gamma\colon[0,1]\rightarrow [0,1]$ such that $p_0 = p_1\circ \gamma$. So let $\gamma$ be defined by: $$\gamma(t) = \begin{cases}\frac{t}{2} & t\in[0,\frac1{2}] \\ t-\frac1{4} & t\in[\frac1{2}, \frac3{4}]\\ 2t-1 & t\in[\frac3{4}, 1]\end{cases}.$$

Let us check that it indeed does the job: For $t\leq \frac1{2}$, $p_1(\gamma(t)) = p_1(\frac{t}{2}) = f(4\frac{t}{2}) = f(2t) = p_0(t)$, for $\frac1{2}\leq t\leq \frac{3}{4}$, $p_1(\gamma(t)) = p_1(t-\frac{1}{4}) = g(4(t-\frac{1}{4})-1) = g(4t-2) = p_0(t)$ and finally, for $t\geq \frac{3}{4}$, we have $p_1(\gamma(t)) = p_1(2t-1) = h(4t-2-1) = h(4t-3) = p_0(t)$.

Finally, note that the graph of the function $\gamma$ above is precisely the graph drawn in Hatcher's book.

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@FPE- Your parametrization for $f.(g.h)$ is correct by verification. But how do you arrive upon a parametrization for the concatenation of an arbitrary number of paths? Is there an algorithm? –  Ayush Khaitan Jul 26 '14 at 8:58
The parameterization of $f\cdot g$ is defined by $\begin{cases} f(2t) & t\in[0,\frac1{2}]\\g(2t-1) & t\in[\frac1{2},1]\end{cases}$. To find the parameterization of $f\cdot(g\cdot h)$, first find that of $g\cdot h$ (using the definition), and then the whole parameterization. It is some kind of inductive process: decompose the function until you come to only a binary composition, which you know how to parameterize using the definition! –  zarathustra Jul 26 '14 at 9:01
That's what I thought too. Thanks! –  Ayush Khaitan Jul 26 '14 at 9:06
@AyushKhaitan: If you consider that your question has been answered, please accept Martin's or my answer by clicking on the small "tick" icon on the left. Thanks! –  zarathustra Jul 26 '14 at 9:21
I refer to the last part of your question: verifying that $p_0$ and $p_1$ are path homotopic. Shouldn't you have $p_1(\gamma(t))=f(2t)$ for $t\in[0,\frac{1}{4}]$ and $g(2t-1)$ for $t\in[\frac{1}{4},\frac{1}{2}]$? –  Ayush Khaitan Jul 26 '14 at 11:11

In general, if $f$ is a path and $\phi : I \to I$ is a continuous map such that $\phi(0)=0$ and $\phi(1)=1$, then $f \circ \phi$ is homotopic to $f$. This is explained by Hatcher.

Now consider three paths $f,g,h$ with $f(1)=g(0)$, $g(1)=h(0)$, so that the paths $(fg)h$ and $f(gh)$ are well-defined.

Consider the map $\phi : I \to I$ defined by

$\phi(t) = \left\{\begin{array}{c} t/2 & 0 \leq t \leq 1/2 \\ t - 1/4 & 1/2 \leq t \leq 3/4 \\ 2t-1 & 3/4 \leq t \leq 1\end{array} \right..$

Now compute $f(gh) = (fg)h \circ \phi$. Hence, $f(gh)$ is homotopic to $(fg)h$.

You can find a different (but also graphical) proof in May's Concise course (p.6).

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