Math4202 Topology II (Lecture 10)

(1) If $f\simeq_p f_1$, $g\simeq_p g_1$, then $f*g\simeq_p f_1*g_1$.

Let $F$ be homotopy between $f$ and $f_1$, $G$ be homotopy between $g$ and $g_1$.

We can define

$$F*G:[0,1]\times [0,1]\to X,\quad F*G(s,t)=\left(F(-,t)*G(-,t)\right)(s)=\begin{cases} F(2s,t) & 0\leq s\leq \frac{1}{2}\\ G(2s-1,t) & \frac{1}{2}\leq s\leq 1 \end{cases}$$

$F*G$ is a homotopy between $f*g$ and $f_1*g_1$.

We can check this by enumerating the cases from definition of homotopy.

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(2) $([f]*[g])*[h]=[f]*([g]*[h])$.

For $f*(g*h)$, along the interval $[0,\frac{1}{2}]$ we map $x_1\to x_2$, then along the interval $[\frac{1}{2},\frac{3}{4}]$ we map $x_2\to x_3$, then along the interval $[\frac{3}{4},1]$ we map $x_3\to x_4$.

For $(f*g)*h$, along the interval $[0,\frac{1}{4}]$ we map $x_1\to x_2$, then along the interval $[\frac{1}{4},\frac{1}{2}]$ we map $x_2\to x_3$, then along the interval $[\frac{1}{2},1]$ we map $x_3\to x_4$.

We can construct the homotopy between $f*(g*h)$ and $(f*g)*h$ as follows.

Let $f((4-2t)s)$ for $F(s,t)$,

when $t=0$, $F(s,0)=f(4s)\in f*(g*h)$, when $t=1$, $F(s,1)=f(2s)\in (f*g)*h$.

…

We make the linear maps between $f*(g*h)$ and $(f*g)*h$ continuous, then $f*(g*h)\simeq_p (f*g)*h$. With our homotopy constructed above

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(3) $e_{x_0}*f\simeq_p f\simeq_p f*e_{x_1}$.

We can construct the homotopy between $e_{x_0}*f$ and $f$ as follows.

$$H(s,t)=\begin{cases} x_0 & t\geq 2s\\ f(2s-t) & t\leq 2s \end{cases}$$

or you may induct from $f(\frac{s-t/2}{1-t/2})$ if you like.

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(4) $f*\bar{f}=e_{x_0},\quad \bar{f}*f=e_{x_1}$.

Note that we don’t need to reach $x_1$ every time.

$f_t=f(ts)$ $s\in[0,\frac{1}{2}]$.

$\bar{f}_t=\bar{f}(1-ts)$ $s\in[\frac{1}{2},1]$.