- Thread starter
- #1

Prove that $\gamma\gamma^-$ is null homotopic in U.

So I need to show that $\int_{\gamma\gamma^-}f(z)dz = 0$. How can I show this though? Not all closed curves are necessarily zero.

- Thread starter dwsmith
- Start date

- Thread starter
- #1

Prove that $\gamma\gamma^-$ is null homotopic in U.

So I need to show that $\int_{\gamma\gamma^-}f(z)dz = 0$. How can I show this though? Not all closed curves are necessarily zero.

- Moderator
- #2

- Feb 7, 2012

- 2,785

Rather than using the property $\int_{\gamma\gamma^-}f(z)dz = 0$, I think it would be easier to construct an explicit homotopy from $\gamma\gamma^-$ to the constant path at $z_0.$

Prove that $\gamma\gamma^-$ is null homotopic in U.

So I need to show that $\int_{\gamma\gamma^-}f(z)dz = 0$. How can I show this though? Not all closed curves are necessarily zero.

Given $\gamma:[0,1]\to U$, with $\gamma^-(t) = \gamma(1-t)$, define $H_s$ for $s\in[0,1]$ by $$H_s(t) = \begin{cases}\gamma^-(2st)&(t\leqslant1/2),\\ \gamma(2st)&(t\geqslant1/2).\end{cases}$$ Then $H_1=\gamma\gamma^-$ and $H_0$ is the constant path at $z_0.$

Of course, you need to check that, for each $s$, $H_s$ is a path in $U$, and that the map $s\mapsto H_s$ is indeed a homotopy.

- Thread starter
- #3

How did you come up with this piece wise defined map?Rather than using the property $\int_{\gamma\gamma^-}f(z)dz = 0$, I think it would be easier to construct an explicit homotopy from $\gamma\gamma^-$ to the constant path at $z_0.$

Given $\gamma:[0,1]\to U$, with $\gamma^-(t) = \gamma(1-t)$, define $H_s$ for $s\in[0,1]$ by $$H_s(t) = \begin{cases}\gamma^-(2st)&(t\leqslant1/2),\\ \gamma(2st)&(t\geqslant1/2).\end{cases}$$ Then $H_1=\gamma\gamma^-$ and $H_0$ is the constant path at $z_0.$

Of course, you need to check that, for each $s$, $H_s$ is a path in $U$, and that the map $s\mapsto H_s$ is indeed a homotopy.

- Thread starter
- #4

Would this map work as well?How did you come up with this piece wise defined map?

$$

F(s,t)=\begin{cases}\gamma(2s,t), & 0\leq s\leq 1/2\\

\gamma^-(2s-1,t), & 1/2\leq s\leq 1

\end{cases}

$$

- Thread starter
- #5

Let

$$

\gamma\gamma^- = \begin{cases}

\gamma(2t), & 0\leq t\leq\frac{1}{2}\\

\gamma(2 - 2t), & \frac{1}{2}\leq t\leq 1

\end{cases}.

$$

Then the path $\gamma\gamma^-$ follows $\gamma$ and then $\gamma^-$ back to $z_0$.

Define the homotopy

$$

K(t,s) = \begin{cases}

\gamma(2ts), & 0\leq t\leq\frac{1}{2}\\

\gamma(2s - 2ts), & \frac{1}{2}\leq t\leq 1.

\end{cases}

$$

Then $K(0,2) = K(1,s) = z_0$, $K(t,0) = c$ (the constant path), and

$$

K(t,1) = \begin{cases}

\gamma(2ts), & 0\leq t\leq\frac{1}{2}\\

\gamma(2 - 2t), & \frac{1}{2}\leq t\leq 1

\end{cases}

= \gamma\gamma^-.

$$

We can observe that $K$ is continuous.

Now $[\gamma][\gamma^-] = [\gamma\gamma^-] = [c]$.

Thus the product of the classes of $\gamma$ and $\gamma^-$ is the constant path, i.e. null homotopic.

How is this?

- Moderator
- #6

- Feb 7, 2012

- 2,785

Looks good.