Let $X$ be path-connected and let $b\in X$. Show that every path in $X$ is homotopic with endpoints fixed to a path passing through $b$.

Question

$X$ is path-connected and $b\in x$; show every path in $X$ is homotopic with endpoints fixed to a path passing through $b$.

This is the hint in the book: Let $\gamma$ be a path from x to y. If $\alpha$ is any path from $x$ to $b$, then the path $\alpha(\alpha^{-1}\gamma)$ passes through $b$, and it's homotopic to $\gamma$ with endpoints fixed.

I don't understand why $\alpha(\alpha^{-1}\gamma)$ is homotopic to $\gamma$ with endpoints fixed?

Answer 1

Let $\gamma$ be any path with endpoints $x = \gamma(0)$ and $y=\gamma(1)$. Let $\alpha$ be a path from $y$ to $b$, i.e., $\alpha(0)=y$ and $\alpha(1)=b$. Let $\bar{\alpha}$ denote the path $\alpha(1-t)$.

Now $\gamma*\alpha*\bar{\alpha}$ is a path from x to y (here $*$ denotes the usual definition of path concatenation). Since $\alpha*\bar{\alpha}$ is homotopic relative to $\{0,1\}$ to the constant path, we have that $$\gamma*\alpha*\bar{\alpha}\simeq \gamma*c_{y}\text{ rel}\{0,1\}$$

Now the constant path has the property that given any path $\gamma$, $$\gamma * c_{\gamma(1)}\simeq \gamma \text{ rel}\{0,1\} $$

Since homotopy is an equivalance relation, we have that $$\gamma*\alpha*\bar{\alpha}\simeq \gamma\text{ rel}\{0,1\}$$

Answer 2

Let's take a look at constraints for the homotopy $H : [0,1] \to [0,1]$.

At the bottom of the homotopy, $$H(s,0) = \alpha (\alpha^{-1} \gamma(s)) = \begin{cases} \alpha(2s) & \quad\text{if $0 \le s \le 1/2$} \\ \alpha^{-1}(4s-2) & \quad\text{if $1/2 \le s \le 3/4$} \\ \gamma(4s-3) & \quad\text{if $3/4 \le s \le 1$} \end{cases} $$ At the top of the homotopy, $$H(s,1) = \gamma(s), \quad 0 \le s \le 1 $$ And, of course, $H(0,t)$ and $H(1,t)$ are constant.

How to interpolate this over all of $[0,1] \times [0,1]$?

On the bottom edge of the square $[0,1] \times [0,1]$ where $t=0$,

• color the subinterval $3/4 \le s \le 1$ red,
• color the subinterval $1/2 \le s \le 3/4$ green,
• color the subinterval $0 \le s \le 1/2$ blue.

On the top edge where $t=1$, color the entire interval $0 \le s \le 1$ red.

Now subdivide the square $[0,1] \times [0,1]$ into a red trapezoid, a green triangle, and a blue triangle. Define $H$ separately on each of these three polygons. I'll describe how $H$ is defined using on the red trapezoid using $\gamma$, since that's the most complicated one (in brief, for the green triangle you do something similar with $\alpha^{-1}$, and with the blue triangle something similar with $\alpha$).

The red trapezoid intersects the level $t$ along the interval $3(1-t)/4 \le s \le 1$. As $t$ is held fixed and as $s$ advances from $3(1-t)/4$ to $1$, you scale $s$ linearly so that the result advances from $0$ to $1$ and then plug into $\gamma$ to get $$H(s,t) = \gamma\left(\frac{s - 3(1-t)/4}{1 - 3(1-t)/4} \right) \quad \text{if}\quad 0 \le t \le 1, \quad 3(1-t)/4 \le s \le 1 $$